Chapter 9 Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 – Straight Lines

NCERT Solutions for Class 11 Maths Chapter 9 (Straight Lines) transition students from basic coordinate geometry to the algebraic representation of lines. While earlier classes focused on plotting points, this chapter explores the properties of lines, such as Slope, Parallelism, and Perpendicularity. It is the foundational chapter for Coordinate Geometry, which is essential for Calculus and Physics.

The NCERT Solutions for Class 11 Maths Chapter 9 by ALLEN are designed by expert faculty to simplify coordinate geometry through clear derivations and systematic problem-solving. The solutions focus on the various forms of a line's equation—such as Point-Slope form, Slope-Intercept form, and Normal form—making it easier for students to choose the most efficient method for a given problem.

These solutions provide a rigorous framework for finding the angle between lines and the distance of a point from a line, topics that are consistently tested in engineering entrance exams.

1.0Class 11 Maths Chapter 9: Key Concepts

This chapter focuses on the algebraic description of lines and the relationship between different lines in a plane. Key lessons include:

Slope of a Line: Understanding the measure of steepness, defined as $m = tan θ or m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} .$
Conditions for Parallelism and Perpendicularity:
Lines are parallel if $m_{1} = m_{2} .$
- Lines are perpendicular if $m_{1} \cdot m_{2} = - 1.$
Angle between Two Lines: Using the formula $tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}} .$
Various Forms of the Equation of a Line:
- Horizontal and Vertical lines: y = a or x = b.
- Point-Slope form: $y - y_{1} = m (x - x_{1}) .$
- Slope-Intercept form: y = mx + c.
- Two-Point form: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1}) .$
- Intercept form: $\frac{x}{a} + \frac{y}{b} = 1.$
- Normal form: $x cos ω + y sin ω = p .$
General Equation of a Line: Transforming Ax + By + C = 0 into other forms.
Distance Formulas:
- Distance of a Point from a Line: $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}} .$
- Distance between Two Parallel Lines: $d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}} .$

Also Read: Intercept Form

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 9

EXERCISE - 9.1

Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, -5) and (- 4, - 2). Also, find its area. Sol. The given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC. Accordingly,

$area (ABCD) = area (△ ABC) + area (△ ACD)$

We know that the area of a triangle whose vertices are $(x_{1}, y_{1}), (x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ is

$\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$

Therefore, area of $△ ABC$

$= \frac{1}{2} ∣ - 4 (7 + 5) + 0 (- 5 - 5) + 5 (5 - 7) ∣ = \frac{1}{2} ∣ - 4 (12) + 5 (- 2) ∣ = \frac{1}{2} - 48 - 10) ∣ = \frac{1}{2} - 58) ∣ = \frac{1}{2} \times 58 = 29 sq. units$

Area of $△ ACD$

$= \frac{1}{2} ∣ - 4 (- 5 + 2) + 5 (- 2 - 5) + (- 4) (5 + 5) ∣ = \frac{1}{2} ∣ - 4 (- 3) + 5 (- 7) - 4 (10) ∣ = \frac{1}{2} ∣12 - 35 - 40∣ = \frac{1}{2} ∣ - 63∣ = \frac{63}{2} sq.units$

Thus, area $(ABCD) = (29 + \frac{63}{2})$

$= (\frac{58 + 63}{2}) = \frac{121}{2} sq.units$

The base of an equilateral triangle with side 2 a lies along they y -axis such that the mid-point of the base is at the origin. Find vertices of the triangle. Sol. Let ABC be the given equilateral triangle with side 2a. Accordingly, $AB = BC = CA = 2 a$ Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin. i.e., $BO = OC = a$ , where O is the origin. Now, it is clear that the coordinates of point C are ( $0, a$ ), while the coordinates of point B are (0, -a). It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular. Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to $△ AOC$ , we obtain

$(AC)^{2} = (OA)^{2} + (OC)^{2}$

$\Rightarrow (2 a)^{2} = (OA)^{2} + a^{2}$ $\Rightarrow 4 a^{2} - a^{2} = (OA)^{2}$ $\Rightarrow (OA)^{2} = 3 a^{2}$ $\Rightarrow OA = \pm 3 a$ ∴ Coordinates of point $A = (\pm 3 a, 0)$

Thus, the vertices of the given equilateral triangle are (0, a), (0, -a), and $(3 a, 0)$ or (0,

a) , (0, -a), and $(- 3 a, 0)$ .

Find the distance between $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis. Sol. The given points are $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ . (i) When PQ is parallel to the y -axis, $x_{1} = x_{2}$ In this case, distance between P and Q

$= (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} = (y_{2} - y_{1})^{2} = ∣ y_{2} - y_{1} ∣$

(ii) When PQ is parallel to the x -axis, $y_{1} = y_{2}$ In this case, distance between P and Q

$= (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} = (x_{2} - x_{1})^{2} = ∣ x_{2} - x_{1} ∣$

Find a point on the x-axis, which is equidistant from the points $(7, 6)$ and $(3, 4)$ .

Sol. Let ( $a, 0$ ) be the point on the x axis that is equidistant from the points $(7, 6)$ and (3, 4).

Accordingly,

$(7 - a)^{2} + (6 - 0)^{2} = (3 - a)^{2} + (4 - 0)^{2}$

$\Rightarrow a^{2} - 14 a + 85 = a^{2} - 6 a + 25$

On squaring both sides, we obtain

$\Rightarrow \Rightarrow \Rightarrow a^{2} - 14 a + 85 = a^{2} - 6 a + 25 - 14 a + 6 a = 25 - 85 - 8 a = - 60 a = \frac{60}{8} = \frac{15}{2}$

Thus, the required point on the x-axis is $(\frac{15}{2}, 0)$ .

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points $P (0, - 4)$ and $B (8, 0)$ . Sol. The coordinates of the mid-point of the line segment joining the points $P (0, - 4)$ and $B (8, 0)$ are $(\frac{0 + 8}{2}, \frac{- 4 + 0}{2}) = (4, - 2)$ It is known that the slope (m) of a non-vertical line passing through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}, x_{2} \neq = x_{1} .$

Therefore, the slope of the line passing through $(0, 0)$ and $(4, - 2)$ is

$\frac{- 2 - 0}{4 - 0} = \frac{- 2}{4} = - \frac{1}{2}$

Hence, the required slope of the line is $- \frac{1}{2}$ .

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle. Sol. The vertices of the given triangle are A (4, 4), B (3, 5), and C (-1, -1).

It is known that the slope (m) of a non-vertical line passing through the points $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}, x_{2} \neq = x_{1}$

∴ Slope of $AB (m_{1}) = \frac{5 - 4}{3 - 4} = - 1$ ∴ Slope of $BC (m_{2}) = \frac{- 1 - 5}{- 1 - 3} = \frac{- 6}{- 4} = \frac{3}{2}$ ∴ Slope of $CA (m_{3}) = \frac{4 + 1}{4 + 1} = \frac{5}{5} = 1$

It is observed that $m_{1} m_{3} = - 1$ This shows that line segments AB and CA are perpendicular to each other i.e., the given triangle is right-angled at A (4, 4). Thus, the points (4, 4), (3, 5), and (-1, -1) are the vertices of a right-angled triangle.

Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise. Sol. If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is $9 0^{\circ} + 3 0^{\circ} = 12 0^{\circ}$ .

ques-7-exer-9.1-maths-class-11 chapter-9

Thus, the slope of the given line is $tan 12 0^{\circ} = tan (18 0^{\circ} - 6 0^{\circ})$

$= - tan 6 0^{\circ} = - 3$

Without using distance formula, show that points (-2, -1), (4, 0), $(3, 3)$ and (-3, 2) are vertices of a parallelogram. Sol. Let points (-2, -1), (4, 0), (3, 3), and (-3, 2) be respectively denoted by A, B, C, and D.

Slope of $AB = \frac{0 + 1}{4 + 2} = \frac{1}{6}$ Slope of $CD = \frac{2 - 3}{- 3 - 3} = \frac{- 1}{- 6} = \frac{1}{6}$ ⇒ Slope of AB = Slope of CD $\Rightarrow AB$ and CD are parallel to each other. Now, slope of $BC = \frac{3 - 0}{3 - 4} = \frac{3}{- 1} = - 3$ Slope of $AD = \frac{2 + 1}{- 3 + 2} = \frac{3}{- 1} = - 3$ ⇒ Slope of $BC =$ Slope of AD $\Rightarrow BC$ and AD are parallel to each other. Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram. Thus, points (-2, -1), (4, 0), (3, 3), and (-3, 2) are the vertices of a parallelogram.

Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2). Sol. The slope of the line joining the points (3, -1) and (4, -2) is

$m = \frac{- 2 - ( - 1 )}{4 - 3} = - 2 + 1 = - 1$

Now, the inclination ( $θ$ ) of the line joining the points (3, -1) and (4, -2) is given by $tan θ = - 1$ $\Rightarrow θ = (9 0^{\circ} + 4 5^{\circ}) = 13 5^{\circ}$

Thus, the angle between the x-axis and the line joining the points $(3, - 1)$ and (4, -2) is 135°.

The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{1}{3}$ , find the slopes of the lines. Sol. Let $m_{1}$ and m be the slopes of the two given lines such that $m_{1} = 2 m, m_{2} = m$ We know that if $θ$ is the angle between the lines $l_{1}$ and $l_{2}$ with slopes $m_{1}$ and $m_{2}$ , then $tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$ It is given that the tangent of the angle between the two lines is $\frac{1}{3}$ . $∴ \frac{1}{3} = \frac{m - 2 m}{1 + ( 2 m ) m}$ $\Rightarrow \frac{1}{3} = \frac{- m}{1 + 2 m ^{2}}$ $\Rightarrow \frac{1}{3} = \frac{- m}{1 + 2 m ^{2}}$ or $\frac{1}{3} = - (\frac{- m}{1 + 2 m ^{2}}) = \frac{m}{1 + 2 m ^{2}}$

Case I

$\Rightarrow \frac{1}{3} = \frac{- m}{1 + 2 m ^{2}} \Rightarrow 1 + 2 m^{2} = - 3 m \Rightarrow 2 m^{2} + 3 m + 1 = 0 \Rightarrow 2 m (m + 1) + 1 (m + 1) = 0 \Rightarrow (m + 1) (2 m + 1) = 0 \Rightarrow m = - 1 or m = \frac{- 1}{2}$

If $m = - 1$ , then the slopes of the lines are -1 and -2. If $m = - \frac{1}{2}$ , then the slopes of the lines are $- \frac{1}{2}$ and -1 .

Case II

$\Rightarrow \frac{1}{3} = \frac{m}{1 + 2 m ^{2}}$ $\Rightarrow 2 m^{2} + 1 = 3 m$ $\Rightarrow 2 m^{2} - 3 m + 1 = 0$ $\Rightarrow 2 m^{2} - 2 m - m + 1 = 0$ $\Rightarrow 2 m (m - 1) - 1 (m - 1) = 0$ $\Rightarrow (m - 1) (2 m - 1) = 0$ $\Rightarrow m = 1$ or $m = \frac{1}{2}$

If $m = 1$ , then the slopes of the lines are 1 and 2. If $m = \frac{1}{2}$ , then the slopes of the lines are $\frac{1}{2}$ and 1. Hence, the slopes of the lines are -1 and -2 or $- \frac{1}{2}$ and -1 or 1 and 2 or $\frac{1}{2}$ and 1.

A line passes through $(x_{1}, y_{1})$ and $(h, k)$ . If slope of the line is m, show that $k - y_{1} = m (h - x_{1})$ . Sol. The slope of the line passing through $(x_{1}, y_{1})$ and (h, k) It is given that the slope of the lines is m.

$m = \frac{y _{2} +- y _{1}}{x _{2} - x _{1}}, x_{1} \neq = x_{2}$

$∴ \frac{k - y _{1}}{h - x _{1}} = m$ $\Rightarrow k - y_{1} = m (h - x_{1})$ Hence, $k - y_{1} = m (h - x_{1})$ . Hence proved

EXERCISE - 9.2

In Q. 1 to 8, find the equation of the line which satisfy the given condition :

Write the equations for the x and y-axes. Sol. The y-coordinate of every point on the x-axis is 0. Therefore, the equation of the x -axis is $x = 0$ . The x-coordinate of every point on the y-axis is 0. Therefore, the equation of the y -axis is $y = 0$ .
Passing through the point $(- 4, 3)$ with slope $\frac{1}{2}$ . Sol. We know that the equation of the line passing through point $(x_{0}, y_{0})$ , whose slope is m, is $(y - y_{0}) = m (x - x_{0})$ Thus, the equation of the line passing through point (-4, 3), whose slope is $\frac{1}{2}$ , is

$(y - 3) = \frac{1}{2} (x + 4) 2 (y - 3) = x + 4 2 y - 6 = x + 4$

i.e. $x - 2 y + 10 = 0$

Passing though $(0, 0)$ with slope m. Sol. We know that the equation of the line passing through point $(x_{0}, y_{0})$ , whose slope is m is $(y - y_{0}) = m (x - x_{0})$ . Thus, the equation of the line passing through point ( 0,0 ), whose slope is $m$ , is

$(y - 0) = m (x - 0)$

i.e. $y = mx$

Passing through $(2, 23)$ and is inclined with the x-axis at an angle of 75°. Sol. The slope of the line that inclines with the xaxis at an angle of 75° is $m = tan 7 5^{\circ}$

$\Rightarrow m = tan (4 5^{\circ} + 3 0^{\circ}) = \frac{tan 4 5 ^{\circ} + tan 3 0 ^{\circ}}{1 - tan 4 5 ^{\circ} \cdot tan 3 0 ^{\circ}} = \frac{1 + \frac{1}{3}}{1 - 1 \cdot \frac{1}{3}} = \frac{\frac{3 + 1}{3}}{\frac{3 - 1}{3}} = \frac{3 + 1}{3 - 1}$

We know that the equation of the line passing through point $(x_{0}, y_{0})$ whose slope is m, is $(y - y_{0}) = m (x - x_{0})$ . Thus, if a line passes through $(2, 23)$ and inclines with the x-axis at an angle of 75°, then the equation of the line is given as

$(y - 23) = \frac{3 + 1}{3 - 1} (x - 2) (y - 23) (3 - 1) = (3 + 1) (x - 2) y (3 - 1) - 23 (3 - 1) = x (3 + 1) - 2 (3 + 1) (3 + 1) x - (3 - 1) y = 23 + 2 - 6 + 23 (3 + 1) x - (3 - 1) y = 4 (3 - 1)$

Intersects the x-axis at a distance of 3 units to the left of origin with slope -2. Sol. It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as $y = m (x - d)$ For the line intersecting the x-axis at a distance of 3 units to the left of the origin, $d = - 3$ . The slope of the line is given as $m = - 2$ Thus, the required equation of the given line is $y = - 2 [x - (- 3)]$ $\Rightarrow y = - 2 x - 6$ i.e., $2 x + y + 6 = 0$
Intersects the y -axis at a distance of 2 units above the origin and makes an angle of 30° with the positive direction of the x-axis. Sol. It is known that if a line with slope m makes y-intercept c, then the equation of the line is given as $y = mx + c$ Here, $c = 2$ and $m = tan 3 0^{\circ} = \frac{1}{3}$ . Thus, the required equation of the given line is

$y = \frac{1}{3} x + 2$

$\Rightarrow y = \frac{x + 2 3}{3} \Rightarrow 3 y = x + 23$ i.e. $x - 3 y + 23 = 0$

Passing through the points (-1, 1) and (2, -4). Sol. It is known that the equation of the line passing through points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

$y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$

Therefore, the equation of the line passing through the points (-1, 1) and (2, -4) is

$(y - 1) = \frac{- 4 - 1}{2 + 1} (x + 1)$

$\Rightarrow (y - 1) = \frac{- 5}{3} (x + 1) \Rightarrow 3 (y - 1) = - 5 (x + 1)$ $\Rightarrow 3 y - 3 = - 5 x - 5$ i.e. $5 x + 3 y + 2 = 0$

The vertices of $△ PQR$ are P(2, 1), Q(-2, 3) and R(4, 5). Find equation of the median through the vertex R. Sol. It is given that the vertices of $Δ PQR$ are P (2, 1), Q (-2, 3), and R (4, 5). Let RL be the median through vertex R. Accordingly, L is the mid-point of PQ. By mid-point formula, the coordinates of point L are given by $(\frac{2 - 2}{2}, \frac{1 + 3}{2}) = (0, 2)$

It is known that the equation of the line passing through points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is.

$y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$

Therefore, the equation of RL can be determined by substituting $(x_{1}, y_{1}) = (4, 5)$ and $(x_{2}, y_{2}) = (0, 2)$ .

$Hence, y - 5 = \frac{2 - 5}{0 - 4} (x - 4)$

$\Rightarrow y - 5 = \frac{- 3}{- 4} (x - 4)$ $\Rightarrow 4 (y - 5) = 3 (x - 4)$ $\Rightarrow 4 y - 20 = 3 x - 12$ $\Rightarrow 3 x - 4 y + 8 = 0$ Thus, the required equation of the median through vertex R is $3 x - 4 y + 8 = 0$ .

Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points $(2, 5)$ and (-3, 6). Sol. The slope of the line joining the points

$(2, 5) and (- 3, 6) is m = \frac{6 - 5}{- 3 - 2} = \frac{1}{- 5}$

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other. Therefore, slope of the line perpendicular to the line through the points $(2, 5)$ and $(- 3, 6) =$

$- \frac{1}{m} = - \frac{1}{( \frac{- 1}{5} )} = 5$

Now, the equation of the line passing through point (-3, 5), whose slope is 5, is

$(y - 5) = 5 (x + 3) y - 5 = 5 x + 15$

i.e. $5 x - y + 20 = 0$

A line perpendicular to the line segment joining the points $(1, 0)$ and $(2, 3)$ divides it in the ratio $1 : n$ . Find the equation of the line. Sol. According to the section formula, the coordinates of the point that divides the line segment joining the points $(1, 0)$ and $(2, 3)$ in the ratio 1: n is given by

$(\frac{n ( 1 ) + 1 ( 2 )}{1 + n}, \frac{n ( 0 ) + 1 ( 3 )}{1 + n}) = (\frac{n + 2}{n + 1}, \frac{3}{n + 1})$

The slope of the line joining the points $(1, 0)$ and $(2, 3)$ is $m = \frac{3 - 0}{2 - 1} = 3$ We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other. Therefore, slope of the line that is perpendicular to the line joining the points $(1, 0)$ and $(2, 3) = - \frac{1}{m} = - \frac{1}{3}$ Now, the equation of the line passing through $(\frac{n + 2}{n + 1}, \frac{3}{n + 1})$ and whose slope is $- \frac{1}{3}$ given by $(y - \frac{3}{n + 1}) = \frac{- 1}{3} (x - \frac{n + 2}{n + 1})$

$\Rightarrow 3 [(n + 1) y - 3] = - [x (n + 1) - (n + 2)] \Rightarrow 3 (n + 1) y - 9 = - (n + 1) x + n + 2 \Rightarrow (n + 1) x + 3 (1 + n) y = n + 11$

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3). Sol. The equation of a line in the intercept form is

$\frac{x}{a} + \frac{y}{b} = 1$

Here, a and b are the intercepts on x and y axes respectively. It is given that the line cuts off equal intercepts on both the axes. This means that $a = b$ . Accordingly, equation (i) reduces to

$\frac{x}{a} + \frac{y}{a} = 1$

$\Rightarrow x + y = a$

Since the given line passes through point (2, 3), equation (2) reduces to

$2 + 3 = a \Rightarrow a = 5$

On substituting the value of a in equation (2), we obtain $x + y = 5$ , which is the required equation of the line.

Find equation of the line passing through the point $(2, 2)$ and cutting off intercepts on the axes whose sum is 9. Sol. The equation of a line in the intercept form is

$\frac{x}{a} + \frac{y}{b} = 1$

Here, a and b are the intercepts on x and y axes respectively. It is given that

$a + b = 9$

$\Rightarrow b = 9 - a$

From equations (1) and (2), we obtain

$\frac{x}{a} + \frac{y}{9 - a} = 1$

It is given that the line passes through point (2, 2). Therefore, equation (3) reduces to

$\frac{2}{a} + \frac{2}{9 - a} = 1$

$\Rightarrow 2 (\frac{1}{a} + \frac{1}{9 - a}) = 1$ $\Rightarrow 2 (\frac{9 - a + a}{a ( 9 - a )}) = 1$ $\Rightarrow \frac{18}{9 a - a ^{2}} = 1$ $\Rightarrow 18 = 9 a - a^{2}$ $\Rightarrow a^{2} - 9 a + 18 = 0$ $\Rightarrow a^{2} - 6 a - 3 a + 18 = 0$ $\Rightarrow a (a - 6) - 3 (a - 6) = 0$ $\Rightarrow a (a - 6) (a - 3) = 0$ $\Rightarrow a = 6$ or $a = 3$ If $a = 6$ and $b = 9 - 6 = 3$ , then the equation of the line is

$\frac{x}{6} + \frac{y}{3} = 1$

$\Rightarrow x + 2 y - 6 = 0$ If $a = 3$ and $b = 9 - 3 = 6$ , then the equation of the line is

$\frac{x}{3} + \frac{y}{6} = 1$

$\Rightarrow 2 x + y - 6 = 0$

Find equation of the line through the point $(0, 2)$ making an angle $\frac{2 π}{3}$ with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin. Sol. The slope of the line making an angle $\frac{2 π}{3}$ with the positive x-axis is

$m = tan (\frac{2 π}{3}) = - 3$

Now, the equation of the line passing through point $(0, 2)$ and having a slope $- 3$ is

$(y - 2) = - 3 (x - 0) y - 2 = - 3 x$

i.e. $3 x + y - 2 = 0$ The slope of line parallel $3 x + y - 2 = 0$ is $- 3$ to line. It is given that the line parallel to line $3 x + y - 2 = 0$ crosses the y-axis 2 units below the origin i.e., it passes through point (0, -2). Hence, the equation of the line passing through point (0,-2) and having a slope $- 3$ is

$y - (- 2) = - 3 (x - 0)$

$\Rightarrow y + 2 = - 3 x$ $\Rightarrow 3 x + y + 2 = 0$

The perpendicular from the origin to a line meets it at the point ( $- 2, 9$ ), find the equation of the line. Sol. The slope of the line joining the origin $(0, 0)$ and point $(- 2, 9)$ is

$m_{1} = \frac{9 - 0}{- 2 - 0} = - \frac{9}{2}$

Accordingly, the slope of the line perpendicular to the line joining the origin and point (-2, 9) is

$m_{2} = - \frac{1}{m _{1}} = \frac{1}{( - \frac{9}{2} )} = \frac{2}{9}$

Now, the equation of the line passing through point ( $- 2, 9$ ) and having a slope $m_{2}$ is

$(y - 9) = \frac{2}{9} (x + 2)$

$\Rightarrow 9 y - 81 = 2 x + 4$ i.e. $2 x - 9 y + 85 = 0$

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if $L = 124.942$ when $C = 20$ and $L = 125.134$ when $C = 110$ , express L in terms of C. Sol. It is given that when $C = 20$ , the value of L is 124.942 , whereas when $C = 110$ , the value of L is 125.134. Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C. Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., $(20, 124.942)$ and $(110, 125.134)$ in the XY-plane. Therefore, the linear relation between L and C is the equation of the line passing through points (20, 124.942) and (110, 125.134).

$(L - 124.942) = \frac{125.134 - 124.942}{110 - 20} (C - 20) L - 124.942 = \frac{0.192}{90} (C - 20)$

i.e. $L = \frac{0.192}{90} (C - 20) + 124.942$ which is the required linear relation.

The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre? Sol. The relationship between selling price and demand is linear. Assuming selling price per litre along the x-axis and demand along the y-axis, we have two points i.e., $(14, 980)$ and $(16, 1220)$ in the XY plane that satisfy the linear relationship between selling price and demand. Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through points (14, 980) and (16, 1220).

$y - 980 = \frac{1220 - 980}{16 - 14} (x - 14)$

$\Rightarrow y - 980 = \frac{240}{2} (x - 14)$

$\Rightarrow y = 980 = 120 (x - 14)$ i.e. $y = 120 (x - 14) + 980$ When $x = ₹17/$ litre .

$y = 120 (17 - 14) + 980$

$\Rightarrow 7 = 120 \times 3 + 980 = 360 + 980 = 1340$ Thus, the owner of the milk store could sell 1340 litres of milk weekly at ₹17/litre.

P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$ Sol. Let AB be the line segment between the axes and let $P (a, b)$ be its mid-point.

exer-9.2-ques-17-maths-ncert-class-11-chap-9

Let the coordinates of A and B be (0, y) and $(x, 0)$ respectively. Since $P (a, b)$ is the mid-point of AB, $(\frac{0 + x}{2}, \frac{y + 0}{2}) = (a, b)$ $\Rightarrow (\frac{x}{2}, \frac{y}{2}) = (a, b)$ $\Rightarrow \frac{x}{2} = a$ and $\frac{y}{2} = b$ $∴ x = 2 a$ and $y = 2 b$ Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0). The equation of the line passing through points (0, 2b) and (2a, 0) is $(y - 2 b) = \frac{( 0 - 2 b )}{( 2 a - 0 )} (x - 0)$ $\Rightarrow y - 2 b = \frac{- 2 b}{2 a} (x)$ $\Rightarrow a (y - 2 b) = - bx$ $\Rightarrow ay - 2 ab = - bx$ i.e. $b x + a y = 2 ab$ On dividing both sides by ab , we obtain

$\frac{bx}{ab} + \frac{ay}{ab} = \frac{2 ab}{ab}$

$\Rightarrow \frac{x}{a} + \frac{y}{b} = 2$ Thus, the equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$ .

Point $R (h, k)$ divides a line segment between the axes in the ratio $1 : 2$ . Find equation of the line. Sol. Let AB be the line segment between the axes such that point $R (h, k)$ divides AB in the ratio 1 : 2.

class-11-maths-chap-9-ncert-exer-9.2-ques-18

Let the respective coordinates of A and B be (x, 0) and (0, y). Since point $R (h, k)$ divides AB in the ratio 1: 2, according to the section formula,

$(h, k) = (\frac{1 \times 0 + 2 \times x}{1 + 2}, \frac{1 \times y + 2 \times 0}{1 + 2})$

$\Rightarrow (h, k) = (\frac{2 x}{3}, \frac{y}{3})$ $\Rightarrow h = \frac{2 x}{3}$ and $k = \frac{y}{3}$ $\Rightarrow x = \frac{3 h}{2}$ and $y = 3 k$ Therefore, the respective coordinates of A and $B$ are $(\frac{3 h}{2}, 0)$ and (0, 3k). Now, the equation of line AB passing through points $(\frac{3 h}{2}, 0)$ and (0, 3k) is

$(y - 0) = \frac{3 k - 0}{0 - \frac{3 h}{2}} (x - \frac{3 h}{2})$

$\Rightarrow y = - \frac{2 k}{h} (x - \frac{3 h}{2})$ $\Rightarrow hy = - 2 kx + 3 hk$ i.e. $2 kx + hy = 3 hk$ Thus, the required equation of the line is $2 kx + hy = 3 hk$

By using the concept of equation of a line, prove that the three points (3, 0), (-2, -2) and $(8, 2)$ are collinear. Sol. In order to show that points (3, 0), (-2, -2), and $(8, 2)$ are collinear, it suffices to show that the line passing through points (3, 0) and (-2, -2) also passes through point (8, 2). The equation of the line passing through points $(3, 0)$ and (-2, -2) is

$(y - 0) = \frac{( - 2 - 0 )}{( - 2 - 3 )} (x - 3)$

$\Rightarrow y = \frac{- 2}{- 5} (x - 3)$ $\Rightarrow 5 y = 2 x - 6$ i.e. $2 x - 5 y = 6$ It is observed that at $x = 8$ and $y = 2$ , L.H.S. $= 2 \times 8 - 5 \times 2 = 16 - 10 = 6 =$ R.H.S. Therefore, the line passing through points ( 3 , 0 ) and (-2, -2) also passes through point (8, 2). Hence, points (3, 0), (-2, -2), and $(8, 2)$ are collinear.

EXERCISE - 9.3

Reduce the following equations into slopeintercept form and find their slopes and the yintercepts.

$(i) x + 7 y = 0 (ii) 6 x + 3 y - 5 = 0 (iii) y = 0$

Sol. (i) The given equation is $x + 7 y = 0$ . It can be written as $y = - \frac{1}{7} x + 0$ This equation is of the form $y = mx + c$ , where $m = - \frac{1}{7}$ and $c = 0$ . Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept are $- \frac{1}{7}$ and 0 respectively. (ii) The given equation is $6 x + 3 y - 5 = 0$ . It can be written as

$y = \frac{1}{3} (- 6 x + 5) \Rightarrow y = - 2 x + \frac{5}{3}$

This equaion is of the form $y = mx + c$ , where $m = - 2$ and $c = \frac{5}{3}$ . Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are -2 and $\frac{5}{3}$ respectively.

(iii) The given equation is $y = 0$ . It can be written as $y = 0. x + 0$ This equation is of the form $y = mx + c$ , where $m = 0$ and $c = 0$ . Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

Reduce the following equations into intercept form and find their intercepts on the axes. (i) $3 x + 2 y - 12 = 0$ (ii) $4 x - 3 y = 6$ (iii) $3 y + 2 = 0$ Sol. (i) The given equation is $3 x + 2 y - 12 = 0$ . It can be written as

$3 x + 2 y = 12$

$\Rightarrow \frac{3 x}{12} + \frac{2 y}{12} = 1$ i.e. $\frac{x}{4} + \frac{y}{6} = 1$ This equation is of the form $\frac{x}{a} + \frac{y}{b} = 1$ where $a = 4$ and $b = 6$ . Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively. (ii) The given equation is $4 x - 3 y = 6$ . It can be written as

$\frac{4 x}{6} - \frac{3 y}{6} = 1$

$\Rightarrow \frac{2 x}{3} - \frac{y}{2} = 1$ i.e. $\frac{x}{( \frac{3}{2} )} + \frac{y}{( - 2 )} = 1$ This equation is of the form $\frac{x}{a} + \frac{y}{b} = 1$ where $a = \frac{3}{2}$ and $b = - 2$ . Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are $\frac{3}{2}$ and -2 respectively.

(iii) The given equation is $3 y + 2 = 0$ . It can be written as $0 x + 3 y = - 2$ The equation is of the form $\frac{x}{a} + \frac{y}{b} = 1$ , where a $= 0$ and $b = - \frac{2}{3}$ . Therefore, equation (3) is in the intercept form, where the intercept on the y -axis $- \frac{2}{3}$ is and it has no intercept on the x-axis.

Find the distance of the point $(- 1, 1)$ from the line $12 (x + 6) = 5 (y - 2)$ . Sol. The given equation of the line is

$12 (x + 6) = 5 (y - 2) .$

$\Rightarrow 12 x + 72 = 5 y - 10$ $\Rightarrow 12 x - 5 y + 82 = 0$ On comparing equation (1) with general equation of line $Ax + By + C = 0$ , we obtain A $= 12, B = - 5$ , and $C = 82$ . It is known that the perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_{1}, y_{1})$ is given by

$d = \frac{∣ Ax _{1} + By _{1} + C ∣}{A ^{2} + B ^{2}} .$

The given point is $(x_{1}, y_{1}) = (- 1, 1)$ . Therefore, the distance of point (-1, 1) from the given line

$= \frac{∣12 ( - 1 ) + ( - 5 ) ( 1 ) + 82∣}{( 12 ) ^{2} + ( - 5 ) ^{2}} = \frac{∣ - 12 - 5 + 82∣}{169} = \frac{∣65∣}{13} = 5 units$

Find the points on the x-axis, whose distances from the line $\frac{x}{3} + \frac{y}{4} = 1$ are 4 units. Sol. The given equation of line is $\frac{x}{3} + \frac{y}{4} = 1$

$or 4 x + 3 y - 12 = 0$

On comparing equation (1) with general equation of line $Ax + By + C = 0$ , we obtain $A = 4, B = 3$ , and $C = - 12$ .

Let $(a, 0)$ be the point on the x-axis whose distance from the given line is 4 units. It is known that the perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_{1}, y_{1})$ is given by $d = \frac{∣ Ax _{1} + By _{1} + C ∣}{A ^{2} + B ^{2}}$

Therefore, $4 = \frac{∣4 a + 3 \times 0 - 12∣}{4 ^{2} + 3 ^{2}}$

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow 4 = \frac{∣4 a - 12∣}{5} \Rightarrow ∣4 a - 12∣ = 20 \pm (4 a - 12) = 20 (4 a - 12) = 20 4 a = 20 + 12 a = 8 or - 2 or or 4 a = - 12) = 20 + 12$

Thus, the required points on the x-axis are (-2, 0 ) and (8, 0).

Find the distance between parallel lines

(i) $15 x + 8 y - 34 = 0$ and $15 x + 8 y + 31 = 0$ (ii) $l (x + y) + p = 0$ and $l (x + y) - r = 0$

Sol. It is known that the distance (d) between parallel lines $Ax + By + C_{1} = 0$ and $Ax + By + C_{2} = 0$ is given by

$d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$

(i) The given parallel lines are $15 x + 8 y - 34 = 0$ and $15 x + 8 y + 31 = 0$ . Here, $A = 15, B = 8, C_{1} = - 34$ , and $C_{2} = 31$ . Therefore, the distance between the parallel lines is

$d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}} \Rightarrow d = \frac{∣ - 34 - 31∣}{( 15 ) ^{2} + ( 8 ) ^{2}}$

$\Rightarrow d = \frac{∣ - 65∣}{17} \Rightarrow \frac{65}{17}$ units (ii) The given parallel lines are

$l (x + y) + p = 0 and l (x + y) - r = 0. l x + l y + p = 0 and l x + l y - r = 0$

Here, $A = l, B = l, C_{1} = p$ , and $C_{2} = - r$ . Therefore, the distance between the parallel lines is

$d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$

$\Rightarrow d = \frac{∣ p + r ∣}{l ^{2} + l ^{2}} \Rightarrow d = \frac{∣ p + r ∣}{2 l ^{2}}$ $\Rightarrow d = \frac{∣ p + r ∣}{l 2} \Rightarrow d = \frac{1}{2} \frac{p + r}{l}$ units

Find equation of the line parallel to the line 3x $- 4 y + 2 = 0$ and passing through the point ( -2, 3). Sol. The equation of the given line is

$or or 3 x - 4 y + 2 = 0 y = \frac{3 x}{4} + \frac{2}{4} y = \frac{3}{4} x + \frac{1}{2}$

which is of the form $y = mx + c$ $∴$ Slope of the given line $= \frac{3}{4}$ It is known that parallel lines have the same slope. ∴ Slope of the other line, $m = \frac{3}{4}$ Now, the equation of the line that has a slope of $\frac{3}{4}$ and passes through the point ( $- 2, 3$ ) is

$(y - 3) = \frac{3}{4} {x - (- 2)}$

$\Rightarrow 4 y - 12 = 3 x + 6$ i.e. $3 x - 4 y + 18 = 0$

Find equation of the line perpendicular to the line $x - 7 y + 5 = 0$ and having x-intercept 3. Sol. The given equation of line is $x - 7 y + 5 = 0$ . or $y = \frac{1}{7} x + \frac{5}{7}$ which is of the form $y = mx + c$

$∴ Slope of the given line = \frac{1}{7}$

The slope of the line perpendicular to the line having a slope of is $\frac{1}{7}$ is

$m = - \frac{1}{( \frac{1}{7} )} = - 7$

The equation of the line with slope -7 and xintercept 3 is given by

$y = m (x - d)$

$\Rightarrow y = - 7 (x - 3) \Rightarrow y = - 7 x + 21 \Rightarrow 7 x + y = 21$

Find angles between the lines $3 x + y = 1$ and $x + 3 y = 1$ Sol. The given lines are $3 x + y = 1$ and

$x + 3 y y = 1. = - 3 x + 1$

and

$y = - \frac{1}{3} x + \frac{1}{3}$

The slope of line (1) is $m_{1} = - 3$ , while the slope of line (2) is $m_{2} = - \frac{1}{3}$ The acute angle i.e., $θ$ between the two lines is given by

$tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$

$\Rightarrow tan θ = \frac{- 3 + \frac{1}{3}}{1 + ( - 3 ) ( - \frac{1}{3} )} = \frac{\frac{- 3 + 1}{3}}{1 + 1} = \frac{- 2}{2 \times 3}$

$\Rightarrow tan θ = \frac{1}{3} \Rightarrow θ = 3 0^{\circ}$

Thus, the angle between the given lines is either $3 0^{\circ}$ or $18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ}$ .

The line through the points $(h, 3)$ and $(4, 1)$ intersects the line $7 x - 9 y - 19 = 0$ , at right angle. Find the value of h. Sol. The slope of the line passing through points (h, $3)$ and $(4, 1)$ is

$m_{1} = \frac{1 - 3}{4 - h} = \frac{- 2}{4 - h}$

The slope of line $7 x - 9 y - 19 = 0$ or

$y = \frac{7}{9} x - \frac{19}{9} is m_{2} = \frac{7}{9}$

It is given that the two lines are perpendicular.

$∴ m_{1} \times m_{2} = - 1 \Rightarrow (\frac{- 2}{4 - h}) \times (\frac{7}{9}) = - 1 \Rightarrow \frac{- 14}{36 - 9 h} = - 1 \Rightarrow 14 = 36 - 9 h \Rightarrow 9 h = 36 - 14 \Rightarrow h = \frac{22}{9}$

Thus, the vlaue of h is $\frac{22}{9}$ .

Prove that the line through the point $(x_{1}, y_{1})$ and parallel to the line $Ax + By + C = 0$ is $A (x - x_{1}) + B (y - y_{1}) = 0$ Sol. The slope of line $Ax + By + C = 0$

$or y = (\frac{- A}{B}) x + (\frac{- C}{B}) is m = - \frac{A}{B}$

It is known that parallel lines have the same slope. $∴$ Slope of the other line $= m = - \frac{A}{B}$

The equation of the line passing through point

$(x_{1}, y_{1}) and having a slope m = - \frac{A}{B} is y - y_{1} = m (x - x_{1})$

$\Rightarrow y - y_{1} = - \frac{A}{B} (x - x_{1})$ $\Rightarrow B (y - y_{1}) = - A (x - x_{1})$ $\Rightarrow A (x - x_{1}) + B (y - y_{1}) = 0$

Hence, the line through point $(x_{1}, y_{1})$ and parallel to line $Ax + By + C = 0$ is $A (x - x_{1}) + B (y - y_{1}) = 0$

Two lines passing through the point $(2, 3)$ intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line. Sol. It is given that the slope of the first line, $m_{1} = 2$ . Let the slope of the other line be $m_{2}$ . The angle between the two lines is $6 0^{\circ}$ . $∴ tan 6 0^{\circ} = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ $\Rightarrow 3 = \frac{2 - m _{2}}{1 + 2 m _{2}}$ $\Rightarrow 3 = \pm (\frac{2 - m _{2}}{1 + 2 m _{2}})$ $\Rightarrow 3 = \frac{2 - m _{2}}{1 + 2 m _{2}}$ or $3 = - (\frac{2 - m _{2}}{1 + 2 m _{2}})$ $\Rightarrow 3 (1 + 2 m_{2}) = 2 - m_{2}$ or $3 (1 + 2 m_{2}) = - (2 - m_{2})$ $\Rightarrow 3 + 23 m_{2} + m_{2} = 2$ or $3 + 23 m_{2} - m_{2} = - 2$ $\Rightarrow 3 + (23 + 1) m_{2} = 2$ or $3 + (23 - 1) m_{2} = - 2$ $\Rightarrow m_{2} = \frac{2 - 3}{( 2 3 + 1 )}$ or $m_{2} = \frac{- ( 2 + 3 )}{( 2 3 - 1 )}$

Case I :

$m_{2} = (\frac{2 - 3}{2 3 + 1})$

The equation of the line passing through point $(2, 3)$ and having a slope of $\frac{( 2 - 3 )}{( 2 3 + 1 )}$ is given by

$(y - 3) = \frac{2 - 3}{2 3 + 1} (x - 2)$

$(23 + 1) y - 3 (23 + 1) = (2 - 3) x - 2 (2 - 3)$ $(3 - 2) x + (23 + 1) y = - 4 + 23 + 63 + 3$ $(3 - 2) x + (23 + 1) y = - 1 + 83$ Case II :

$m_{2} = \frac{- ( 2 + 3 )}{( 2 3 - 1 )}$

The equation of the line passing through point $(2, 3)$ and having a slope of $\frac{- ( 2 + 3 )}{( 2 3 - 1 )}$ is

$(y - 3) = \frac{- ( 2 + 3 )}{( 2 3 - 1 )} (x - 2) (23 - 1) y - 3 (23 - 1) = - (2 + 3) x + 2 (2 + 3) (23 - 1) y + (2 + 3) x = 4 + 23 + 63 - 3 (2 + 3) x + (23 - 1) y = 1 + 83$

In this case, the equation of the other lines is $(2 + 3) x + (23 - 1) y = 1 + 83$ Thus, the required equation of the line is $(3 - 2) x + (23 + 1) y = - 1 + 83$ or $(2 + 3) x + (23 - 1) y = 1 + 83$

Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and (-1, 2). Sol. The right bisector of a line segment bisects the line segment at 90°. The end-points of the line segment are given as A (3, 4) and B (-1, 2).

Accordingly, mid-point of

$AB = (\frac{3 - 1}{2}, \frac{4 + 2}{2}) = (1, 3)$

Slope of $AB = \frac{2 - 4}{- 1 - 3} = \frac{- 2}{- 4} = \frac{1}{2}$

∴ Slope of the line perpendicular to

$AB = - \frac{1}{( \frac{1}{2} )} = - 2$

The equation of the line passing through $(1, 3)$ and having a slope of -2 is

$(y - 3) = - 2 (x - 1)$

$\Rightarrow y - 3 = - 2 x + 2$ $\Rightarrow 2 x + y = 5$ Thus, the required equation of the line is $2 x + y = 5$ .

Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line $3 x - 4 y - 16 = 0$ . Sol. Let (a, b) be the coordinates of the foot of the perpendicular from the point (-1, 3) to the line $3 x - 4 y - 16 = 0$ .

Slope of the line joining $(- 1, 3)$ and $(a, b)$ ,

$m_{1} = \frac{b - 3}{a + 1}$

Slope of the line $3 x - 4 y - 16 = 0$ or

$y = \frac{3}{4} x - 4, m_{2} = \frac{3}{4}$

Since these two lines are perpendicular,

$m_{1} m_{2} = - 1$

$∴ (\frac{b - 3}{a + 1}) \times (\frac{3}{4}) = - 1$

$\Rightarrow \frac{3 b - 9}{4 a + 4} = - 1 \Rightarrow 3 b - 9 = - 4 a - 4 \Rightarrow 4 a + 3 b = 5$

Point $(a, b)$ lies on line $3 x - 4 y = 16$

$∴ 3 a - 4 b = 16$

On solving equations (1) and (2), we obtain

$a = \frac{68}{25} and b = - \frac{49}{25}$

Thus, the required coordinates of the foot of the perpendicular are $(\frac{68}{25}, - \frac{49}{25})$

The perpendicular from the origin to the line y $= mx + c$ meets it at the point (-1, 2). Find the values of m and c. Sol. The given equation of line is $y = mx + c$ . It is given that the perpendicular from the origin meets the given line at (-1, 2). Therefore, the line joining the points $(0, 0)$ and $(- 1, 2)$ is perpendicular to the given line. ∴ Slope of the line joining $(0, 0)$ and (-1, 2) The slope of the given line is m. $∴ m \times - 2 = - 1$ [The two lines are perpendicular]

$\Rightarrow m = \frac{1}{2}$

Since point (-1, 2) lies on the given line, it satisfies the equation $y = mx + c$ .

$∴ \Rightarrow 2 = m (- 1) + c 2 = \frac{1}{2} (- 1) + c$

$\Rightarrow c = 2 + \frac{1}{2} = \frac{5}{2}$ Thus, the respective values of m and c are $\frac{1}{2}$ and $\frac{5}{2}$ .

If p and q are the lengths of perpendiculars from the origin to the lines $x cos θ - y sin θ = k cos 2 θ$ and $x sec θ + y cosec θ = k$ , respectively, prove that $p^{2} + 4 q^{2} = k^{2}$ Sol. The equations of given lines are

$x cos θ - y sin θ = k cos 2 θ x sec θ + y cosec θ = k$

The perpendicular distance (d) of a line $A x + B y + C = 0$ from a point $(x_{1}, y_{1})$ is given by $d = \frac{∣ Ax _{1} + By _{1} + C ∣}{A ^{2} + B ^{2}}$ On comparing equation (1) to the general equation of line i.e., $Ax + By + C = 0$ , we obtain $A = cos θ$ , $B = - sin θ$ , and $C = - k cos 2 θ$ . It is given that p is the length of the perpendicular from $(0, 0)$ to line (1).

$∴ p = \frac{∣ A ( 0 ) + B ( 0 ) + C ∣}{A ^{2} + B ^{2}} = \frac{∣ C ∣}{A ^{2} + B ^{2}}$

$= \frac{∣ - k c o s 2 θ ∣}{c o s ^{2} θ + s i n ^{2} θ} = ∣ - k cos 2 θ ∣$

On comparing equation (2) to the general equation of line i.e., $Ax + By + C = 0$ , we obtain $A = sec θ, B = cosec θ$ , and $C = - k$ . It is given that q is the length of the perpendicular from $(0, 0)$ to line (2). $∴ q = \frac{∣ A ( 0 ) + B ( 0 ) + C ∣}{A ^{2} + B ^{2}} = \frac{∣ C ∣}{A ^{2} + B ^{2}}$

$= \frac{∣ - k ∣}{s e c ^{2} θ + cosec ^{2} θ}$

From (3) and (4), we have

$p^{2} + 4 q^{2} = (∣ - k cos 2 θ ∣)^{2} + 4 (\frac{∣ - k ∣}{sec ^{2} θ + cosec ^{2} θ})^{2} = k^{2} cos^{2} 2 θ + \frac{4 k ^{2}}{( sec ^{2} θ + cosec ^{2} θ )} = k^{2} cos^{2} 2 θ + \frac{4 k ^{2}}{( \frac{1}{c o s ^{2} θ} + \frac{1}{s i n ^{2} θ} )} = k^{2} cos^{2} 2 θ + \frac{4 k ^{2}}{( \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n ^{2} θ c o s ^{2} θ} )} = k^{2} cos^{2} 2 θ + \frac{4 k ^{2}}{( \frac{1}{s i n ^{2} θ c o s ^{2} θ} )}$

$= k^{2} cos^{2} 2 θ + 4 k^{2} sin^{2} θ cos^{2} θ$ $= k^{2} cos^{2} 2 θ + k^{2} (2 sin θ cos θ)^{2}$ $= k^{2} (cos^{2} 2 θ + sin^{2} 2 θ) = k^{2}$ Hence, we proved that $p^{2} + 4 q^{2} = k^{2}$ .

In the triangle ABC with vertices $A (2, 3), B$ $(4, - 1)$ and $C (1, 2)$ , find the equation and length of altitude from the vertex A. Sol. Let AD be the altitude of triangle ABC from vertex A. Accordingly, $AD ⊥ BC$

The equation of the line passing through point $(2, 3)$ and having a slope of 1 is $(y - 3) = 1 (x - 2)$ $\Rightarrow x - y + 1 = 0 \Rightarrow y - x = 1$ Therefore, equation of the altitude from vertex $A = y - x = 1$ . Length of AD = Length of the perpendicular from $A (2, 3)$ to BC The equation of BC is

$(y + 1) = \frac{2 + 1}{1 - 4} (x - 4)$

$\Rightarrow (y + 1) = - 1 (x - 4)$ $\Rightarrow y + 1 = - x + 4$ $\Rightarrow x + y - 3 = 0$

The perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_{1}, y_{1})$ is given by $d = \frac{∣ Ax _{1} + By _{1} + C ∣}{A ^{2} + B ^{2}}$ On comparing equation (1) to the general equation of line $Ax + By + C = 0$ , we obtain $A = 1, B = 1$ , and $C = - 3$ .

∴ Length of $AD = \frac{∣1 \times 2 + 1 \times 3 - 3∣}{1 ^{2} + 1 ^{2}}$

$= \frac{∣2∣}{2} = \frac{2}{2} = 2 units$

Thus, the equation and the length of the altitude from vertex A are $y - x = 1$ and $2$ units respectively.

If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b , then show that $: \frac{1}{p ^{2}} = \frac{1}{a ^{2}} + \frac{1}{b ^{2}}$ Sol. It is known that the equation of a line whose intercepts on the axes are a and b is

$\frac{x}{a} + \frac{y}{b} = 1$

$\Rightarrow b x + a y = ab$

$\Rightarrow bx + ay - ab = 0$

The perpendicular distance (d) of a lines $Ax + By + C = 0$ from a point $(x_{1}, y_{1})$ is given by $d = \frac{∣ Ax _{1} + By _{1} + C ∣}{A ^{2} + B ^{2}}$ On comparing equation (1) to the general equation of line $Ax + By + C = 0$ , we obtain $A = b, B = a$ , and $C = - ab$ . Therefore, if p is the length of the perpendicular from point $(x_{1}, y_{1}) = (0, 0)$ to line (1), we obtain

$p = \frac{∣ A ( 0 ) + B ( 0 ) - ab ∣}{b ^{2} + a ^{2}}$

$\Rightarrow p = \frac{∣ - ab ∣}{a ^{2} + b ^{2}}$ On squaring both sides, we obtain

$p^{2} = \frac{( - ab ) ^{2}}{a ^{2} + b ^{2}}$

$\Rightarrow p^{2} (a^{2} + b^{2}) = a^{2} b^{2}$

$\Rightarrow \frac{a ^{2} + b ^{2}}{a ^{2} b ^{2}} = \frac{1}{p ^{2}} \Rightarrow \frac{1}{p ^{2}} = \frac{1}{a ^{2}} + \frac{1}{b ^{2}}$

Hence, we showed that $\frac{1}{p ^{2}} = \frac{1}{a ^{2}} + \frac{1}{b ^{2}}$ .

MISCELLANEOUS EXERCISE

Find the values of k for which the line $(k - 3) x - (4 - k^{2}) y + k^{2} - 7 k + 6 = 0$ is (a) Parallel to the x -axis, (b) Parallel to the y-axis, (c) Passing through the origin. Sol. The given equation of line is

$(k - 3) x - (4 - k^{2}) y + k^{2} - 7 k + 6 = 0$

(a) If the given line is parallel to the x-axis, then Slope of the given line = Slope of the x -axis The given line can be written as

$(4 - k^{2}) y = (k - 3) x + k^{2} - 7 k + 6 = 0 y = \frac{( k - 3 )}{( 4 - k ^{2} )} x + \frac{k ^{2} - 7 k + 6}{( 4 - k ^{2} )}, which is of the$

form $y = mx + c$ . Slope of the given line $= \frac{( k - 3 )}{( 4 - k ^{2} )}$ Slope of the x -axis $= 0$ $∴ \frac{( k - 3 )}{( 4 - k ^{2} )} = 0$ $\Rightarrow k - 3 = 0 \Rightarrow k = 3$ Thus, if the given line is parallel to the x-axis, then the value of k is 3. (b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined. The slope of the given line is $\frac{( k - 3 )}{( 4 - k ^{2} )}$ . Now, $\frac{( k - 3 )}{( 4 - k ^{2} )}$ is undefined at $k^{2} = 4$

$k^{2} = 4 \Rightarrow k = \pm 2$

Thus, if the given line is parallel to the y-axis, then the value of k is ±2. (c) If the given line is passing through the origin, then point $(0, 0)$ satisfies the given equation of line.

$(k - 3) (0) - (4 - k^{2}) (0) + k^{2} - 7 k + 6 = 0$

$\Rightarrow k^{2} - 7 k + 6 = 0 \Rightarrow k^{2} - 6 k - k + 6 = 0$ $\Rightarrow (k - 6) (k - 1) = 0 \Rightarrow k = 1$ or 6 Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Find the values of $θ$ and p , if the equation $x cos θ + y sin θ = p$ is the normal form of the line $3 x + y + 2 = 0$ . Sol. The equation of the given line is

$3 x + y + 2 = 0.$

This equation can be reduced as

$3 x + y + 2 = 0 \Rightarrow - 3 x - y = 2$

On dividing both sides by

$(- 3)^{2} + (- 1)^{2} = 2$

we obtain $- \frac{3}{2} x - \frac{1}{2} y = \frac{2}{2}$

$\Rightarrow (- \frac{3}{2}) x + (- \frac{1}{2}) y = 1$

Comparing equation (1) to

$x cos θ + y sin θ = p,$

we obtain

$cos θ = - \frac{3}{2}, sin θ = - \frac{1}{2} and p = 1$

Since the values of $sin θ$ and $cos θ$ are negative,

$θ = π + \frac{π}{6} = \frac{7 π}{6}$

Thus, the respective values of $θ$ and p are $\frac{7 π}{6}$ and 1.

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and -6, respectively. Sol. Let the intercepts cut by the given lines on the axes be a and b. It is given that

$a + b = 1 ab = - 6$

On solving equations (1) and (2), we obtain $a = 3$ and $b = - 2$ or $a = - 2$ and $b = 3$ It is known that the equation of the line whose intercepts on the axes are a and b is

$\frac{x}{a} + \frac{y}{b} = 1 or bx + ay - ab = 0$

Case I : $a = 3$ and $b = - 2$ In this case, the equation of the line is $- 2 x + 3 y + 6 = 0$ , i.e., $2 x - 3 y = 6$ . Case II : $a = - 2$ and $b = 3$ In this case, the equation of the line is $3 x - 2 y + 6 = 0$ , i.e., $- 3 x + 2 y = 6$ . Thus, the required equation of the lines are 2x $- 3 y = 6$ and $- 3 x + 2 y = 6$ .

What are the points on the y-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units. Sol. Let (0, b) be the point on the y-axis whose distance from line is 4 units. The given line can be written as

$4 x + 3 y - 12 = 0$

On comparing equation (1) to the general equation of line $Ax + By + C = 0$ , we obtain A $= 4, B = 3$ , and $C = - 12$ . It is known that the perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_{1}, y_{1})$ is given by $d = \frac{∣ Ax _{1} + By _{1} + C ∣}{A ^{2} + B ^{2}}$ . Therefore, if $(0, b)$ is the point on the y-axis whose distance from line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units, then:

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow 3 b 4 = \frac{∣4 ( 0 ) + 3 ( b ) - 12∣}{4 ^{2} + 3 ^{2}} 20 = ∣3 b - 12∣ 20 = (3 b - 12) 3 b = 20 = \pm (3 b - 12) b = \frac{∣3 b - 12∣}{5} or \Rightarrow \Rightarrow or or b = - \frac{8}{3}$

Thus, the required points are $(0, \frac{32}{3})$ and (0,- $\frac{8}{3}$ )

Find the perpendicular distance from the origin to the line joining the points $(cos θ, sin θ)$ and $(cos ϕ, sin ϕ)$ Sol. The equation of the line joining the points $(cos θ, sin θ)$ and $(cos ϕ, sin ϕ)$ is given by

$y - sin θ = \frac{s i n ϕ - s i n θ}{c o s ϕ - c o s θ} (x - cos θ)$

$\Rightarrow y (cos ϕ - cos θ) - sin θ (cos ϕ - cos θ)$ $= x (sin ϕ - sin θ) - cos θ (sin ϕ - sin θ)$ $\Rightarrow x (sin θ - sin ϕ) + y (cos ϕ - cos θ) + cos θ sin ϕ -$ $cos θ sin θ - sin θ cos ϕ + sin θ cos θ = 0$ $\Rightarrow x (sin θ - sin ϕ) + y (cos ϕ - cos θ)$

$+ sin (ϕ - θ) = 0$

$Ax + By + C = 0$ , where $A = sin θ - sin ϕ, B = cos ϕ - cos θ$ and $C = sin (ϕ - θ)$

It is known that the perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_{1}, y_{1})$ is given by $d = \frac{∣ Ax _{1} + By _{1} + C ∣}{A ^{2} + B ^{2}}$ . Therefore, the perpendicular distance (d) of the given line from point $(x_{1}, y_{1}) = (0, 0)$ is

$d = \frac{∣ ( sin θ - sin ϕ ) ( 0 ) + ( cos ϕ - cos θ ) ( 0 ) + sin ( ϕ - θ ) ∣}{( sin θ - sin ϕ ) ^{2} + ( cos ϕ - cos ϕ ) ^{2}} = \frac{∣ sin ( ϕ - θ ) ∣}{sin ^{2} θ + sin ^{2} ϕ - 2 sin θ sin ϕ + cos ^{2} ϕ + cos ^{2} θ - 2 cos ϕ cos θ} = \frac{∣ ( sin ϕ - θ ) ∣}{( sin ^{2} θ + cos ^{2} θ ) + ( sin ^{2} ϕ + cos ^{2} ϕ ) - 2 ( sin θ sin ϕ + cos ϕ + cos θ )} = \frac{∣ sin ( ϕ - θ ) ∣}{1 + 1 - 2 ( cos ( ϕ - θ ))} = \frac{∣ sin ( ϕ - θ ) ∣}{2 ( 1 - cos ( ϕ - θ ))} = \frac{∣ sin ( ϕ - θ ) ∣}{2 ( 2 sin ^{2} ( \frac{ϕ - θ}{2} ) )} = \frac{∣ sin ( ϕ - θ ) ∣}{2 sin \frac{( ϕ - θ )}{2}}$

Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines $x - 7 y + 5 = 0$ and 3x $+ y = 0$ . Sol. The equation of any line parallel to the $y$ -axis is of the form

$x = a$

The two given lines are

$x - 7 y + 5 = 0 3 x + y = 0$

On solving equations (2) and (3), we obtain

$x = - \frac{5}{22} and y = \frac{15}{22} .$

Therefore, $(- \frac{5}{22}, \frac{15}{22})$ is the point of intersection of lines (2) and (3). Since line $x = a$ passes through point $(- \frac{5}{22}, \frac{15}{22})$ . So, $a = - \frac{5}{22}$ Thus, the required equation of the line is $x = - \frac{5}{22}$

Find the equation of a line drawn perpendicular to the line $\frac{x}{4} + \frac{y}{6} = 1$ through the point, where it meets the y-axis. Sol. The equation of the given line is $\frac{x}{4} + \frac{y}{6} = 1$ . This equation can also be written as $3 x + 2 y - 12 = 0$ $y = \frac{- 3}{2} x + 6$ , which is of the form $y = mx + c$ $∴$ Slope of the given line $= \frac{- 3}{2}$ ∴ Slope of line perpendicular to the given

$line = - \frac{1}{( - \frac{3}{2} )} = \frac{2}{3}$

Let the given line intersect the y-axis at (0, y). On substituting x with 0 in the equation of the given line, we obtain $\frac{y}{6} = 1 \Rightarrow y = 6$

$∴$ The given line intersects the y-axis at (0, 6). The equation of the line that has a slope of $\frac{2}{3}$ and passes through point $(0, 6)$ is

$(y - 6) = \frac{2}{3} (x - 0)$

$\Rightarrow 3 y - 18 = 2 x$

$\Rightarrow 2 x - 3 y + 18 = 0$

Thus, the required equation of the line is $2 x - 3 y + 18 = 0$ .

Find the area of the triangle formed by the lines $y - x = 0, x + y = 0$ and $x - k = 0$ . Sol. The equations of the given lines are

$y - x = 0 x + y = 0 x - k = 0$

The point of intersection of lines (1) and (2) is given by $x = 0$ and $y = 0$ The point of intersection of lines (2) and (3) is given by $x = k$ and $y = - k$ The point of intersection of lines (3) and (1) is given by $x = k$ and $y = k$ Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, -k), and (k, k). We know that the area of a triangle whose vertices are $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is

$\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$

Therefore, area of the triangle formed by the three given lines

$= \frac{1}{2} ∣0 (- k - k) + k (k - 0) + k (0 + k) ∣ = \frac{1}{2} k^{2} + k^{2} = \frac{1}{2} 2 k^{2} = k^{2} sq.units$

Find the value of p so that the three lines $3 x + y - 2 = 0, px + 2 y - 3 = 0$ and $2 x - y - 3 = 0$ may intersect at one point. Sol. The equations of the given lines are

$3 x + y - 2 = 0 p x + 2 y - 3 = 0 2 x - y - 3 = 0$

On solving equations (1) and (3), we obtain $x = 1$ and $y = - 1$ Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).

$p (1) + 2 (- 1) - 3 = 0$

$\Rightarrow p - 2 - 3 = 0 \Rightarrow p = 5$ Thus, the required value of p is 5 .

If three lines whose equations are

$y = m_{1} x + c_{1}, y = m_{2} x + c_{2} and y = m_{3} x + c_{3}$

concurrent, then show that

$m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

Sol. The equations of the given lines are

$y = m_{1} x + c_{1} y = m_{2} x + c_{2} y = m_{3} x + c_{3}$

On subtracting equation (1) from (2), we obtain

$0 = (m_{2} - m_{1}) x + (c_{2} - c_{1})$

$\Rightarrow (m_{1} - m_{2}) x = (c_{2} - c_{1})$ $\Rightarrow x = \frac{c _{2} - c _{1}}{m _{1} - m _{2}}$ On substituting this value of x in (1), we obtain

$\Rightarrow \Rightarrow y = m_{1} (\frac{c _{2} - c _{1}}{m _{1} - m _{2}}) + c_{1} y = \frac{m _{1} c _{2} - m _{1} c _{1}}{m _{1} - m _{2}} + c_{1} y = \frac{m _{1} c _{2} - m _{1} c _{1} + m _{1} c _{1} - m _{2} c _{1}}{m _{1} - m _{2}}$

$\Rightarrow y = \frac{m _{1} c _{2} - m _{2} c _{1}}{m _{1} - m _{2}}$ $∴ (\frac{c _{2} - c _{1}}{m _{1} - m _{2}}, \frac{m _{1} c _{2} - m _{2} c _{1}}{m _{1} - m _{2}})$ is the point of intersection of lines (1) and (2). It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

$\frac{m _{1} c _{2} - m _{2} c _{1}}{m _{1} - m _{2}} = m_{3} (\frac{c _{2} - c _{1}}{m _{1} - m _{2}}) + c_{3}$

$\Rightarrow \frac{m _{1} c _{2} - m _{2} c _{1}}{m _{1} - m _{2}} = \frac{m _{3} c _{2} - m _{3} c _{1} + c _{3} m _{1} - c _{3} m _{2}}{m _{1} - m _{2}}$ $\Rightarrow m_{1} c_{2} - m_{2} c_{1} - m_{3} c_{2} + m_{3} c_{1} - c_{3} m_{1} + c_{3} m_{2} = 0$ $\Rightarrow m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

Find the equation of the lines through the point $(3, 2)$ which make an angle of $4 5^{\circ}$ with the line $x - 2 y = 3$ . Sol. Let the slope of the required line be $m_{1}$ . The given line can be written as $y = \frac{1}{2} x - \frac{3}{2}$ , which is of the form $y = mx + c$ ∴ Slope of the given line $= m_{2} = \frac{1}{2}$ It is given that the angle between the required line and line $x - 2 y = 3$ is 45°. We know that if $θ$ is the acute angle between lines $l_{1}$ and $l_{2}$ with slopes $m_{1}$ and $m_{2}$ , then

$tan θ = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}} ∴ tan 4 5^{\circ} = \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}}$

$\Rightarrow 1 = \frac{\frac{1}{2} - m _{1}}{1 + \frac{m _{1}}{2}}$

$\Rightarrow 1 = \frac{( \frac{1 - 2 m _{1}}{2} )}{2 + \frac{m _{1}}{2}}$

$\Rightarrow 1 = \frac{1 - 2 m _{1}}{2 + m _{1}}$

$\Rightarrow 1 = \pm (\frac{1 - 2 m _{1}}{2 + m _{1}})$

$\Rightarrow 1 = \frac{1 - 2 m _{1}}{2 + m _{1}}$

$or 1 = - (\frac{1 - 2 m _{1}}{2 + m _{1}})$

$\Rightarrow 2 + m_{1} = 1 - 2 m_{1}$ or $2 + m_{1} = - 1 + 2 m_{1}$ $\Rightarrow m_{1} = - \frac{1}{3}$ or $m_{1} = 3$

Case I : $m_{1} = 3$ The equation of the line passing through $(3, 2)$ and having a slope of 3 is:

$y - 2 = 3 (x - 3) y - 2 = 3 x - 9 3 x - y = 7$

Case II: $m_{1} = - \frac{1}{3}$ The equation of the line passing through $(3, 2)$ and having a slope of $- \frac{1}{3}$ is:

$y - 2 = - \frac{1}{3} (x - 3)$

$\Rightarrow 3 y - 6 = - x + 3$ $\Rightarrow x + 3 y = 9$ Thus, the equations of the lines are $3 x - y = 7$ and $x + 3 y = 9$ .

Find the equation of the line passing through the point of intersection of the lines $4 x + 7 y -$ $3 = 0$ and $2 x - 3 y + 1 = 0$ that has equal intercepts on the axes. Sol. Let the equation of the line having equal intercepts on the axes be

$\frac{x}{a} + \frac{y}{a} = 1 or x + y = a$

On solving equations $4 x + 7 y - 3 = 0$ and $2 x -$ $3 y + 1 = 0$ , we obtain

$x = \frac{1}{13} and y = \frac{5}{13} .$

$∴ (\frac{1}{13}, \frac{5}{13})$ is the point of intersection of the two given lines. Since equation (1) passes through $(\frac{1}{13}, \frac{5}{13})$ point,

$\frac{1}{13} + \frac{5}{13} = a \Rightarrow a = \frac{6}{13}$

Equation (1) becomes

$x + y = \frac{6}{13} i.e. 13 x + 13 y = 6$

Thus, the required equation of the line is $13 x + 13 y = 6$ .

Show that the equation of the line passing through the origin and making an angle $θ$ with the line $y = mx + c$ is $\frac{y}{x} = \frac{m \pm t a n θ}{1 \mp m t a n θ}$ . Sol. Let the equation of the line passing through the origin be $y = m_{1} x$ or $m_{1} = \frac{y}{x}$ . If this line makes an angle of $θ$ with line $y = mx + c$ , then angle $θ$ is given by $∴ tan θ = \frac{m _{1} - m}{1 + m _{1} m}$ $\Rightarrow tan θ = \frac{\frac{y}{x} - m}{1 + \frac{y}{x} m} \Rightarrow tan θ = \pm (\frac{\frac{y}{x} - m}{1 + \frac{y}{x} m})$ $\Rightarrow tan θ = \frac{\frac{y}{x} - m}{1 + \frac{y}{x} m}$ or $tan θ = - (\frac{\frac{y}{x} - m}{1 + \frac{y}{x} m})$

Case I :

$\Rightarrow \Rightarrow \Rightarrow tan θ = \frac{\frac{y}{x} - m}{1 + \frac{y}{x} m} tan θ + \frac{y}{x} m or tan θ + \frac{y}{x} - m m + tan θ = \frac{y}{x} (1 - m tan θ) \frac{y}{x} = \frac{m + tan θ}{1 - m tan θ}$

Case II :

$tan θ = - (\frac{\frac{y}{x} - m}{1 + \frac{y}{x} m})$

Therefore, the required line is given by $\frac{y}{x} = \frac{m \pm t a n θ}{1 \mp m t a n θ}$

In what ratio, the line joining (-1, 1) and $(5, 7)$ is divided by the line $x + y = 4$ ? Sol. The equation of the line joining the points $(- 1, 1)$ and $(5, 7)$ is given by

$y - 1 = \frac{7 - 1}{5 + 1} (x + 1)$

$\Rightarrow y - 1 = \frac{6}{6} (x + 1)$

$\Rightarrow x - y + 2 = 0$

The equation of the given line is

$x + y - 4 = 0$

The point of intersection of lines (1) and (2) is given by $x = 1$ and $y = 3$ Let point $(1, 3)$ divide the line segment joining (-1, 1) and $(5, 7)$ in the ratio 1 : k. Accordingly, by section formula,

$(1, 3) = (\frac{k ( - 1 ) + 1 ( 5 )}{1 + k}, \frac{k ( 1 ) + 1 ( 7 )}{1 + k})$

$\Rightarrow (1, 3) = (\frac{- k + 5}{1 + k}, \frac{k + 7}{1 + k})$ $\Rightarrow \frac{- k + 5}{1 + k} = 1, \frac{k + 7}{1 + k} = 3$ $∴ \frac{- k + 5}{1 + k} = 1 \Rightarrow - k + 5 = 1 + k$ $\Rightarrow 2 k = 4 \Rightarrow k = 2$ Thus, the line joining the points $(- 1, 1)$ and $(5, 7)$ is divided by line $x + y = 4$ in the ratio 1 : 2.

Find the distance of the line $4 x + 7 y + 5 = 0$ from the point $(1, 2)$ along the line $2 x - y = 0$ . Sol. The given lines are

$2 x - y = 0 4 x + 7 y + 5 = 0$

A $(1, 2)$ is a point on line (1). Let B be the point of intersection of lines (1) and (2).

ncert-class-11-maths-chap-9-ques-15-mis-exer

On solving equations (1) and (2), we obtain $x = \frac{- 5}{18}$ and $y = \frac{- 5}{9}$

∴ Coordinates of point B are $(\frac{- 5}{18}, \frac{- 5}{9})$

By using distance formula, the distance between points A and B can be obtained as

$AB = (1 + \frac{5}{18})^{2} + (2 + \frac{5}{9})^{2} = (\frac{23}{18})^{2} + (\frac{23}{9})^{2} = (\frac{23}{2 \times 9})^{2} + (\frac{23}{9})^{2} = (\frac{23}{9})^{2} (\frac{1}{2})^{2} + (\frac{23}{9})^{2} = (\frac{23}{9})^{2} (\frac{1}{4} + 1) = \frac{23}{9} \frac{5}{4} = \frac{23}{9} \times \frac{5}{2} = \frac{23 5}{18} units$

Thus, the required distance is $\frac{23 5}{18}$ units

Find the direction in which a straight line must be drawn through the point (-1, 2) so that its point of intersection with the line x $+ y = 4$ may be at a distance of 3 units from this point. Sol. Let $y = mx + c$ be the line through point (-1, 2). Accordingly, $2 = m (- 1) + c$ .

$\Rightarrow \Rightarrow ∴ 2 = - m + c c = m + 2 y = m x + m + 2$

The given line is

$x + y = 4$

On solving equations (1) and (2), we obtain

$∴ x = \frac{2 - m}{m + 1} and y = \frac{5 m + 2}{m + 1} (\frac{2 - m}{m + 1}, \frac{5 m + 2}{m + 1}) is the point of$

intersection of lines (1) and (2). Since this point is at a distance of 3 units from point (-1, 2), according to distance formula,

$(\frac{2 - m}{m + 1} + 1)^{2} + (\frac{5 m + 2}{m + 1} - 2)^{2} = 3$

$\Rightarrow (\frac{2 - m + m + 1}{m + 1})^{2} + (\frac{5 m + 2 - 2 m - 2}{m + 1})^{2} = 3^{2}$ $\Rightarrow \frac{9}{( m + 1 ) ^{2}} + \frac{9 m ^{2}}{( m + 1 ) ^{2}} = 9$ $\Rightarrow \frac{1 + m ^{2}}{( m + 1 ) ^{2}} = 1$ $\Rightarrow 1 + m^{2} = m^{2} + 1 + 2 m$ $\Rightarrow 2 m = 0$ $\Rightarrow m = 0$ Thus, the slope of the required line must be zero i.e., the line must be parallel to the xaxis.

Find the image of the point $(3, 8)$ with respect to the line $x + 3 y = 7$ assuming the line to be a plane mirror. Sol. The equation of the given line is

$x + 3 y = 7$

Let point B (a, b) be the image of point A(3, 8). Accordingly, line (1) is the perpendicular bisector of AB.

mis-exer-ncert-class-11-chap-9-ques-17-maths

Slope of $AB = \frac{b - 8}{a - 3}$ , while the slope of line

$(1) = - \frac{1}{3}$

Since line (1) is perpendicular to AB ,

$(\frac{b - 8}{a - 3}) \times (- \frac{1}{3}) = - 1 \Rightarrow b - 8 = 3 a - 9 \Rightarrow \frac{b - 8}{3 a - 9} = 1 \Rightarrow 3 a - b = 1.$

Mid point of $AB = (\frac{a + 3}{2}, \frac{b + 8}{2})$ The mid-point of line segment AB will also satisfy line (1). Hence, from equation (1), we have

$(\frac{a + 3}{2}) + 3 (\frac{b + 8}{2}) = 7$

$\Rightarrow a + 3 + 3 b + 24 = 14$

$\Rightarrow a + 3 b = - 13$

On solving equations (2) and (3), we obtain a $= - 1$ and $b = - 4$ . Thus, the image of the given point with respect to the given line is (-1, -4).

If the lines $y = 3 x + 1$ and $2 y = x + 3$ are equally inclined to the line $y = mx + 4$ , find the value of m.

Sol. The equations of the given lines are

$y = 3 x + 1 2 y = x + 3 y = m x + 4$

Slope of line (1), $m_{1} = 3$ Slope of line (2), $m_{2} = \frac{1}{2}$ Slope of line (3), $m_{3} = m$ It is given that lines (1) and (2) are equally inclined to line (3). This means that the angle between lines (1) and (3) equals the angle between lines (2) and (3). $∴ \frac{m _{1} - m _{3}}{1 + m _{1} m _{3}} = \frac{m _{2} - m _{3}}{1 + m _{2} m _{3}}$ $\Rightarrow \frac{3 - m}{1 + 3 m} = \frac{\frac{1}{2} - m}{1 + \frac{1}{2} m}$ $\Rightarrow \frac{3 - m}{1 + 3 m} = \frac{1 - 2 m}{m + 2}$ $\Rightarrow \frac{3 - m}{1 + 3 m} = \pm (\frac{1 - 2 m}{m + 2})$ $\Rightarrow \frac{3 - m}{1 + 3 m} = \frac{1 - 2 m}{m + 2}$ or $\frac{3 - m}{1 + 3 m} = - (\frac{1 - 2 m}{m + 2})$ If $\frac{3 - m}{1 + 3 m} = \frac{1 - 2 m}{m + 2}$ then $(3 - m) (m + 2) = (1 - 2 m) (1 + 3 m)$ $\Rightarrow - m^{2} + m + 6 = 1 + m - 6 m^{2}$ $\Rightarrow 5 m^{2} + 5 = 0$ $\Rightarrow (m^{2} + 1) = 0$ $\Rightarrow m = - 1$ , which is not real Hence, this case is not possible If $\frac{3 - m}{1 + 3 m} = - (\frac{1 - 2 m}{m + 2})$ then $\Rightarrow (3 - m) (m + 2) = - (1 - 2 m) (1 + 3 m)$ $\Rightarrow - m^{2} + m + 6 = - (1 + m - 6 m^{2})$ $\Rightarrow 7 m^{2} - 2 m - 7 = 0$ $\Rightarrow m = \frac{2 \pm 4 - 4 ( 7 ) ( - 7 )}{2 ( 7 )}$ $\Rightarrow m = \frac{2 \pm 2 1 + 49}{14}$ $\Rightarrow m = \frac{1 \pm 5 2}{7}$ Thus, the required value of m is $\frac{1 \pm 5 2}{7}$

If sum of the perpendicular distances of a variable point $P (x, y)$ from the lines $x + y - 5 = 0$ and $3 x - 2 y + 7 = 0$ is always 10. Show that P must move on a line Sol. The equations of the given lines are

$x + y - 5 = 0 3 x - 2 y + 7 = 0$

The perpendicular distances of $P (x, y)$ from lines (1) and (2) are respectively given by which is the equation of a line.

$d_{1} = \frac{∣ x + y - 5∣}{( 1 ) ^{2} + ( 1 ) ^{2}} and d_{2} = \frac{∣3 x - 2 y + 7∣}{( 3 ) ^{2} + ( - 2 ) ^{2}}$

i.e. $d_{1} = \frac{∣ x + y - 5∣}{2}$ and $d_{2} = \frac{∣3 x - 2 y + 7∣}{13}$ It is given that $d_{1} + d_{2} = 10$

$∴ \frac{∣ x + y - 5∣}{2} + \frac{∣3 x - 2 y + 7∣}{13} = 10$

$\Rightarrow 13 ∣ x + y - 5∣ + 2 ∣3 x - 2 y + 7∣ - 1026 = 0$ $\Rightarrow 13 (x + y - 5) + 2 (3 x - 2 y + 7) - 1026 = 0$ [Assuming $(x + y - 5)$ and $(3 x - 2 y + 7)$ are positive] $\Rightarrow 13 x + 13 y - 513 + 32 x - 22 y + 72 - 1026 = 0$ $\Rightarrow x (13 + 32) + y (13 - 22) + (72 - 513 - 1026) = 0$ which is the equation of a line. Similarly, we can obtain the equation of line for any signs of $(x + y - 5)$ and $(3 x - 2 y + 7)$ Thus, point P must move on a line.

Find equation of the line which is equidistant from parallel lines $9 x + 6 y - 7 = 0$ and $3 x + 2 y +$ $6 = 0$ . Sol. The equations of the given lines are

$9 x + 6 y - 7 = 0 3 x + 2 y + 6 = 0$

Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (h, k) from line (1) is given by

$d_{1} = \frac{∣9 h + 6 k - 7∣}{( 9 ) ^{2} + ( 6 ) ^{2}} = \frac{∣9 h + 6 k - 7∣}{117} = \frac{∣9 h + 6 k - 7∣}{3 13}$

The perpendicular distance of P (h, k) from line (2) is given by

$d_{2} = \frac{∣3 h + 2 k + 6∣}{( 3 ) ^{2} + ( 2 ) ^{2}} = \frac{∣3 h + 2 k + 6∣}{13}$

Since $P (h, k)$ is equidistant from lines (1) and

$(2), d_{1} = d_{2} ∴ \frac{∣9 h + 6 k - 7∣}{3 13} = \frac{∣3 h + 2 k + 6∣}{13}$

$\Rightarrow ∣9 h + 6 k - 7∣ = 3∣3 h + 2 k + 6∣$ $\Rightarrow ∣9 h + 6 k - 7∣ = \pm 3 (3 h + 2 k + 6)$ $\Rightarrow 9 h + 6 k - 7 = 3 (3 h + 2 k + 6)$ or $9 h + 6 k - 7 = - 3 (3 h + 2 k + 6)$ The case $9 h + 6 k - 7 = 3 (3 h + 2 k + 6)$ is not possible as

$9 h + 6 k - 7 = 3 (3 h + 2 k + 6)$

$\Rightarrow - 7 = 18$ (which is absurd) $∴ 9 h + 6 k - 7 = - 9 h - 6 k - 18$ $\Rightarrow 18 h + 12 k + 11 = 0$ Thus, the required equation of the line is $18 h + 12 k + 11 = 0$

A ray of light passing through the point $(1, 2)$ reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Sol.

ques-21-maths-mis-exer-class-11-chap-9-ncert

Let the coordinates of point A be $(a, 0)$ . Draw a line (AL) perpendicular to the x-axis. We know that angle of incidence is equal to angle of reflection. Hence, let $∠ BAL = ∠ CAL = ϕ$

Let $∠ CAX = θ$

$∴ ∠ OAB ∴ ∠ BAX = 18 0^{\circ} - (θ + 2 ϕ) = 18 0^{\circ} - [θ + 2 (9 0^{\circ} - θ)] = 18 0^{\circ} - θ - 18 0^{\circ} + 2 θ = θ = 18 0^{\circ} - θ$

Now, slope of line $AC = \frac{3 - 0}{5 - a}$

$\Rightarrow tan θ = \frac{3}{5 - a}$

Slope of line $AB = \frac{2 - 0}{1 - a}$

$\Rightarrow tan (18 0^{\circ} - θ) = \frac{2}{1 - a} \Rightarrow - tan θ = \frac{2}{1 - a} \Rightarrow tan θ = \frac{2}{a - 1}$

From equations (1) and (2), we obtain

$\Rightarrow \frac{3}{5 - a} = \frac{2}{a - 1} 3 a - 3 = 10 - 2 a \Rightarrow a = \frac{13}{5}$

Thus, the coordinates of point A are $(\frac{13}{5}, 0)$ .

Prove that the product of the lengths of the perpendiculars drawn from the points $(a^{2} - b^{2}, 0)$ and $(- a^{2} - b^{2}, 0)$ to the line $\frac{x}{a} cos θ + \frac{y}{b} sin θ = 1$ is $b^{2}$ . Sol. The equation of the given line is

$\frac{x}{a} cos θ + \frac{y}{b} sin θ = 1$

or $bx cos θ + ay sin θ - ab = 0$ Length of the perpendicular from point $(a^{2} - b^{2}, 0)$ to line (1) is

$p_{1} = \frac{b cos θ ( a ^{2} - b ^{2} ) + a sin θ ( 0 ) - ab}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ} = \frac{b cos θ a ^{2} - b ^{2} - ab}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ}$

Length of the perpendicular from point $(- a^{2} - b^{2}, 0)$ to line (2) is

$p_{2} = \frac{b cos θ ( - a ^{2} - b ^{2} ) + a sin θ ( 0 ) - ab}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ} = \frac{b cos θ a ^{2} - b ^{2} + ab}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ}$

On multiplying equations (2) and (3), we obtain

$p_{1} p_{2} = = \frac{b cos θ a ^{2} - b ^{2} - ab ( b cos θ a ^{2} - b ^{2} + ab )}{( b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ ) ^{2}} = \frac{( b cos θ a ^{2} - b ^{2} - ab ) ( b cos θ a ^{2} - b ^{2} + ab )}{( b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ )} = \frac{( b cos θ a ^{2} - b ^{2} ) ^{2} - ( ab ) ^{2}}{( b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ )} = \frac{b ^{2} cos ^{2} θ ( a ^{2} - b ^{2} ) - a ^{2} b ^{2}}{( b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ )} = \frac{a ^{2} b ^{2} cos ^{2} θ - b ^{4} cos ^{2} θ - a ^{2} b ^{2}}{( b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ )} = \frac{b ^{2} a ^{2} cos ^{2} θ - b ^{2} cos ^{2} θ - a ^{2}}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ} \frac{b ^{2} a ^{2} cos ^{2} θ - b ^{2} cos ^{2} θ - a ^{2} sin ^{2} θ - a ^{2} cos ^{2} θ}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ} = \frac{b ^{2} - ( b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ )}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ} = \frac{b ^{2} ( b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ )}{b ^{2} cos ^{2} θ + a ^{2} sin ^{2} θ} = b^{2}$

Hence proved

A person standing at the junction (crossing) of two straight paths represented by the equations $2 x - 3 y + 4 = 0$ and $3 x + 4 y - 5 = 0$ wants to reach the path whose equation is $6 x - 7 y + 8 =$ 0 in the least time. Find equation of the path that he should follow. Sol. The equations of the given lines are

$2 x - 3 y + 4 = 0 3 x + 4 y - 5 = 0 6 x - 7 y + 8 = 0$

The person is standing at the junction of the paths represented by lines (1) and (2). On solving equations (1) and (2), we obtain

$x = - \frac{1}{17} and y = \frac{22}{17} .$

Thus, the person is standing at point $(- \frac{1}{17}, \frac{22}{17})$ The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point $(- \frac{1}{17}, \frac{22}{17})$ .

$- \frac{1}{m} = \frac{1}{( \frac{6}{7} )} = - \frac{7}{6}$

The equation of the line passing through $(- \frac{1}{17}, \frac{22}{17})$ and having a slope of $- \frac{7}{6}$ is given by

$(y - \frac{22}{17}) = - \frac{7}{6} (x + \frac{1}{17})$

$\Rightarrow 6 (17 y - 22) = - 7 (17 x + 1)$ $\Rightarrow 102 y - 132) = - 119 x - 7$ $\Rightarrow 119 x + 102 y = 125$ Hence, the path that the person should follow is $119 x + 102 y = 125$ .

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Explore Class 11 Maths NCERT Solutions chapter-wise, with solved exercises, important formulas, and easy explanations to help understand and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

4.0Class 11 Maths Chapter 9 Straight Lines: Exercise-Wise Questions and Topics

Exercise	Number of Questions	Important Topics Covered
Exercise 9.1	11 Questions & Solutions	Slope, intercepts, and equations of a straight line
Exercise 9.2	19 Questions & Solutions	Point-slope, two-point, and slope-intercept forms
Exercise 9.3	17 Questions & Solutions	Distance and midpoint formulas
Miscellaneous Exercise	23 Questions & Solutions	Mixed questions on slope, intercepts, line equations, distance, and midpoint

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 9 (Straight Lines)

Clear Understanding of Coordinate Geometry: Learn how points, slopes, intercepts, and straight lines are represented on the Cartesian plane through clear explanations and graphical examples.
Different Forms of the Equation of a Line: Understand the slope-intercept, point-slope, two-point, intercept, and normal forms of a straight line and learn when to use each form.
Step-by-Step Slope Calculations: Learn how to find the slope of a line from two points and understand the relationship between slope and the angle of inclination, including cases involving obtuse angles.
Intercepts and Graphical Representation: Understand how to find the (x)-intercept and (y)-intercept of a line and represent the equation correctly on a coordinate plane.
Distance from a Point to a Line: Learn how to calculate the perpendicular distance between a point and a straight line using the appropriate formula and clear calculation steps.
Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts with accurate mathematical methods and alignment with the latest NCERT syllabus.
Complete NCERT Exercise Coverage: All questions from the NCERT exercises, including the miscellaneous exercise, are covered with step-by-step solutions to help students understand and apply the concepts of Straight Lines.

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Frequently Asked Questions

Why are NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines important for students?

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Are NCERT Solutions for Class 11 Maths Chapter 9 by ALLEN useful for competitive exams?

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NCERT Solutions Class 11 Maths Chapter 9 – Straight Lines

1.0Class 11 Maths Chapter 9: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 9

EXERCISE - 9.1

EXERCISE - 9.2

EXERCISE - 9.3

MISCELLANEOUS EXERCISE

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

4.0Class 11 Maths Chapter 9 Straight Lines: Exercise-Wise Questions and Topics

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 9 (Straight Lines)

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