Point and Straight Lines

Points and straight lines are fundamental entities in mathematics, serving as cornerstones of analytical geometry. They facilitate geometric constructions and algebraic reasoning, bridging the gap between abstract concepts and tangible representations. Comprehending their characteristics and how they interact is essential for solving problems across various mathematical domains, making them indispensable tools for both novice learners and seasoned mathematicians alike.

1.0Distance Formula Between Two Points

Distance Between the Points P (x₁, y₁) and Q (x₂, y₂) is-

P Q=$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

2.0Section Formula of a Straight Line

The coordinates of a point that divides the line segment connecting the points (x₁, y₁) and (x₂, y₂) within itself, in the ratio m:n, can be determined.

$\left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\right)$

Especially when m = n, the coordinates of the midpoint of the line segment that connects the points (x₁, y₁) and (x₂, y₂) are determined.

$\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2}\right)$

3.0Area of the Triangle

The formula to calculate the Area A of the triangle with vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is- 

Area of the triangle: - $\frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|$

4.0Straight Line

A straight line is a foundational geometric concept in mathematics, representing the shortest path between two points in a two-dimensional or three-dimensional space. Visually, a straight line extends infinitely in both directions, maintaining a consistent trajectory without any curvature or deviation. Algebraically, a straight line can be described by various equations, such as slope-intercept form, point-slope form, or general form. Straight lines play an integral role in geometry, trigonometry, calculus, and physics, serving as the basis for understanding motion, defining geometric shapes, and solving mathematical problems. 

5.0Standard Form of Equations of a Straight Line

Slope Intercept Form

The slope-intercept form of a straight line is a common way to represent a line algebraically. It is given by the equation:

y = mx + b 

Where:

• m is the slope of the line, representing the rate of change of y with respect to x.
• b is the y-intercept of the line, indicating the point where the line intersects the y-axis.

In this form:

• The slope m determines the steepness and direction of the line. If m > 0, the line slopes upward to the right; if m < 0, the line slopes downward to the right; and if m = 0, the line is horizontal.
• The y-intercept b denotes the value of y at x = 0, representing the point where the line crosses the y-axis.

Point Slope Form

The point-slope form of the equation of a straight line is another common way to represent a line algebraically. It is given by the equation:

y – y₁ = m (x – x₁)

Where:

• m is the slope of the line.
• (x₁, y₁) is a point on the line.

In this form:

• This form directly incorporates a known point on the line and the slope, making it convenient for quickly writing the equation of a line given these two pieces of information.

Two Point Form

The two-point form of the equation of a straight line provides a way to find the equation when you know two distinct points on the line. It is given by the equation:

$\mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right)$

Where:

• The points (x₁, y₁) and (x₂, y₂) represent two separate points along the line.

In this form:

• The equation describes the ratio of the changes in y to the changes in x between the two points, which remains constant along the entire line.

Intercept Form

The intercept form of the equation of a straight line is a useful representation that provides insights into the intercepts made by the line on the coordinate axes. It is given by the equation:

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

Where:

• a is the x-intercept of the line, which is the point where the line intersects the x-axis.
• b is the y-intercept of the line, which is the point where the line intersects the y-axis.

In this form:

• The equation expresses the line as the sum of two fractions, where each fraction represents the distance from a point on the line to the corresponding intercept.
• When x = 0, the first term becomes 0, yielding the value of the y-intercept b. Similarly, when y = 0, the second term becomes 0, resulting in the value of the x-intercept a.

6.0Distance Between a Point and a Straight line

The distance between a point (x₀​, y₀​) and a straight-line Ax + By + C = 0 is given by the formula:

$\mathrm{d}=\frac{|{\mathrm{Ax}}_0+{\mathrm{By}}_0+\mathrm{C}|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}$

In this context, A, B, and C refer to the coefficients of the line's equation and (x₀​, y₀​) is the coordinates of the point.

7.0Distance Between Two Parallel Lines

1. The distance between two parallel lines ax + by + c₁ = and ax + by + c₂ = 0 is $\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$ (Note: The coefficients of x & y in both equations should be same)
2. The area of the parallelogram= $\frac{{\mathrm{p}}_1{\mathrm{p}}_2}{\sin \theta }$, where p₁ & p₂ are distances between two pairs of opposite sides & θ is the angle between any two adjacent sides. Note that area of the parallelogram bounded by the lines y= m₁x + c₁,y =m₁x+c₂ and y = m₂x+ d₁,y = m₂x+ d₂ is given by $\left|\frac{\left({\mathrm{c}}_1-{\mathrm{c}}_2\right)\left({\mathrm{d}}_1-{\mathrm{d}}_2\right)}{{\mathrm{m}}_1-{\mathrm{m}}_2}\right|$.

8.0Foot of Perpendicular

This gives the point where the perpendicular from a point cuts the line.

Let Point P(x₁, y₁), Line L: ax + by + c = 0 then, if foot of Perpendicular N is (x, y), then

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\frac{\left(a{x}_1+b{y}_1+c\right)}{a^2+b^2}$

9.0Examples Based on Point and Straight Lines

Example 1: Find the distance between the two parallel lines 2x + 3y – 4 = 0 and 2x +3 y + 5 = 0.

Solution: a = 2, b = 3, c₁ = –4, c₂ = 5

d=$\frac{|c_2-c_1|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\Rightarrow d=\frac{|5-(-4)|}{\sqrt{2^2+3^2}}\Rightarrow d=\frac{|9|}{\sqrt{4+9}}$

d=$\frac{9}{\sqrt{13}}\Rightarrow d=\frac{9\sqrt{13}}{13}$

So, the distance between two parallel lines is $\frac{9\sqrt{13}}{13}$.

Example 2: Find the distance between the point P (3, 4) and the straight line 2x –3y + 16 = 0.

Solution: A = 2, B= – 3, C = 16, Given point (3, 4)

$\mathrm{d}=\frac{|{\mathrm{Ax}}_0+{\mathrm{By}}_0+\mathrm{c}|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\mathrm{d}=\frac{|2(3)-3(4)+16|}{\sqrt{2^2+(-3{)}^2}}$

$d=\frac{|6-12+16|}{\sqrt{4+9}}\Rightarrow d=\frac{|10|}{\sqrt{13}}\Rightarrow d=\frac{10\sqrt{13}}{13}$

The distance between the point P (3,4) and the straight line 2x–3y + 16 = 0 is $\frac{10\sqrt{13}}{13}$ units.

Example 3: Find the distance between the point P (3, –2) and Q (–1, 4).

Solution: Using the distance formula

d=$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$\Rightarrow \mathrm{d}=\sqrt{(-1-3{)}^2+(4-(-2){)}^2}\Rightarrow \mathrm{d}=\sqrt{(-4{)}^2+(4+2{)}^2}$

$d=\sqrt{16+36}\Rightarrow d=\sqrt{52}\Rightarrow d=2\sqrt{13}$

Example 4: The coordinates of two points A and B are (–1,3) and (5, –2) respectively. If the point P divides the line segment joining A and B internally in the ratio 2: 3, then find the coordinates of point P.

Solution: Using the section formula, the coordinates of point P can be found:

$\mathrm{P}=\left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\right)$

$\mathrm{P}=\left(\frac{2(5)+3(-1)}{5},\frac{2(-2)+3(3)}{5}\right)$

$\mathrm{P}=\left(\frac{10-3}{5},\frac{-4+9}{5}\right)=\left(\frac{7}{5},\frac{5}{5}\right)=\left(\frac{7}{5},1\right)$

Example 5: The coordinates of the midpoint of the line segment joining the point (x₁, y₁) and (x₂, y₂) is (3,4). If one of the points is (1, –3), find the other point.

Solution: Using section formula

$\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2}\right)=(\mathrm{x},\mathrm{y})$

$\Rightarrow \left(\frac{1+{\mathrm{x}}_2}{2},\frac{-3+{\mathrm{y}}_2}{2}\right)$=(3,4)

$\Rightarrow \frac{1+{\mathrm{x}}_2}{2}=3,\frac{-3+{\mathrm{y}}_2}{2}$=4

⇒ 1 +x₂ = 6, –3 + y₂ = 8

⇒ x₂ = 5, y₂ = 11

So, the other point is (5,11)

10.0Sample Question on Point and Straight Lines

Q. How do you find the distance between a point and a straight line?

Ans. The distance d between a point (x₀, y₀) and a straight-line Ax + By + C = 0 is given by the formula:

$\mathrm{d}=\frac{|{\mathrm{Ax}}_0+{\mathrm{By}}_0+\mathrm{C}|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}$

This formula is derived from the concept of perpendicular distance.