NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry: Exercise 7.1
NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 is a foundational exercise of coordinate geometry. The chapter is a perfect mix of solving geometrical problems with the use of algebra. The exercise provided here is explained with a detailed breakdown of some important concepts connected with Coordinate Geometry, especially the distance formula of two coordinate points. This exercise is prepared according to the CBSE syllabus and for examination purposes. So, let’s dive into NCERT Solutions Class 10 maths chapter 7 exercise 7.1 PDF online.
1.0Download NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 : Free PDF
2.0Introduction to Coordinate Geometry
Coordinate Geometry (also referred to as Analytical Geometry) is a mathematical branch that applies algebraic techniques to the study of geometric problems. It makes use of a coordinate system, often the Cartesian coordinate system, in which every point in a plane is assigned an ordered pair of numbers (x, y). These are the coordinates of the point and define its location in relation to two perpendicular axes, typically the x-axis, also known as abscissa (horizontal) and the y-axis or ordinate (vertical). In coordinate geometry, the cartesian plane is divided into four quadrants namely (I, II, III, IV) with specific sign convention.
3.0Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 Overview: Key Concepts
Distance Formula
The distance formula is used to calculate the distance between two points on the coordinate plane. It is derived from the Pythagorean Theorem and holds true for points in any quadrant. The distance that we are talking about is basically the difference between any two coordinates. The formula is applicable to all points in the coordinate plane regardless of their quadrants. The formula to find the distance between any two coordinates, say, A(x1,y1) and B (x2,y2) can be expressed as:
Distance from the Axes
The exercise involves finding the distance between two coordinates, if the points are given on the coordinate axes.
- Points on the x-axis: Points on the x-axis are those in which the y-coordinate is zero. These points are on the horizontal axis of the coordinate plane. The form of such points is (x,0), where x can be any real number. For instance, (3,0) and (−2,0) are points on the x-axis.
- Points on the y-axis: Points on the y-axis are those for which the x-coordinate is zero. These points are on the vertical axis of the coordinate plane. The general form of such points is (0,y), where y can be any real number. For instance, (0,4) and (0,−5) are points on the y-axis.
Distance from the Origin
The distance from the origin is defined as the length of the straight line between a point and the origin (0,0) of the coordinate plane. In order to determine this distance, the distance formula is used. For a point (x,y), the distance from the origin is:
Collinearity of Points
The collinearity of the points is used to check if the points lie on the same straight line or are collinear. The collinearity of two or more than two coordinates can be checked by comparing the distances between these points. These distances can be calculated using the distance formula. If the sum of the two points' distances is equal to the distance between the other two points, then the points are collinear.
For example, consider three coordinates, A, B, and C; these points are collinear if they satisfy this condition:
4.0NCERT Class 10 Maths Chapter 7 Exercise 7.1 : Detailed Solutions
1. Find the distance between the following pairs of points:
(a) (2, 3), (4, 1)
(b) (-5, 7), (-1, 3)
(c) (a, b), (-a, -b)
Sol.
(a) The given points are: A(2, 3), B(4, 1). Required distance:
AB = BA = √((x₂ - x₁)² + (y₂ - y₁)²)
AB = √((4 - 2)² + (1 - 3)²) = √(2² + (-2)²)
= √(4 + 4) = √8 = 2√2 units
(b) Here x₁ = -5, y₁ = 7 and x₂ = -1, y₂ = 3
Therefore, the required distance:
= √((x₂ - x₁)² + (y₂ - y₁)²)
= √((-1 - (-5))² + (3 - 7)²)
= √((-1 + 5)² + (-4)²)
= √(16 + 16) = √32 = √(2 × 16)
= 4√2 units
(c) Here x₁ = a, y₁ = b and x₂ = -a, y₂ = -b
Therefore, the required distance:
= √((x₂ - x₁)² + (y₂ - y₁)²)
= √((-a - a)² + (-b - b)²)
= √((-2a)² + (-2b)²) = √(4a² + 4b²)
= √(4(a² + b²)) = 2√(a² + b²) units
2. Find the distance between the points (0, 0) and (36, 15).
Sol. Let the points be A(0, 0) and B(36, 15)
Therefore, AB = √((36 - 0)² + (15 - 0)²)
= √(36² + 15²) = √(1296 + 225)
= √1521 = √39² = 39
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Sol. The given points are:
A(1, 5), B(2, 3) and C(-2, -11).
Let us calculate the distance: AB, BC and CA by using the distance formula.
AB = √((2 - 1)² + (3 - 5)²) = √(1² + (-2)²)
= √(1 + 4) = √5 units
BC = √((-2 - 2)² + (-11 - 3)²)
= √((-4)² + (-14)²) = √(16 + 196) = √212
= 2√53 units
CA = √((-2 - 1)² + (-11 - 5)²)
= √((-3)² + (-16)²) = √(9 + 256) = √265
= √(5 × 53) units
From the above we see that: AB + BC ≠ CA
Hence the above stated points A(1, 5), B(2, 3) and C(-2, -11) are not collinear.
4. Check whether (5,-2), (6,4) and (7,-2) are the vertices of an isosceles triangle.
Sol. Let the points be A(5,-2), B(6,4) and C(7,-2).
Therefore, AB = √((6-5)² + (4-(-2))²)
= √(1² + 6²) = √(1 + 36) = √37
BC = √((7-6)² + (-2-4)²)
= √(1² + (-6)²) = √(1 + 36) = √37
AC = √((7-5)² + (-2-(-2))²)
= √(2² + 0²) = √(4 + 0) = 2
We have AB = BC ≠ AC.
Therefore, triangle ABC is an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using the distance formula, find which of them is correct.
Solution:
Let the number of horizontal columns represent the x-coordinates, and the vertical rows represent the y-coordinates. Therefore, the points are: A(3, 4), B(6, 7), C(9, 4), and D(6, 1).
Calculate the distances between the points:
AB = √((6 - 3)² + (7 - 4)²) = √(3² + 3²) = √(9 + 9) = √18 = 3√2
BC = √((9 - 6)² + (4 - 7)²) = √(3² + (-3)²) = √(9 + 9) = √18 = 3√2
CD = √((6 - 9)² + (1 - 4)²) = √((-3)² + (-3)²) = √(9 + 9) = √18 = 3√2
AD = √((6 - 3)² + (1 - 4)²) = √(3² + (-3)²) = √(9 + 9) = √18 = 3√2
Since AB = BC = CD = AD, all four sides are equal.
Calculate the lengths of the diagonals:
AC = √((9 - 3)² + (4 - 4)²) = √(6² + 0²) = 6
BD = √((6 - 6)² + (1 - 7)²) = √(0² + (-6)²) = 6
Since BD = AC, both diagonals are also equal.
Therefore, ABCD is a square. Thus, Champa is correct, as ABCD is a square.
6. Name the quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solutions:
(i) Points: A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)
Determine distances: AB, BC, CD, DA, AC, and BD.
AB = √((1 + 1)² + (0 + 2)²) = √(4 + 4) = √8 = 2√2
BC = √((-1 - 1)² + (2 - 0)²) = √(4 + 4) = √8 = 2√2
CD = √((-3 + 1)² + (0 - 2)²) = √(4 + 4) = √8 = 2√2
DA = √((-1 + 3)² + (-2 - 0)²) = √(4 + 4) = √8 = 2√2
Since AB = BC = CD = DA, the sides of the quadrilateral are equal. ...(1)
AC = √((-1 + 1)² + (2 + 2)²) = √(0 + 16) = 4
BD = √((-3 - 1)² + (0 - 0)²) = √(16 + 0) = 4
Diagonal AC = Diagonal BD. ...(2)
From (1) and (2), we conclude that ABCD is a square.
(ii) Points: A(-3, 5), B(3, 1), C(0, 3), D(-1, -4)
AB = √((3 - (-3))² + (1 - 5)²) = √(6² + (-4)²) = √(36 + 16) = √52 = 2√13
BC = √((0 - 3)² + (3 - 1)²) = √(9 + 4) = √13
CD = √((-1 - 0)² + (-4 - 3)²) = √((-1)² + (-7)²) = √(1 + 49) = √50
DA = √((-3 - (-1))² + (5 - (-4))²) = √((-2)² + 9²) = √(4 + 81) = √85
AC = √((0 - (-3))² + (3 - 5)²) = √(3² + (-2)²) = √(9 + 4) = √13
BD = √((-1 - 3)² + (-4 - 1)²) = √((-4)² + (-5)²) = √(16 + 25) = √41
We see that √13 + √13 = 2√13, i.e., AC + BC = AB.
Therefore, A, B, and C are collinear. Thus, ABCD is not a quadrilateral.
(iii) Points: A(4, 5), B(7, 6), C(4, 3), D(1, 2)
AB = √((7 - 4)² + (6 - 5)²) = √(3² + 1²) = √10
BC = √((4 - 7)² + (3 - 6)²) = √((-3)² + (-3)²) = √18
CD = √((1 - 4)² + (2 - 3)²) = √((-3)² + (-1)²) = √10
DA = √((1 - 4)² + (2 - 5)²) = √(9 + 9) = √18
AC = √((4 - 4)² + (3 - 5)²) = √(0 + (-2)²) = 2
BD = √((1 - 7)² + (2 - 6)²) = √(36 + 16) = √52
Since AB = CD and BC = DA, the opposite sides of the quadrilateral are equal.
Also, AC ≠ BD, meaning the diagonals are unequal.
Therefore, ABCD is a parallelogram.
7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Sol. We know that any point on the x-axis has its ordinate = 0.
Let the required point be P(x, 0).
Let the given points be A(2, -5) and B(-2, 9).
Therefore, AP = √((x - 2)² + 5²) = √(x² - 4x + 4 + 25) = √(x² - 4x + 29)
BP = √((x - (-2))² + (-9)²) = √((x + 2)² + (-9)²) = √(x² + 4x + 4 + 81) = √(x² + 4x + 85)
Since A and B are equidistant from P,
Therefore, AP = BP
=> √(x² - 4x + 29) = √(x² + 4x + 85)
=> x² - 4x + 29 = x² + 4x + 85
=> x² - 4x - x² - 4x = 85 - 29
=> -8x = 56
=> x = 56 / -8 = -7
Therefore, the required point is (-7, 0).
8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Sol. Distance between P(2, -3) and Q(10, y) = 10 units
=> √((10 - 2)² + (y + 3)²) = 10
=> 64 + (y + 3)² = 100
=> (y + 3)² = 36
=> y² + 6y + 9 = 36
=> y² + 6y - 27 = 0
=> y² + 9y - 3y - 27 = 0
=> y(y + 9) - 3(y + 9) = 0
=> (y + 9)(y - 3) = 0
=> y + 9 = 0 or y - 3 = 0
=> y = -9 or 3
Hence, there can be two values of y which are -9 and 3.
9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Sol. Here,
QP = √[(5 - 0)² + ((-3) - 1)²] = √(5² + (-4)²) = √(25 + 16) = √41
QR = √[(x - 0)² + (6 - 1)²] = √(x² + 5²) = √(x² + 25)
Given that QP = QR, therefore:
√41 = √(x² + 25)
Squaring both sides, we have:
x² + 25 = 41
x² + 25 - 41 = 0
x² - 16 = 0
x² = 16
x = ±√16 = ±4
Thus, the point R is (4, 6) or (-4, 6).
Now,
QR = √[(±4 - 0)² + (6 - 1)²] = √(16 + 25) = √41
And PR = √[(±4 - 5)² + (6 + 3)²]
PR = √[(-4 - 5)² + (6 + 3)²] or √[(4 - 5)² + (6 + 3)²]
PR = √[(-9)² + 9²] or √(1 + 81)
PR = √(2 × 9²) or √82
PR = 9√2 or √82
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Sol. Let A(3, 6) and B(-3, 4) be the given points. Point P(x, y) is equidistant from the points A and B.
Therefore, PA = PB
√[(x - 3)² + (y - 6)²] = √[(x + 3)² + (y - 4)²]
Squaring both sides:
(x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²
(x² - 6x + 9) + (y² - 12y + 36) = (x² + 6x + 9) + (y² - 8y + 16)
-6x - 12y + 45 = 6x - 8y + 25
12x + 4y - 20 = 0
3x + y - 5 = 0
5.0Benefits of Studying Class 10 Maths NCERT Solutions Chapter 7 Exercise 7.1
- Strengthened Foundational Concepts
- Improved Problem-Solving Skills
- Exam Preparation and Confidence
- Enhanced Understanding of Key Topics
- Easy Access and Free Download
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