NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry - Exercise 7.1

Solution:

We'll use the distance formula: PQ = [(x-x) + (y-y)]

(i) For points (2, 3) and (4, 1):
Distance = [(4-2) + (1-3)]
= [2 + (-2)]
= [4 + 4]
= 8
= 22 units

(ii) For points (-5, 7) and (-1, 3):
Distance = [(-1-(-5)) + (3-7)]
= [(4) + (-4)]
= [16 + 16]
= 32
= 42 units

(iii) For points (a, b) and (-a, -b):
Distance = [(-a-a) + (-b-b)]
= [(-2a) + (-2b)]
= [4a + 4b]
= (4(a+b))
= 2(a+b) units

Solution:

For points (0, 0) and (36, 15):
Distance = [(36-0) + (15-0)]
= [36 + 15]
= [1296 + 225]
= 1521
= 39 units

This distance of 39 units represents the distance between towns A and B as discussed in Section 7.2.

Solution:

Three points are collinear if they lie on the same straight line. We can check this using the distance formula.

Let A(1, 5), B(2, 3) and C(-2, -11) be the points.

AB = [(2-1) + (3-5)] = [1 + 4] = 5 units

BC = [(-2-2) + (-11-3)] = [16 + 196] = 212 14.56 units

CA = [(-2-1) + (-11-5)] = [9 + 256] = 265 16.28 units

Checking: AB + BC 2.24 + 14.56 16.80 units CA

Since the sum of distances AB + BC is not equal to CA, the points are not collinear.

Solution:

An isosceles triangle has at least two equal sides. So we need to find the distances between all points and check if any two distances are equal.

Let A(5, -2), B(6, 4) and C(7, -2) be the vertices of the triangle.

AB = [(6-5) + (4-(-2))] = [1 + 36] = 37 units

BC = [(7-6) + (-2-4)] = [1 + 36] = 37 units

CA = [(7-5) + (-2-(-2))] = [4 + 0] = 4 = 2 units

Since AB = BC = 37 units, the triangle has two equal sides.

Therefore, (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Solution:

Assuming the positions based on the figure:
Let the points be A(3, 4), B(6, 7), C(9, 4) and D(6, 1)

Finding all four sides:

AB = [(6-3) + (7-4)] = [9 + 9] = 18 = 32 units

BC = [(9-6) + (4-7)] = [9 + 9] = 18 = 32 units

CD = [(6-9) + (1-4)] = [9 + 9] = 18 = 32 units

DA = [(3-6) + (4-1)] = [9 + 9] = 18 = 32 units

Finding the diagonals:

AC = [(9-3) + (4-4)] = [36 + 0] = 6 units

BD = [(6-6) + (1-7)] = [0 + 36] = 6 units

All sides are equal (AB = BC = CD = DA = 32 units) and diagonals are equal (AC = BD = 6 units).

These are the properties of a square, which has all sides equal and diagonals equal.

Therefore, ABCD is a square, and Champa is correct.

Solution:

(i) For points A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0):
AB = [(1-(-1)) + (0-(-2))] = [4 + 4] = 8 units
BC = [(-1-1) + (2-0)] = [4 + 4] = 8 units
CD = [(-3-(-1)) + (0-2)] = [4 + 4] = 8 units
DA = [(-1-(-3)) + (-2-0)] = [4 + 4] = 8 units

AC = [(-1-(-1)) + (2-(-2))] = [0 + 16] = 4 units
BD = [(-3-1) + (0-0)] = [16 + 0] = 4 units

All sides are equal (AB = BC = CD = DA = 8 units) and diagonals are equal (AC = BD = 4 units).

Therefore, ABCD is a square.

(ii) For points A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4):
First, let's check if these points form a quadrilateral by finding the area of triangle ABC:

Area of triangle ABC = |(-3)(1-3) + 3(3-5) + 0(5-1)|
= |-3(-2) + 3(-2) + 0|
= |6 - 6 + 0|
= |0| = 0

Since the area of triangle ABC is 0, points A, B, and C are collinear.

Therefore, these points do not form a quadrilateral as three of them lie on the same straight line.

(iii) For points A(4, 5), B(7, 6), C(4, 3) and D(1, 2):
AB = [(7-4) + (6-5)] = [9 + 1] = 10 units
BC = [(4-7) + (3-6)] = [9 + 9] = 18 units
CD = [(1-4) + (2-3)] = [9 + 1] = 10 units
DA = [(4-1) + (5-2)] = [9 + 9] = 18 units

AC = [(4-4) + (3-5)] = [0 + 4] = 2 units
BD = [(1-7) + (2-6)] = [36 + 16] = 52 units

Opposite sides are equal (AB = CD = 10 units and BC = DA = 18 units) but diagonals are not equal.

This forms a parallelogram, not a rectangle or square because the diagonals are not equal.

Solution:

Any point on the x-axis has its y-coordinate as 0. So let P(x, 0) be the point on the x-axis that is equidistant from A(2, -5) and B(-2, 9).

Since P is equidistant from A and B:
PA = PB
[(x-2) + (0-(-5))] = [(x-(-2)) + (0-9)]
[(x-2) + 25] = [(x+2) + 81]

Squaring both sides:
(x-2) + 25 = (x+2) + 81
x - 4x + 4 + 25 = x + 4x + 4 + 81
-4x + 29 = 4x + 85
-8x = 56
x = -7

Therefore, the point on the x-axis is (-7, 0), which is equidistant from (2, -5) and (-2, 9).

Solution:

Given: PQ = 10 units

Using the distance formula:
[(10-2) + (y-(-3))] = 10
[8 + (y+3)] = 10
[64 + (y+3)] = 10

Squaring both sides:
64 + (y+3) = 100
(y+3) = 36
(y+3) = 6

Taking square root on both sides:
y+3 = 6

Therefore:
y = -3 + 6 = 3, or y = -3 - 6 = -9

The required values of y are 3 or -9.

Solution:

Since Q is equidistant from P and R:
QP = QR

First, let's find QP:
QP = [(5-0) + (-3-1)] = [25 + 16] = 41

Setting QP = QR:
[(x-0) + (6-1)] = 41
[x + 25] = 41

Squaring both sides:
x + 25 = 41
x = 16
x = 4

Therefore, x can be either 4 or -4.

Finding QR:
QR = [(4-0) + (6-1)] = [16 + 25] = 41 units

Finding PR:
When x = 4: PR = [(5-4) + (-3-6)] = [1 + 81] = 82 units
When x = -4: PR = [(5-(-4)) + (-3-6)] = [81 + 81] = 162 = 92 units

Solution:

Let P(x, y) be a point that is equidistant from A(3, 6) and B(-3, 4).

Since P is equidistant from A and B:
PA = PB
[(x-3) + (y-6)] = [(x-(-3)) + (y-4)]
[(x-3) + (y-6)] = [(x+3) + (y-4)]

Squaring both sides:
(x-3) + (y-6) = (x+3) + (y-4)

Expanding both sides:
x - 6x + 9 + y - 12y + 36 = x + 6x + 9 + y - 8y + 16

Cancelling out common terms:
-6x + y - 12y + 45 = 6x + y - 8y + 25

Simplifying:
-6x - 12y + 45 = 6x - 8y + 25
-12x - 4y + 20 = 0
Dividing by -4:
3x + y - 5 = 0

Therefore, the required relation is 3x + y - 5 = 0, or equivalently, y = 5 - 3x.

This relation represents a straight line, and any point (x, y) that satisfies this equation will be equidistant from (3, 6) and (-3, 4).