# MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Pdf, These solutions are solved subject experts from the latest edition books.

## MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Coordinate Geometry Class 10 Exercise 7.1 Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:
(i) Here x1 = 2, y1 = 3 and x2 = 4, y2 = 1
∴ The required distance

(ii) Here x1 = -5, y1 = 7 and x2 = -1, y2 = 3
∴ The required distance

(iii) Here x1 = a, y1 = b and x2 = -a, y2 = -b
∴ The required distance

Ncert Class 10 Maths 7.1 Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns /I and B discussed below as following: ‘A town 6 is located 36 km east and 15 km north of the town A’.
Solution:
Part-I
Let the points be A(0, 0) and B(36, 15)

Part-II
We have A(0, 0) and B(36,15) as the positions of two towns.

Class 10 Maths Chapter 7 Exercise 7.1 Solutions Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let the points be A( 1, 5), B(2, 3) and C(-2, -11). A, B and C are collinear, if
AB + BC = AC or AC + CB = AB or BA + AC = BC

Here, AB + BC ≠ AC, AC + CB ≠ AB, BA + AC ≠ BC
∴ A, B and C are not collinear.

Chapter 7 Maths Class 10 Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let the points be A(5, -2), 6(6, 4) and C(7, -2).

We have AB = BC ≠ AC.
∴ ∆ABC is an isosceles triangle.

Class 10th Maths Exercise 7.1 Question 5.
In a classroom, 4 friends are seated at the points A B, Cand D as shown in the figure. 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 distance formula, find which of them is correct.

Solution:
Let the horizontal columns represent the x-coordinates whereas the vertical rows represent the y-coordinates.
∴ The points are A(3, 4), B(6, 7), C(9, 4) and D(6, 1)

i.e., BD = AC ⇒ Both the diagonals are also equal.
∴ ABCD is a square.
Thus, Champa is correct.

Class 10th Maths Chapter 7.1 Question 6.
Name the type of 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)
Solution:
(i) Let the points be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) of a quadrilateral ABCD.

⇒ AB = BC = CD = AD i.e., all the sides are equal.
Also, AC and BD (the diagonals) are equal.
∴ ABCD is a square.

(ii) Let the points be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).

We see that $$\sqrt{13}+\sqrt{13}=2 \sqrt{13}$$
i. e., AC + BC = AB
⇒ A, B and C are collinear. Thus, ABCD is not a quadrilateral.

(iii) Let the points be A(4, 5), B(7, 6), C(4, 3) and D(1, 2).

Since, AB = CD, BC = DA i.e., opposite sides of the quadrilateral are equal.
And AC ≠ BD ⇒ Diagonals are unequal.
∴ ABCD is a parallelogram.

Exercise 7.1 Class 10 Question 7 Question 7.
Find the point on x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
We know that any point on 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)

∴ The required point is (-7, 0)

Maths Class 10 Chapter 7 Question 8.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution:
The given points are P(2, -3) and Q(10, y).

Squaring both sides, y2 + 6y + 73 = 100
⇒ y2 + 6y – 27 = 0
⇒ y2 – 3y + 9y – 27 = 0
⇒ (y – 3)(y + 9) = 0
Either y – 3 = 0 ⇒ y = 3
or y + 9 = 0 ⇒ y = -9
∴ The required value of y is 3 or -9.

Class 10 Ka Math Chapter 7.1 Question 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.
Solution:

Squaring both sides, we have x2 + 25 = 41
⇒ x2 + 25 – 41 = 0
⇒ x2 – 16 = 0 2 x = $$\pm \sqrt{16}$$ = ±4
Thus, the point R is (4, 6) or (-4, 6)
Now,

Class 10th Coordinate Geometry Exercise 7.1 Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Solution:
Let the points be A(x, y), B(3, 6) and C(-3, 4).
∴ AB = $$\sqrt{(3-x)^{2}+(6-y)^{2}}$$
And AC = $$\sqrt{[(-3)-x]^{2}+(4-y)^{2}}$$
Since, the point (x, y) is equidistant from (3, 6) and (-3, 4).
∴ AB = AC
⇒ $$\sqrt{(3-x)^{2}+(6-y)^{2}}=\sqrt{(-3-x)^{2}+(4-y)^{2}}$$
Squaring both sides,
(3 – x)2 + (6 – y)2 = (-3 – x)2 + (4 – y)2
⇒ (9 + x2 – 6x) + (36 + y2 – 12y)
⇒ (9 + x2 + 6x) + (16 + y2 – 8y)
⇒ 9 + x2 – 6x + 36 + y2 – 12y – 9 – x2 – 6x – 16 – y2 + 8y = 0
⇒ – 6x – 6x + 36 – 12y – 16 + 8y = 0
⇒ – 12x – 4y + 20 = 0
⇒ -3x – y + 5 = 0
⇒ 3x + y – 5 = 0 which is the required relation between x and y.