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

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 x_{1} = 2, y_{1} = 3 and x_{2} = 4, y_{2} = 1

∴ The required distance

(ii) Here x_{1} = -5, y_{1} = 7 and x_{2} = -1, y_{2} = 3

∴ The required distance

(iii) Here x_{1} = a, y_{1} = b and x_{2} = -a, y_{2} = -b

∴ The required distance

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.

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.

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.

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.

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.

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)

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, y^{2} + 6y + 73 = 100

⇒ y^{2} + 6y – 27 = 0

⇒ y^{2} – 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.

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 x^{2} + 25 = 41

⇒ x^{2} + 25 – 41 = 0

⇒ x^{2} – 16 = 0 ^{2} x = \(\pm \sqrt{16}\) = ±4

Thus, the point R is (4, 6) or (-4, 6)

Now,

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 + x^{2} – 6x) + (36 + y^{2} – 12y)

⇒ (9 + x^{2} + 6x) + (16 + y^{2} – 8y)

⇒ 9 + x^{2} – 6x + 36 + y^{2} – 12y – 9 – x^{2} – 6x – 16 – y^{2} + 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.