**KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry 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) If A(x_{1}, y_{1}) = A(2, 3)

B(x_{2}, y_{2}) = B(4, 1), then

ii) Let P(x_{1}, y_{1}) = P(-5, 7)

Q(x_{2}, y_{2}) = Q(-1, 3)

iii) A (a, b), B (-a, -b)

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:

Whether the points A (1, 5), B (2, 3) and C (-2, -11) are collinear?

Here, AB + BC = AC, then it is straight line.

But \(\sqrt{5}+\sqrt{221} \neq \sqrt{261}\)

∴ These points are non-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, C, and D as shown in the following 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 the distance formula, find which of them is correct.

Solution:

A(3, 4), B (6, 7), C (9, 4), D (6, 1)

Four sides of quadrilateral ABCD are equal to each other.

∵ AB = BC = CD = DA = \(3 \sqrt{2}\)

Diagonal AC = Diagonal BD

∴ ABCD is a square.

∴ Champa is correct among the two.

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 the x-axis which is equidistant from (2, -5) and (-2, 9).

Solution:

we have PA = PB

∴ PA^{2} = PB^{2}

x^{2} – 4x + 29 = x^{2} + 4x + 85

-4x – 4x = 85 – 29

-8x = 56

8x = -56

∴ x = \(\frac{-56}{8}\)

∴ x = – 7

∴ Coordinates of P are (x, 0), it means (-7, 0).

∴ Required point (x, 0) = (-7, 0)

∴ x = -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, 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 value of x. Also, find the distances QR and PR.

Solution:

Point Q is equidistant from P and R, then PQ = QR, x = ?

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.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter Chapter 7 Coordinate Geometry Ex 7.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter Chapter 7 Coordinate Geometry Exercise 7.1, drop a comment below and we will get back to you at the earliest