**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 A and B discussed in Section 7.2?

Solution:

The distance between the points A(0, 0) and B(36, 15):

∴ AB = 39

∴ The distance between the two towns A and B discussed in section 7.2 is

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:

Here, AB = BC.

∴ If Two sides of the triangle ABC are equal to each other then it 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 1 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) A (1, -2), B (1, 0), C (-1, 2), D (-3, 0)

In the quadrilateral ABCD,

diagonal AC = diagonal BD = 4 cm., and

AB = BC = CD = DA = \(\sqrt{8}\) units.

i. e., Here all four sides are equal to each other then ABCD is a square.

ii) P (-3, 5), Q (3, 1), R (0, 3), S (-1, -4)

As per Distance formula,

Here all four sides of quadrilateral PQRS are not equal to each other. Hence PQRS is a quadrilateral.

iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)

As per the Distance formula,

In the quadrilateral ABCD

AB = DC = \(\sqrt{10}\)

BC = AD = \(\sqrt{18}\)

Diagonal AC ≠ Diagonal BD

Adjacent sides are equal to each other and diagonals are not equal to each other. This 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:

If the distance between P and Q, PQ = 10 units, then y =?

By Squaring on both sides,

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

y^{2} + 6y + 73 – 100 = 0

y^{2} + 6y – 27 = 0

y^{2} + 9y – 3y – 27 = 0

y (y + 9) – 3(y + 9) = 0

(y + 9) (y – 3) = 0

If y + 9 = 0 then OR y – 3 = 0 then

y = -9 y = 0

∴ y = -9 OR +3.

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:

A (3, 6) = A (x_{1}, y_{1}) AP = PB.

P(x, y)

Let B(-3, 4) = B(x_{2}, y_{2}).

Here, AP = PB

∴ AP^{2} = PB^{2}

x^{2} – 6x + y^{2} – 12y + 45 = x^{2} + y^{2} + 6x – 8y + 25

-6x – 6x – 12y + 8y = 25 – 45

– 12x – 4y = -20

12x + 4y = 20

∴ 3x + y = 5

∴ 3x + y – 5 = 0

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