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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

**9th Maths Coordinate Geometry Exercise 5.2 Question 1.**

Find the distance between the following pairs of points.

(i) (1, 2) and (4, 3)

(ii) (3, 4) and (- 7, 2)

(ii) (a, b) and (c, b)

(iv) (3, -9) and (-2, 3)

Solution:

We know that distance,

**9th Samacheer Maths Solution Ex 5.2 Question 2.**

Determine whether the given set of points in each case are collinear or not.

(i) (7, -2), (5, 1), (3, 4)

(ii) (a, -2), (a, 3), (a, 0)

Solution:

(i) Let the points be A (7, -2), B (5, 1) and C (3, 4). By the distance formula.

∴ Hence the points are collinear.

(ii) Let the points be A (a, -2), B (a, 3) and C (a, 0).

∴ Hence the points are collinear.

**Chapter 5 Coordinate Geometry Answers Question 3.**

Show that the following points taken in order form an isosceles triangle.

(i) A (5, 4), B(2, 0), C (-2, 3)

(ii) A (6, -4),B (-2, -4), C (2, 10)

Solution:

(i) Let the points be A (5, 4), B (2, 0) and C (-2, 3)

Here AB + BC > CA and AB = BC. ∴ ∆ ABC is an isosceles triangle.

(ii) Let the points be A (6, -4), B (-2, -4) and C (2, 10).

Here BC + BA > CA and BC = CA. Two sides are equal, so ∆ ABC is an isosceles triangle

**Coordinate Geometry Solutions Ex 5.2 Question 4.**

Show that the following points taken in order form an equilateral triangle in each case.

(i) A(2, 2), B(-2, -2), C(\(-2 \sqrt{3}\), \(2 \sqrt{3}\))

(ii) A(\(\sqrt{3}\) ,2), B (0, 1), C(0, 3)

Solution:

(i) Let the points be A (2, 2) B (-2, -2) and C(\(-2 \sqrt{3}\), \(2 \sqrt{3}\))

All the 3 sides of ∆ABC are equal, Hence ∆ABC is an equilateral triangle.

(ii) Let the points be A (\(\sqrt{3}\), 2), B (0, 1) and C (0, 3).

All the 3 sides of ∆ABC are equal. Hence ∆ABC is an equilateral triangle.

**Class 9 Maths Exercise 5.2 Solution Question 5.**

Show that the following points taken in order form the vertices of a parallelogram.

(i) A(-3, 1), B(-6, -7), C (3, -9) and D(6, -1)

(ii) A (-7, -3), B(5, 10), C(15, 8) and D(3, -5)

Solution:

(i) Let A, B, C and D represent the points (-3, 1), (-6, -7) (3, -9) and (6, -1) respectively.

The opposite sides are equal. Hence ABCD is a parallelogram.

(ii) Let A, B, C and D represent the points (-7, -3), (5, 10) (15, 8) and (3, -5)

The opposite sides are equal. Hence ABCD is a parallelogram.

**Exercise 5.2 Class 9 Question 6.**

Verify that the following points taken in order form the vertices of a rhombus.

(i) A(3, -2), B (7, 6),C (-1, 2) and D (-5, -6)

(ii) A (1, 1), B(2, 1),C (2, 2) and D(1, 2)

Solution:

(i) Let the points be A(3, -2), B (7, 6), C (-1, 2) and D (-5, -6)

∴ All the four sides of quadrilateral ABCD are equal. Hence ABCD is a rhombus.

**9th Maths 5.2 Question 7.**

If A(-1, 1), B(1, 3) and C(3, a) are points and if AB = BC, then find ‘a’

Solution:

**Maths Solutions For Class 9 Samacheer Kalvi Ex 5.2 Question 8.**

The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units, what are the coordinates of A?

Solution:

Co-ordinates of A are (-5, -5) or (9, 9)

**9th Maths Book Ex 5.2 Question 9.**

The point (x, y) is equidistant from the points (3, 4) and (-5, 6). Find a relation between x and y.

Solution:

P(x, y) is equidistant from the points A(3, 4) and B(-5, 6)

**Geometry 5.2 9th Maths Ex 5.2 Question 10.**

Let A(2, 3) and B(2, -4) be two points. If P lies on the x-axis, such that AP = \(\frac{3}{7}\)AB, find the coordinates of P.

Solution:

**9th Maths Guide Ex 5.2 Question 11.**

Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, -4) and (5, -6)

Solution:

Therefore S is the centre of the circle, passing through A, B and C.

**5 Coordinate Geometry Ex 5.2 Question 12.**

The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.

Solution: