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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

Question 1.

What is the slope of a line whose inclination with positive direction of x -axis is

(i) 90°

(ii) 0°

Solution:

(i) θ = 90°

m = tan θ = tan 90° = ∝ (undefined)

(ii) m = tan θ = tan 0° = 0

Question 2.

What is the inclination of a line whose slope is (i) 0

Solution:

(i) Slope = 0

tan θ = 0

tan 0 = 0

∴ θ = 0°

(ii) Slope = 1

tan θ = 1

tan 45° = 1

∴ θ = 45°

angle of inclination is 45°

Question 3.

Find the slope of a line joining the points

(i) (5, \(\sqrt{5})\)) with origin

(ii) (sin θ, -cos θ) and (-sin θ, cos θ)

(i) (5, \(\sqrt{5})\)) with origin (0, 0)

Solution:

Question 4.

What is the slope of a line perpendicular to the line joining A(5, 1) and P where P is the mid-point of the segment joining (4, 2) and (-6,4).

Solution:

P is the mid point of the segment joining (4, 2) and (-6, 4)

Question 5.

Show that the given points are collinear (-3, -4), (7, 2) and (12, 5)

Solution:

The verticles are A(-3, -4), B(7, 2) and C (12, 5)

Slope of AB = Slope of BC

∴ The points A, B and C lie on the same line.

∴ They are collinear.

Question 6.

If the three points (3, -1), (a, 3), (1, -3) are collinear, find the value of a.

Solution:

Slope of AB = slope of BC.

Question 7.

The line through the points (-2, a) and (9, 3) has slope –\(\frac{1}{2}\). Find the value of a.

Solution:

A line joining the points (-2, a) and (9, 3) has slope m = –\(\frac{1}{2}\).

Question 8.

The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24) . Find the value of x.

Solution:

Question 9.

Show that the given points form a right angled triangle and check whether they satisfies pythagoras theorem

(i) A(1, -4), B(2, -3) and C(4, -7)

(ii) L(0, 5), M(9, 12) and N(3, 14)

Solution:

Question 10.

Show that the given points form a parallelogram : A(2.5, 3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5)

Solution:

∴ The given points form a parallelogram.

Question 11.

If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.

Solution:

A(2, 2), B(-2, -3), C(1, -3), D(x, y)

Since ABCD forms a parallelogram, slope of opposite sides are equal and diagonals bisect each other.

Mid point of BD = Mid point of AC

Question 12.

Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7). Show that ABCD is a trapezium.

Solution:

A (3, -4), B (9, -4), C (5, -7) and D (7, -7)

If only one pair of opposite sides of a quadrilateral are parallel, then it is said to be a trapezium.

∴ One pair of opposite sides are parallel.

∴ ABCD is a trapezium.

Question 13.

A quadrilateral has vertices at A(- 4, -2), B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram

Solution:

In a parallelogram diagonals bisect each other. Opposite sides are parallel as their slopes are equal the mid points of the diagonals are the same.

∴ Mid points of the sides of a quadrilateral form a parallelogram.

Question 14.

PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS =2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.

Solution: