# 10th Maths Book Back Coordinate Geometry Ex 5.2

## Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 5 – Coordinate Geometry Ex 5.2 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 4 Geometry consist of 5 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Relations and Functions Ex 5.2 Answers in English.

### 10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Chapter 5 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below:

### Exercise 5.2 Coordinate Geometry

1.What is the slope of a line whose inclination with the 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

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°

3.Find the slope of a line joining the points
(i) (5, 5–√)) with origin
(ii) (sin θ, -cos θ) and (-sin θ, cos θ)
(i) (5, 5–√)) with origin (0, 0)
Solution:

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 midpoint of the segment joining (4, 2) and (-6, 4)

5. Show that the given points are collinear (-3, -4), (7, 2), and (12, 5)
Solution:
The vertices are A(-3, -4), B(7, 2), and C (12, 5)

The slope of AB = Slope of BC
∴ The points A, B, and C lie on the same line.
∴ They are collinear.

6. If the three points (3, -1), (a, 3), (1, -3) are collinear, find the value of a.
Solution:
The slope of AB = slope of BC.

7. The line through the points (-2, a) and (9, 3) has a slope –12. Find the value of a.
Solution:
A line joining the points (-2, a) and (9, 3) has a slope m = –12.

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:

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:

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.

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

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.

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.