# 10th Maths Chapter 5 Coordinate Geometry Exercise 5.2

10th Standard Maths Chapter 5 Coordinate Geometry Exercise 5.2 Guide. TN SSLC Samacheer Kalvi Guide Chapter 5 Exercise 5.2 Book Back Answers & Solutions. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

### 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 –1/2. Find the value of a.

Solution:
A line joining the points (-2, a) and (9, 3) has a slope m = –12.

Solution:

### 9. Show that the given points form a right-angled triangle and check whether they satisfy 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 midpoints of the diagonals are the same.
∴ Midpoints of the sides of a quadrilateral form a parallelogram.