# AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.4

### 10th Class Maths 7th Lesson Coordinate Geometry Ex 7.4 Textbook Questions and Answers

Question 1.
Find the slope of the line joining the two given points.
i) (4,-8) and (5,-2).
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{-2+8}{5-4}$$
= 6

ii) (0, 0) and (√3, 3)
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{3-0}{\sqrt{3}-0}$$
= $$\frac{3}{\sqrt{3}}$$
= $$\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}$$
= √3

iii) (2a, 3b) and (a, -b).
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{-b-3b}{a-2a}$$
= $$\frac{-4b}{-a}$$
= $$\frac{4b}{a}$$

iv) (a, 0) and (0, b).
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{b-0}{0-a}$$
= $$\frac{-b}{a}$$

v) A (-1.4, -3.7), B (-2.4, 1.3).
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{1.3+3.7}{-2.4+1.4}$$
= $$\frac{5.0}{-1}$$
= -5

vi) A (3, -2), B (-6, -2).
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{-2+2}{-6-3}$$
= 0

vii) A (-3$$\frac{1}{2}$$, 3), B (-7, 2$$\frac{1}{2}$$).
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{2 \frac{1}{2}-3}{-7+3 \frac{1}{2}}$$
= $$\frac{-\frac{1}{2}}{-3 \frac{1}{2}}$$
= $$\frac{1}{2}$$ × $$\frac{2}{7}$$
= $$\frac{1}{7}$$
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
= $$\frac{0-4}{4-0}$$
= $$\frac{-4}{4}$$