Table of Contents
Exercise 7.4 Solutions – Class X Mathematics
1. Find the slope of the line passing through the given two points:
The slope \( m \) of a line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \( m = \frac{y_2 – y_1}{x_2 – x_1} \).
(i) (4, -8) and (5, -2)
Here, \( (x_1, y_1) = (4, -8) \), \( (x_2, y_2) = (5, -2) \).
Slope = \( \frac{-2 – (-8)}{5 – 4} = \frac{-2 + 8}{1} = \frac{6}{1} = 6 \).
The slope is 6.
(i) (4, -8) and (5, -2)
Here, \( (x_1, y_1) = (4, -8) \), \( (x_2, y_2) = (5, -2) \).
Slope = \( \frac{-2 – (-8)}{5 – 4} = \frac{-2 + 8}{1} = \frac{6}{1} = 6 \).
The slope is 6.
(ii) (0, 0) and (\( \sqrt{3}, 3 \))
\( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (\sqrt{3}, 3) \).
Slope = \( \frac{3 – 0}{\sqrt{3} – 0} = \frac{3}{\sqrt{3}} = \sqrt{3} \).
The slope is \( \sqrt{3} \).
\( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (\sqrt{3}, 3) \).
Slope = \( \frac{3 – 0}{\sqrt{3} – 0} = \frac{3}{\sqrt{3}} = \sqrt{3} \).
The slope is \( \sqrt{3} \).
(iii) (2a, 3b) and (a, -b)
\( (x_1, y_1) = (2a, 3b) \), \( (x_2, y_2) = (a, -b) \).
Slope = \( \frac{-b – 3b}{a – 2a} = \frac{-4b}{-a} = \frac{4b}{a} \).
The slope is \( \frac{4b}{a} \).
\( (x_1, y_1) = (2a, 3b) \), \( (x_2, y_2) = (a, -b) \).
Slope = \( \frac{-b – 3b}{a – 2a} = \frac{-4b}{-a} = \frac{4b}{a} \).
The slope is \( \frac{4b}{a} \).
(iv) (a, 0) and (0, b)
\( (x_1, y_1) = (a, 0) \), \( (x_2, y_2) = (0, b) \).
Slope = \( \frac{b – 0}{0 – a} = \frac{b}{-a} = -\frac{b}{a} \).
The slope is \( -\frac{b}{a} \).
\( (x_1, y_1) = (a, 0) \), \( (x_2, y_2) = (0, b) \).
Slope = \( \frac{b – 0}{0 – a} = \frac{b}{-a} = -\frac{b}{a} \).
The slope is \( -\frac{b}{a} \).
(v) A(-1.4, -3.7), B(-2.4, 1.3)
\( (x_1, y_1) = (-1.4, -3.7) \), \( (x_2, y_2) = (-2.4, 1.3) \).
Slope = \( \frac{1.3 – (-3.7)}{-2.4 – (-1.4)} = \frac{1.3 + 3.7}{-2.4 + 1.4} = \frac{5}{-1} = -5 \).
The slope is -5.
\( (x_1, y_1) = (-1.4, -3.7) \), \( (x_2, y_2) = (-2.4, 1.3) \).
Slope = \( \frac{1.3 – (-3.7)}{-2.4 – (-1.4)} = \frac{1.3 + 3.7}{-2.4 + 1.4} = \frac{5}{-1} = -5 \).
The slope is -5.
(vi) A(3, -2), B(-6, -2)
\( (x_1, y_1) = (3, -2) \), \( (x_2, y_2) = (-6, -2) \).
Slope = \( \frac{-2 – (-2)}{-6 – 3} = \frac{0}{-9} = 0 \).
The slope is 0 (horizontal line).
\( (x_1, y_1) = (3, -2) \), \( (x_2, y_2) = (-6, -2) \).
Slope = \( \frac{-2 – (-2)}{-6 – 3} = \frac{0}{-9} = 0 \).
The slope is 0 (horizontal line).
(vii) A\( \left(-\frac{3}{2}, 3 \right) \), B\( \left(-7, \frac{1}{2} \right) \)
\( (x_1, y_1) = \left(-\frac{3}{2}, 3 \right) \), \( (x_2, y_2) = \left(-7, \frac{1}{2} \right) \).
Slope = \( \frac{\frac{1}{2} – 3}{-7 – \left(-\frac{3}{2}\right)} = \frac{\frac{1}{2} – 3}{-7 + \frac{3}{2}} = \frac{\frac{1 – 6}{2}}{\frac{-14 + 3}{2}} = \frac{\frac{-5}{2}}{\frac{-11}{2}} = \frac{-5}{-11} = \frac{5}{11} \).
The slope is \( \frac{5}{11} \).
\( (x_1, y_1) = \left(-\frac{3}{2}, 3 \right) \), \( (x_2, y_2) = \left(-7, \frac{1}{2} \right) \).
Slope = \( \frac{\frac{1}{2} – 3}{-7 – \left(-\frac{3}{2}\right)} = \frac{\frac{1}{2} – 3}{-7 + \frac{3}{2}} = \frac{\frac{1 – 6}{2}}{\frac{-14 + 3}{2}} = \frac{\frac{-5}{2}}{\frac{-11}{2}} = \frac{-5}{-11} = \frac{5}{11} \).
The slope is \( \frac{5}{11} \).
(viii) A(0, 4), B(4, 0)
\( (x_1, y_1) = (0, 4) \), \( (x_2, y_2) = (4, 0) \).
Slope = \( \frac{0 – 4}{4 – 0} = \frac{-4}{4} = -1 \).
The slope is -1.
\( (x_1, y_1) = (0, 4) \), \( (x_2, y_2) = (4, 0) \).
Slope = \( \frac{0 – 4}{4 – 0} = \frac{-4}{4} = -1 \).
The slope is -1.
