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Question
Find the angles between the following pair of straight lines:
3x + 4y − 7 = 0 and 4x − 3y + 5 = 0
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Solution
The equations of the lines are
3x + 4y − 7 = 0 ... (1)
4x − 3y + 5 = 0 ... (2)
Let \[m_1 \text { and } m_2\] be the slopes of these lines.
\[m_1 = - \frac{3}{4}, m_2 = \frac{4}{3}\]
\[\because m_1 m_2 = - \frac{3}{4} \times \frac{4}{3}\]
\[ = - 1\]
Hence, the given lines are perpendicular.
Therefore, the angle between them is 90°.
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| Column C1 | Column C2 |
| (a) Through the point (2, 1) is | (i) 2x – y = 4 |
| (b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is |
(ii) x + y – 5 = 0 |
| (c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is |
(iii) x – y –1 = 0 |
| (d) Equally inclined to the axes is | (iv) 3x – 4y – 1 = 0 |
