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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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(ii) `(- 1/3, 11/3), (4/3, 7/3)` |
| (c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are |
(iii) `(1, 12/5), (-3, 16/5)` |
