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Question
Find the coordinates of the foot of the perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.
Solution
Let A (−1, 3) be the given point.
Also, let M (h, k) be the foot of the perpendicular drawn from A (−1, 3) to the line 3x − 4y − 16 = 0
Point M (h, k) lies on the line 3x − 4y − 16 = 0
3h − 4k − 16 = 0 ... (1)
Lines 3x − 4y − 16 = 0 and AM are perpendicular.
\[\therefore\] \[\frac{k - 3}{h + 1} \times \frac{3}{4} = - 1\]
\[\Rightarrow 4h + 3k - 5 = 0\] ... (2)
Solving eq (1) and eq (2) by cross multiplication, we get:
\[\frac{h}{20 + 48} = \frac{k}{- 64 + 15} = \frac{1}{9 + 16}\]
\[ \Rightarrow a = \frac{68}{25}, b = - \frac{49}{25}\]
Hence, the coordinates of the foot of perpendicular are \[\left( \frac{68}{25}, - \frac{49}{25} \right)\].
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