Let us consider the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)

So, let the slope of the line joining (-1, 3) and (a, b) be m_{1}

m_{1 }= \(
\dfrac{(b-3)}{(a+1)}\)

And let the slope of the line 3x – 4y – 16 = 0 be m_{2}

y = \( \dfrac{3}{4}x\) – 4

m_{2} = \( \dfrac{3}{4}\)

Since these two lines are perpendicular, m_{1} × m_{2} = -1

\( \dfrac{(b-3)}{(a+1)}\) × (\( \dfrac{3}{4}\)) = -1

\( \dfrac{(3b-9)}{(4a+4)}\) = -1

3b – 9 = -4a – 4

4a + 3b = 5 …….(1)

Point (a, b) lies on the line 3x – 4y = 16

3a – 4b = 16 ……..(2)

Solving equations (1) and (2), we get

a = \( \dfrac{68}{25}\) and b = \(- \dfrac{49}{25}\)

The co-ordinates of the foot of perpendicular is (\( \dfrac{68}{25}\), \( - \dfrac{49}{25}\))

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