Question

Find the coordinates of the foot of the perpendicular drawn from the point (-1,3) to the line 3x - 4y - 16 = 0.

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

$(\frac{68}{25},\frac{-49}{25})$

Step-by-step explanation:

Equation of line = 3x - 4y - 16 = 0.

$y= \frac{3}{4}x-4$

So, Slope of line = $\frac{3}{4}$

Point of perpendicular = (-1,3)

Let (a,b) be the coordinates of the foot of the perpendicular

So, This point must satisfy the equation of line

3a - 4b - 16 = 0

a = $\frac{16+4b}{3}$

Since we know that the product of slopes of perpendicular lines is -1

So, $\frac{3}{4} \times m = -1\\m = \frac{-4}{3}$

Equation of perpendicular = $y-y_1=m(x-x_1)=b-3=\frac{-4}{3}(a+1)$

$b-3=\frac{-4}{3}(a+1)$

$a=\frac{-3b}{4}+\frac{5}{4}$

Substitute in 1

$3(\frac{-3b}{4}+\frac{5}{4}) - 4b - 16 = 0$

$b=\frac{-49}{25}$

$a=\frac{-3(\frac{-49}{25})}{4}+\frac{5}{4}$

$a=\frac{68}{25}$

Hence he coordinates of the foot of the perpendicular drawn from the point (-1,3) to the line 3x - 4y - 16 = 0 is $(\frac{68}{25},\frac{-49}{25})$