Question

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

Answer 1

Hint: Convert the given equation into slope intercept form, then substitute the obtained slope m in line equation to obtain the perpendicular line equation. Then solve the given equation and the obtained perpendicular line equation to obtain the point of intersection.

Formula Used:
If $y = mx + c$ is the equation of a line then the equation of the line perpendicular to this line is $y = - \dfrac{1}{m}x + {c_1}$.

Complete step by step solution:
The given equation can be written as
 $4y = 3x - 16$
Or,$y = \dfrac{3}{4}x - 4$
Hence, the slope is $m = \dfrac{3}{4}$ .
Now, the equation of the perpendicular line is,
$y = - \dfrac{4}{3}x + a$ ---(1)
This line passes through the point $\left( { - 1,3} \right)$,
So, $3 = - \dfrac{4}{3}( - 1) + a$
$a = 3 - \dfrac{4}{3}$
$ = \dfrac{5}{3}$
Hence, from (1) we have,
$y = - \dfrac{4}{3}x + \dfrac{5}{3}$
$4x + 3y - 5 = 0$
Now, multiply $3x - 4y - 16 = 0$ by 4 and $4x + 3y - 5 = 0$ by 3 and subtract to obtain the value of y.
$12x - 16y - 64 - 12x - 9y + 15 = 0$
$ - 25y - 49 = 0$
$y = - \dfrac{{49}}{{25}}$
Substitute $y = - \dfrac{{49}}{{25}}$ in the equation $4x + 3y - 5 = 0$ to obtain the value of x.
$4x + 3\left( { - \dfrac{{49}}{{25}}} \right) - 5 = 0$
$4x = 5 + \dfrac{{147}}{{25}}$
$x = \dfrac{{272}}{{25 \times 4}} = \dfrac{{68}}{{25}}$
Therefore, the point of intersection is $\left( {\dfrac{{68}}{{25}}, - \dfrac{{49}}{{25}}} \right)$

The required foot point is $\left( {\dfrac{{68}}{{25}}, - \dfrac{{49}}{{25}}} \right)$

Note:To obtain a perpendicular line first convert the given equation to slope intercept form then substitute the obtained slope m by $ - \dfrac{1}{m}$ in line equation $y = mx + c$ then obtain the perpendicular line.