Question

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

Answer 1

Answer:

$\text{The foot of the perpendicular is }(\frac{68}{25},\frac{-49}{25})$

Step-by-step explanation:

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

$\text{The foot of perpendicular from the point ( -1, 3) to the line 3x-4y-16=0 is given by }$

$\boxed{\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}}$

$\text{Here, }$

$(x_1,y_1)=(-1,3)$

$a=3,\,b=-4,\,c=-16$

$\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{-(3(-1)+(-4)(3)+(-16))}{3^2+(-4)^2}$

$\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{-(-3-12-16)}{9+16}$

$\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{-(-31)}{25}$

$\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{31}{25}$

$\text{Now, }$

$\frac{x+1}{3}=\frac{31}{25}$

$x+1=\frac{93}{25}$

$x=\frac{93-25}{25}$

$\implies\:x=\frac{68}{25}$

$\frac{y-3}{-4}=\frac{31}{25}$

$y-3=\frac{-124}{25}$

$y=\frac{-124+75}{25}$

$\implies\:y=\frac{-49}{25}$

$\therefore\text{The foot of the perpendicular is }(\frac{68}{25},\frac{-49}{25})$

Answer 2

Answer:

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

Step-by-step explanation:

We have the straight line 3x-4y-16=0 .........(1)

Arranging this equation into slope-intercept form (i.e. y=mx+c  form, where, m=slope and c= y-intercept), we get the equation as,

y= (3/4)x-4 ......(2)

So, the slope is (3/4).

Let us assume that the slope of the other straight line be M.

Now, we know that the product of slopes of two perpendicular straight lines is -1.

So, M×($\frac{3}{4}$)=-1

⇒ M= -($\frac{4}{3}$)

Hence, the equation of the perpendicular straight line is

y=-($\frac{4}{3}$)x+K [ Where K is the y-intercept of this straight line]

This line passes through the point (-1,3) [Given].

So, we get 3=-($\frac{4}{3}$)(-1)+K

⇒3=($\frac{4}{3}$) +K

⇒K= $\frac{5}{3}$

Now, the equation of the perpendicular straight line is

y=-($\frac{4}{3}$)x+$\frac{5}{3}$

⇒4x+3y-5 =0 ......(3)

Now, Solving equations (1) and (3), we get,

9x+16x-48-20=0

⇒25x-68=0

⇒x= $\frac{68}{25}$

From equation (2), we get,

y=($\frac{3}{4}$)×($\frac{68}{25}$)-4

By simplification, we get,

y=-($\frac{49}{25}$).

Therefore, the foot of the perpendicular is {($\frac{68}{25}$),-($\frac{49}{25}$)} (Answer)