Question

Find the coordinates of the foot of perpendicular drawn from the point(-2,3) on the line 3x-2y+5=0

Answer 1

$\textbf{The foot of perpendicular from the point (-2, 3) to the line 3x-2y+5=0 is given by }$

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

$\text{Here, }$

$(x_1,y_1)=(-2,3)$

$a=3,\,b=-2,\,c=5$

$\frac{x+2}{3}=\frac{y-3}{-2}=\frac{-(3(-2)-2(3)+5)}{3^2+(-2)^2}$

$\frac{x+2}{3}=\frac{y-3}{-2}=\frac{7}{13}$

$\implies\frac{x+2}{3}=\frac{7}{13}$

$\implies\,x+2=\frac{21}{13}$

$\implies\,x=\frac{21}{13}-2$

$\implies\bf\,x=\frac{-5}{13}$

$\text{and}$

$\frac{y-3}{-2}=\frac{7}{13}$

$\implies\,y-3=\frac{-14}{13}$

$\implies\,y=\frac{-14}{13}+3$

$\implies\bf\,y=\frac{25}{13}$

$\therefore\textbf{The foot of the perpendicular is \bf\,(\frac{-5}{13},\;\frac{25}{13} })$

Find more:

Find the coordinates of the foot of the perpendicular drawn from the point (1 -2) to the line y=2x+1

https://brainly.in/question/6182343#

Answer 2

Answer:

(5,6)

Step-by-step explanation:

Given data, Point (2,3) is perpendicular to equation X+Y=11

Rewriting the equation in line equation format ( Y = mX + C)

Y = -X + 11___________________(a)

This shows slope M1 is -1

As the point is perpendicular to this line, M1 x M2 = -1

(-1) x M2 = -1

M2 = 1

Standard equation form, Y = X + C

Substituting the given points (2,3) in above equation.

3 = 2 + C

C = 1

I.e Y = X + 1__________________(b)

Now the coordinate should satisfy both equation (a) & (b)

Thus solving both equations, we can get X = 5, Y= 6

I.e Coordinates (5,6)