Question

Find the foot of perpendicular from the point $(-3,-4)$ on the line $4(x+2)=3(y-4)$.

• A. $(-\frac{31}{5},-\frac{8}{5})$
• B. $(\frac{31}{5},\frac{8}{5})$
• C. $(-\frac{27}{5},-\frac{13}{5})$
• D. $(\frac{27}{5},\frac{13}{5})$
  

Answer 1

The correct option is A $(-\frac{31}{5},-\frac{8}{5})$

Equation of the line is $4(x+2)=3(y-4)$

$4x+8=3y-12$

$4x-3y=-20$

Let the foot of the perpendicular from point $P(-3,-4)$ on the line is $Q(a,b)$

Slope of given line is $\frac{4}{3}$

Then slope of line $PQ$ is $-\frac{3}{4}$ ( $m_1{m}_2=-1$ for perpendicular lines )

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

$4b+16=-3a-9$

$3a+4b=-25$ ......... (1)

Point $Q(a,b)$ also lies on the line $4x-3y=-20$

So,

$4a-3b=-20$ .......... (2)

Solving (1) and (2) we get,

$a=-\frac{31}{5}$, $b=-\frac{8}{5}$