Find the equation of the perpendicular drawn from the point P

Let the equation of line $A B$ be $x-4 y+7=0$

and point $C$ be $(-2,3)$

CD is perpendicular to the line AB, and we need to find:

1) Equation of Perpendicular drawn from point $\mathrm{C}$

2) Coordinates of D

Let the coordinates of point D be (a, b)

Also, point $D(a, b)$ lies on the line $A B$, i.e. point $(a, b)$ satisfy the equation of line $A B$

Putting $x=a$ and $y=b$, in equation, we get

$a-4 b+7=0 \ldots$ (i)

Also, the CD is perpendicular to the line AB

and we know that, if two lines are perpendicular then the product of their slope is equal to -1

$\therefore$ Slope of $A B \times$ Slope of $C D=-1$

$\Rightarrow$ Slope of $\mathrm{CD}=\frac{-1}{\text { Slope of } \mathrm{AB}}$

$=\frac{-1}{\frac{1}{4}}$

Slope of $C D=-4$

Now, Equation of line CD formed by joining the points $C(-2,3)$ and $D(a, b)$ and having the slope $-4$ is

$\mathrm{y}_{2}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)$

$\Rightarrow b-3=(-4)[a-(-2)]$

$\Rightarrow b-3=-4(a+2)$

$\Rightarrow b-3=-4 a-8$

$\Rightarrow 4 a+b+5=0 \ldots$ (ii)

Now, our equations are

$a-4 b+7=0 \ldots(i)$

and $4 a+b+5=0 \ldots$ (ii)

Multiply the eq. (ii) by 4, we get

$16 a+4 b+20=0 \ldots$ (iii)

Adding eq. (i) and (iii), we get

$a-4 b+7+16 a+4 b+20=0$

$\Rightarrow 17 a+27=0$

$\Rightarrow 17 a=-27$

$\Rightarrow a=-\frac{27}{17}$

Putting the value of a in eq. (i), we get

$-\frac{27}{17}-4 b+7=0$

$\Rightarrow \frac{-27-68 b+119}{17}=0$

$\Rightarrow 92-68 b=0$

$\Rightarrow-68 b=-92$

$\Rightarrow \mathrm{b}=\frac{92}{68}$

$\Rightarrow \mathrm{b}=\frac{23}{17}$

Hence, the coordinates of $\mathrm{D}(\mathrm{a}, \mathrm{b})$ is $\left(-\frac{27}{17}, \frac{23}{17}\right)$