# Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1)

$\begin {array} {1 1} (A)\;x−4y+3=0 & \quad (B)\;x+4y+3=0 \\ (C)\;x−4y−3=0 & \quad (D)\; x+4y−3=0 \end {array}$

Thank you very much for the answer.In my book of mathematics points (3,-1) is in front of points(2,3).So I was taking them as x1 and y1.
Thanks again

Toolbox:
• Equation of a line pasing through two points $(x_1,y_1)$ and $(x_2, y_2)$ is $\large\frac{y-y_1}{y_2-y_1}$$= \large\frac{x-x_1}{x_2-x_1} • If two lines are perpendicular, then the product of their slopes is -1. Step 1 Equation of the line passing through the points (2, 3) and (3, -1 ) is \large\frac{y-3}{-1-3}$$= \large\frac{x-2}{3-2}$
$\Rightarrow \large\frac{y-3}{-4}$$=\large\frac{x-2}{1}$
or $y-3=-4(x-2)$
$\Rightarrow y-3 = -4x+8$
(i.e.,) $4x+y=11$
This is the equation of the line.
Step 2
The line which is perpendicular to this line is $x-4y+k=0$
Since it passes through the point (5,2)
$5-4(2)+k=0$
$\Rightarrow k-3=0$