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

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- 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$

or = k = 3$

Hence the equation of the required line

$ x -4y+3=0$

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