Question

The intercept form of the equation of the straight line passing through the point $(4,-3)$ and perpendicular to the line passing through the points $(1,1)$ and $(2,3)$ is

• A. $\frac{x-4}{1}=\frac{y+3}{1}$
• B. $\frac{x}{-2}+\frac{y}{-1}=1$
• C. $\frac{x}{8}-\frac{y}{6}=1$
• D. $\frac{x}{1}+\frac{y}{1}=1$

Answer 1

Answer: (B)

Let $A=(4,-3)$
$B=(1,1)$
$C=(2,3)$
Slope of BC $=\frac{3-1}{2-1}=2$
$\therefore$ Slope of required line $=-\frac{1}{2}$
$[\because$ Lines are perpendicular ]
$\therefore$ Equation of line passing through $A(4,-3)$ and having slope is $\frac{-1}{2}$
$y-y_1=m\left(x-x_1\right)$
$\Rightarrow y+3=\frac{-1}{2}(x-4)$
$\Rightarrow 2y+6=-x+4$
$\Rightarrow x+2y+2=0$
$\Rightarrow x+2y=-2$
$\frac{x}{-2}+\frac{y}{-1}=1$
$[\because$ divided by $-2$ on both sides ]