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 \, 2 y+6 =-x+4 $
$ \Rightarrow \, x+2 y+2 =0 $
$\Rightarrow \, x+2 y =-2$
$\frac{x}{-2}+\frac{y}{-1}=1$
$[\because$ divided by $-2$ on both sides ]