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

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, $\frac{x-4}{1}=\frac{y+3}{1}$, $\frac{x}{-2}+\frac{y}{-1}=1$, $\frac{x}{8}-\frac{y}{6}=1$, $\frac{x}{1}+\frac{y}{1}=1$

MARKS App · Accepted Answer

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

Search any question & find its solution

Looking for more such questions to practice?