The intercept form of the equation of the straight line passing through the point \((4,-3)\) and…
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
\(\frac{x-4}{1}=\frac{y+3}{1}\)
\(\frac{x}{-2}+\frac{y}{-1}=1\)
\(\frac{x}{8}-\frac{y}{6}=1\)
\(\frac{x}{1}+\frac{y}{1}=1\)
Solution
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.