Q158 / 270 · NDA 2013 (Phase 2) · Mathematics
What is the equation of the line which passes through $(4,-5)$ and is perpendicular to $3 x+4 y+5=0$ ?
- Correct answerA$4 x-3 y-31=0$
- B$3 x-4 y-41=0$
- C$4 x+3 y-1=0$
- D$3 x+4 y+8=0$
Solution
Given line is $3 x+4 y+5=0$
or, $y=-\frac{3}{4} x-\frac{5}{4}$
Slope of line $(m)=-\frac{3}{4}$
Product of Slopes of two perpendicular lines $=-1$
Let slope of required line is $m^{\prime}$.
then $m \cdot m^{\prime}=-1$
$\begin{aligned}
\left(-\frac{3}{4}\right) \cdot m^{\prime} & =-1 \\
m^{\prime} & =\frac{4}{3}
\end{aligned}$
From formula, equation of line with slope $(\mathrm{P})$ and passing through point $(\alpha, \beta)$ is
$\begin{aligned}
& (y-\beta)=p .(x-\alpha) \\
& \text { Here, } P=\frac{4}{3},(\alpha, \beta)=(4,-5) \\
& (y-(-5))=\frac{4}{3}(x-4) \\
& \text { or, } 3 y+15=4 x-16 \\
& \text { or, } 4 x-3 y-31=0
\end{aligned}$
Hence, the correct answer is option: 'A'.