Question

Find the equation of a straight line passing through the point of intersection of the lines $3x+y-9=0$ and $4x+3y-7=0$ and perpendicular to the line $5x-4y+1=0$.

• A. $5x-4y+1=0$
• B. $5x+4y-1=0$
• C. $4x+5y-1=0$
• D. $3x+5y-1=0$

Answer 1

Answer: (C)

The equation to the family of lines passing through the point of intersection of the lines
$3x+y-9=0$ and $4x+3y-1=0$ is
$3x+y-9+k(4x+3y-7)=0$
$\Rightarrow (3+4k)x+(1+3k)y-(9+7k)=0\dots (i)$
where $k$ is a parameter.
Slope of $\left(i\right)=-\left(\frac{3+4k}{1+3k}\right)$ and slope of the given line
$5x-4y+1=0$ is $\frac{5}{4}$.
But the line $\left(i\right)$ is perpendicular to the line $5x-4y+1=0$, therefore
$-\left[\frac{3+4k}{1+3k}\right]\times \frac{5}{4}=-1$
$\Rightarrow 8k=-11$
$\Rightarrow k=-\frac{11}{8}$
Putting this value of $k$ in $\left(i\right)$, we get
$\left(3+4\left(-\frac{11}{8}\right)\right)x+\left(1+3\left(-\frac{11}{8}\right)\right)y-\left(9+7\left(-\frac{11}{8}\right)\right)=0$
$\Rightarrow \left(24-44\right)x+\left(8-33\right)y-\left(72-77\right)=0$
$\Rightarrow -20x-25y+5=0$
$\Rightarrow 4x+5y-1=0$, which is the equation of the required line.