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
$3x + y - 9 + k(4x + 3y - 7) = 0$
$\Rightarrow (3 + 4 k)x + (1 + 3 k)y - (9 + 7k) = 0 \quad \ldots(i)$
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.