Question

Find the equation of the straight line passing through the point of intersection of the straight lines $5x-6y=1$ and $3x+2y+5=0$ and is perpendicular to the straight line $3x-5y+11=0$.

• A. $5x+3y+8=0$
• B. $5x-3y+8=0$
• C. $5x+3y-8=0$
• D. None of these

Answer 1

The correct option is A $5x+3y+8=0$

Consider the given equation of the line.

$5x-6y=1$ $...........\left(1\right)$

$3x+2y=-5$ $...........\left(2\right)$

$\left(1\right)+\left(2\right)\times 3$

$5x-6y=1$

$9x+6y=-15$

$\overline{14x+0=-14}$

$x=-1$ sub in $\left(2\right)$

$3(-1)+2y=-5$

$2y=-2$

$y=-1$

Point of intersection of above both lines is

$=(-1,-1)$

Given straight line

$3x-5y+11=0$

Slope of line $m_1=-\frac{3}{-5}=\frac{3}{5}$

Since, the line is perpendicular to the the line, so the slope the line is

$m_2=\frac{-1}{m_1}=\frac{-1}{\frac{3}{5}}=-\frac{5}{3}$

So, the equation of line passes through intersection point

$y+1=-\frac{5}{3}(x+1)$

$3y+3=-5x-5$

$5x+3y+8=0$

Hence, this is the answer.