Question

Find the equation of the line through the point of intersection of the lines $3x-4y+1=0$ & $5x+y-1=0$ and perpendicular to the line $2x-3y=10.$

• A. $69x+46y-25=0$
• B. $69x+46y-22=0$
• C. $69x-46y-25=0$
• D. $69x+46y+22=0$

Answer 1

The correct option is A $69x+46y-25=0$

Point of intersection is $3x-4y+1=0$ and $5x+y-1=0$ is $(\frac{3}{23},\frac{8}{23})$
Slope of line $2x-3y=10$ is $\frac{2}{3}$
Hence slope of line perpendicular to it is $-\frac{3}{2}$
Therefore equation of line with slope $-\frac{3}{2}$ and passing through point $(\frac{3}{23},\frac{8}{23})$ is
$69x+46y-25=0.$