Question

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

Answer 1

The equation of any line through the intersection of lines $5x-6y-1=0$ and

$3x+2y+5=0$ is

$5x-6y-1+k(3x+2y+5)=0...\left(i\right)$

or Slope of this line is $\frac{-(5+3k)}{-6+2k}$

Also, slope of the line $3x-5y+11=0$

is $\frac{3}{5}$

Now, both are perpendicular

So, $\frac{-(5+3k)}{-6+2k}\times \frac{3}{5}=-1$

or $k=45$

Therefore, equation of required line in given by

$5x-6y-1+45(3x+2y+5)=0$

or $5x+3y+8=0$