Question

Find the equation of the straight line that passes through the point (3, 4) and perpendicular to the line 3x + 2y + 5 = 0

Answer 1

The equation of a line perpendicular to 3x + 2y + 5 = 0 is 2x - 3y + $\lambda$ = 0 .........(i)
This passes through the point (3, 4)
$\therefore 3\times 2-3\times 4+\lambda =0\Rightarrow \lambda =6$
Putting $\lambda =6$ in (i) we get $2x-3y+6=0$ which is the required equation