Straight lines

Given the line $2x+3y+5 = 0 \implies y = -\dfrac 23x - \dfrac 53$ . Therefore, the gradient of the line is $-\dfrac 23$ . Let the perpendicular line be $y = mx + c$ . where $m$ and $c$ are the gradient and $x$ -intercept. Then $-\dfrac 23 m = - 1$ , $\implies m = \dfrac 32$ . Since the line passes through the point $(-2,-4)$ , then $-4 = \dfrac 32(-2)+c$ , $\implies c = - 1$ . Therefore, the line is $y = \dfrac 32x-1$ , $\implies \boxed{3x - 2y-2 = 0}$ .

Perpendicular line to the straight line 2x+3y+5=0 is 3x-2y+k=0---(1).As the point passes through (-2,-4),we get from the equation (1)-3.(-2)-2.(-4)+k=0;k=-2. So putting value of k in (1),we get the required equation 3x-2y-2=0.