Find the equation of the straight line perpendicular to the line 2x + 3y = 0 and passing through the point of intersection of the lines x + 3y – 1 = 0 and x – 2y + 4 = 0.

Given lines
$\begin{array}{l}\mathrm{x}+3\mathrm{y}-1=0 ...\left(1\right)\\ \mathrm{x}-2\mathrm{y}+4=0 ...\left(2\right)\end{array}$
Solve (1) & (2)
Point of intersection (x, y) = (–2, 1)
and given line 2x + 3y = 0 ... (3)
Let the eq. of the straight line perpendicular to and passing through (–2, 1) be
$$\begin{gathered} 3\mathrm{x}-2\mathrm{y}+\mathrm{k}=0\Rightarrow 3(-2)-2\left(1\right)+\mathrm{k}=0\Rightarrow \mathrm{k}=8 \\ \therefore \text{ the eq' }\mathrm{n}\text{ of st.line is }3\mathrm{x}-2\mathrm{y}+8=0 \end{gathered}$$