Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).

The equation of the given line is

$3 x-4 y+2=0$

or $y=\frac{3 x}{4}+\frac{2}{4}$

or $y=\frac{3}{4} x+\frac{1}{2}$, which is of the form $y=m x+c$

$\therefore$ Slope of the given line $=\frac{3}{4}$

It is known that parallel lines have the same slope.

$\therefore$ Slope of the other line $=m=\frac{3}{4}$

Now, the equation of the line that has a slope of $\frac{3}{4}$ and passes through the point $(-2,3)$ is

$(y-3)=\frac{3}{4}\{x-(-2)\}$

$4 y-12=3 x+6$

i.e., $3 x-4 y+18=0$