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

The line is 3x – 4y + 2 = 0

So, y = \( \dfrac{3}{4}x \) +\( \dfrac{2}{4} \)

= \( \dfrac{3}{4}x \) + \( \dfrac{1}{2} \)

Which is of the form y = mx + c, where m is the slope of the given line.

The slope of the given line is \( \dfrac{3}{4} \)

We know that parallel line have same slope.

Slope of other line = m = \( \dfrac{3}{4} \)

Equation of line having slope m and passing through (x₁, y₁) is given by

y – y₁ = m (x – x₁)

Equation of line having slope \( \dfrac{3}{4} \) and passing through (-2, 3) is

y – 3 = \( \dfrac{3}{4} \) (x – (-2))

4y – 3 × 4 = 3x + 3 × 2

3x – 4y = 18

The equation is 3x – 4y = 18