Find the equation of the line that is perpendicular to 3x + 2y - 8 = 0 and passes through the mid-point of the line segment joining the points (5, -2) and (2, 2).

Given equation of line,

⇒ 3x + 2y - 8 = 0

Converting it in the form y = mx + c,

⇒ 2y = -3x + 8

⇒ y = $-\dfrac{3}{2}x + 4$

Comparing with y = mx + c we get,

Slope (m₁) = $-\dfrac{3}{2}$

Now, the coordinates of the mid-point of the line segment joining the points (5, -2) and (2, 2) will be

$\Rightarrow \Big(\dfrac{5 + 2}{2}, \dfrac{-2 + 2}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{7}{2}, 0\Big).$

Let's consider the slope of the line perpendicular to the given line be m₂.

Then,

$\Rightarrow m$1 \times m2 = -1 \\[1em] \Rightarrow -\dfrac{3}{2} \times m2 = -1 \\[1em] \Rightarrow m2 = \dfrac{2}{3}.

The equation of the new line with slope m₂ and passing through $(\dfrac{7}{2}, 0)$ can be given by point-slope form i.e.,

y - y₁ = m(x - x₁)

Putting values we get,

$\Rightarrow y - 0 = \dfrac{2}{3}(x - \dfrac{7}{2}) \\[1em] \Rightarrow 3y = 2(x - \dfrac{7}{2}) \\[1em] \Rightarrow 3y = 2x - 7 \\[1em] \Rightarrow 2x - 3y - 7 = 0.$

Hence, the equation of the line is 2x - 3y - 7 = 0.