Points A, B and C divide the line segment joining the points D(8, -4) and E(-12, 16) in four equal parts. Find the equation of the line that passes through point A and is perpendicular to DE.

By formula,

(x, y) = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

From figure,

A divides the line segment ED in the ratio 1 : 3.

$\therefore A = \Big(\dfrac{1 \times -12 + 3 \times 8}{1 + 3}, \dfrac{1 \times 16 + 3 \times -4}{1 + 3}\Big) \\[1em] = \Big(\dfrac{-12 + 24}{4}, \dfrac{16 - 12}{4}\Big) \\[1em] = \Big(\dfrac{12}{4}, \dfrac{4}{4}\Big) \\[1em] = (3, 1).$

By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

Substituting values we get :

$\text{Slope} = \dfrac{16 - (-4)}{-12 - 8} \\[1em] = \dfrac{20}{-20} \\[1em] = -1.$

We know that,

Product of slope of perpendicular lines = -1.

Let slope of line perpendicular to DE be m.

⇒ m × -1 = -1

⇒ m = 1.

By point-slope form,

Equation : y - y₁ = m(x - x₁)

So, equation of line with slope = 1 and passing through A = (3, 1) is

⇒ y - 1 = 1[x - 3]

⇒ y - 1 = x - 3

⇒ y = x - 3 + 1

⇒ y = x - 2

Hence, equation of line passing through A and perpendicular to DE is y = x - 2.