The vertices of a ΔABC are A(2, –11), B(2, 13) and C(–12, 1). Find the equations of its sides.

Given,

Coordinates A(2, −11) and B(2, 13)

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

Substituting values we get,

Slope of AB = $\dfrac{13 - (-11)}{2 - 2} = \dfrac{13 + 11}{0} =\dfrac{24}{0}$

Slope is not defined.

The line AB is a vertical line parallel to the y-axis.

Points have the same x-coordinate, x = 2.

Equation of line AB: x = 2

Given, Points: B(2,13), C(−12,1)

By formula,

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

Substituting values we get,

Slope of BC = $\dfrac{1 - (13)}{-12 - 2} = \dfrac{-12}{-14} =\dfrac{6}{7}$

By point-slope form,

Equation of the line BC, y - y₁ = m(x - x₁)

⇒ y - 13 = $\dfrac{6}{7}$ (x - 2)

⇒ 7(y - 13) = 6(x - 2)

⇒ 7y - 91 = 6x - 12

⇒ 7y - 6x = -12 + 91

⇒ 7y - 6x = 79

Equation of BC: 7y - 6x = 79

Given, Points: C(−12,1), A(2,−11)

By formula,

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

Substituting values we get,

Slope of CA = $\dfrac{-11 - 1}{2 - (-12)} = \dfrac{-12}{14} =-\dfrac{6}{7}$

By point-slope form,

Equation of the line CA, y - y₁ = m(x - x₁)

⇒ y - 1 = $-\dfrac{6}{7}$ (x + 12)

⇒ 7(y - 1) = -6(x + 12)

⇒ 7y - 7 = -6x - 72

⇒ 7y + 6x - 7 + 72 = 0

⇒ 7y + 6x + 65 = 0

Equation of the line CA: 7y + 6x + 65 = 0

Hence, the equation of AB, BC and CA are x = 2, 7y - 6x = 79, 7y + 6x + 65 = 0 respectively.