Question

The vertices of a $$\triangle \mathrm{ABC}$$ are A(2, –11), B(2, 13) and C(–12, 1). Find the equations of its sides.

[Hint. The slope of AB = `(13 + 11)/(2 - 2) = 24/0`, which is not defined.
∴ AB is parallel to y-axis. Its equation is x = а.
Also, it passes through (2, –11). So, a = 2. Hence, x = 2.]

Answer 1

Equation of side $$AB$$:

Both $$A(2, -11)$$ and $$B(2, 13)$$ have the same $$x$$-coordinate, $$x = 2$$.

Thus, $$AB$$ is a vertical line parallel to the $$y$$-axis and its equation is $$x = 2$$ or $$x - 2 = 0$$

Equation of side $$BC$$:

For $$B(2, 13)$$ and $$C(-12, 1)$$, slope is:

$$m_{BC} = \frac{1 - 13}{-12 - 2}$$

$$= \frac{-12}{-14}$$

$$= \frac{6}{7}$$ 

Equation of $$BC$$:

$$y - 13 = \frac{6}{7}(x - 2)$$ 

$$7(y - 13) = 6(x - 2)$$ 

$$7y - 91 = 6x - 12$$ 

$$6x - 7y + 79 = 0$$

Equation of side $$AC$$:

For $$A(2, -11)$$ and $$C(-12, 1)$$, slope is:

$$m_{AC} = \frac{1 - (-11)}{-12 - 2}$$

$$= \frac{12}{-14}$$

$$= -\frac{6}{7}$$

Equation of $$AC$$: 

$$y - (-11) = -\frac{6}{7}(x - 2)$$

$$7(y + 11) = -6(x - 2)$$

$$7y + 77 = -6x + 12$$

$$6x + 7y + 65 = 0$$