Question

The vertices of a $$\triangle \mathrm{ABC}$$ are A(3, 8), B(–1, 2) and C(6, –6).

Find:

1. Slope of BC.
2. Equation of a line perpendicular to BC and passing through A.

Answer 1

i. Coordinates of $$\mathrm{B}(-1, 2)$$ and $$\mathrm{C}(6, -6)$$.

Slope of $$\mathrm{BC}$$:

$$m_{\mathrm{BC}} = \frac{-6 - 2}{6 - (-1)}$$ 

$$= \frac{-8}{7}$$

$$= -\frac{8}{7}$$

ii. Slope of the line perpendicular to $$\mathrm{BC}$$:

$$m = -\frac{1}{m_{\mathrm{BC}}}$$

$$= -\frac{1}{-\frac{8}{7}}$$

$$= \frac{7}{8}$$

Equation of the line passing through $$\mathrm{A}(3, 8)$$:

$$y - 8 = \frac{7}{8}(x - 3)$$

$$8(y - 8) = 7(x - 3)$$

$$8y - 64 = 7x - 21$$

$$7x - 8y + 43 = 0$$