Question

Given that A(5, 4), B(–3, –2) and C(1, –8) are the vertices of a $$\triangle \mathrm{ABC}$$.

Find: 

1. the slope of median AD,
2. the slope of altitude BM.

Answer 1

i. $$\mathrm{AD}$$ is the median from $$\mathrm{A}$$ to $$\mathrm{BC}$$, so $$\mathrm{D}$$ is the mid-point of $$\mathrm{BC}$$. 

Coordinates of $$\mathrm{D}$$: 

$$\mathrm{D} = \left(\frac{-3 + 1}{2}, \frac{-2 + (-8)}{2}\right)$$

$$= \left(\frac{-2}{2}, \frac{-10}{2}\right)$$

$$= (-1, -5)$$

Slope of median $$\mathrm{AD}$$ joining $$\mathrm{A}(5, 4)$$ and $$\mathrm{D}(-1, -5)$$: 

$$m_{\mathrm{AD}} = \frac{-5 - 4}{-1 - 5}$$

$$= \frac{-9}{-6}$$

$$= \frac{3}{2}$$

ii. $$\mathrm{BM}$$ is the altitude on $$\mathrm{AC}$$, so $$\mathrm{BM} \perp \mathrm{AC}$$. 

Slope of $$\mathrm{AC}$$ joining $$\mathrm{A}(5, 4)$$ and $$\mathrm{C}(1, -8)$$: 

$$m_{\mathrm{AC}} = \frac{-8 - 4}{1 - 5}$$

$$= \frac{-12}{-4}$$

$$= 3$$

Since $$\mathrm{BM} \perp \mathrm{AC}$$: 

$$m_{\mathrm{BM}} = -\frac{1}{m_{\mathrm{AC}}}$$

$$= -\frac{1}{3}$$