Question

Prove that the line through \(\mathrm{A}(-2, 6)\) and \(\mathrm{B}(4, 8)\) is perpendicular to the line through \(\mathrm{C}(8, 12)\) and \(\mathrm{D}(4, 24)\).

Answer 1

Given: The points $$A(-2, 6)$$, $$B(4, 8)$$, $$C(8, 12)$$ and $$D(4, 24)$$.

To prove: The line through $$A$$ and $$B$$ is perpendicular to the line through $$C$$ and $$D$$.

Proof:

By theorem, two non-vertical lines with slopes $$m_1$$ and $$m_2$$ are perpendicular to each other, if and only if $$m_1 m_2 = -1$$.

1. Slope of the line through $$A(-2, 6)$$ and $$B(4, 8)$$

$$m_1 = \frac{8 - 6}{4 - (-2)}$$

$$= \frac{2}{6}$$

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

2. Slope of the line through $$C(8, 12)$$ and $$D(4, 24)$$

$$m_2 = \frac{24 - 12}{4 - 8}$$

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

$$= -3$$

3. Product of slopes $$= m_1 \times m_2$$

$$= \frac{1}{3} \times (-3)$$

$$= -1$$

4. Since $$m_1 m_2 = -1$$, the line through $$A$$ and $$B$$ is perpendicular to the line through $$C$$ and $$D$$.

Hence proved.