Question

Without using Pythagoras Theorem, prove that the points \(\mathrm{A}(1, 3), \mathrm{B}(3, -1)\) and \(\mathrm{C}(-5, -5)\) are the vertices of a right-angled triangle.

[Hint: Show that \(\text{(Slope of AB)} \times (\text{slope of BC)} = -1\).]

Answer 1

Given: Points $$A(1, 3)$$, $$B(3, -1)$$ and $$C(-5, -5)$$.

To prove: Points $$A$$, $$B$$ and $$C$$ are the vertices of a right-angled triangle.

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 $$AB$$,

$$m_1 = \frac{-1 - 3}{3 - 1}$$

$$= \frac{-4}{2}$$

$$= -2$$

2. Slope of $$BC$$,

$$m_2 = \frac{-5 - (-1)}{-5 - 3}$$

$$= \frac{-4}{-8}$$

$$= \frac{1}{2}$$

3. $$(\text{Slope of } AB) \times (\text{Slope of } BC) = m_1 \times m_2$$

$$= (-2) \times \frac{1}{2}$$

$$= -1$$

4. Since $$m_1 m_2 = -1$$, line segment $$AB$$ is perpendicular to line segment $$BC$$, i.e., $$AB \perp BC$$

5. $$\angle B = 90^\circ$$, which implies $$\triangle ABC$$ is a right-angled triangle.

Hence proved.