Question

Prove that the lines 2x + 3y + 8 = 0 and 27x – 18y + 10 = 0 are perpendicular to each other.

Answer 1

Given: Equations of two lines, $$2x + 3y + 8 = 0$$ and $$27x - 18y + 10 = 0$$.

To prove: The lines are perpendicular to each other.

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. $$2x + 3y + 8 = 0$$

$$\Rightarrow 3y = -2x - 8$$

$$\Rightarrow y = -\frac{2}{3}x - \frac{8}{3}$$

2. Slope of the first line, $$m_1 = -\frac{2}{3}$$

3. $$27x - 18y + 10 = 0$$

$$\Rightarrow 18y = 27x + 10$$

$$\Rightarrow y = \frac{27}{18}x + \frac{10}{18} = \frac{3}{2}x + \frac{5}{9}$$

4. Slope of the second line, $$m_2 = \frac{3}{2}$$

5. $$m_1 \times m_2 = \left(-\frac{2}{3}\right) \times \left(\frac{3}{2}\right) = -1$$

6. Since $$m_1 m_2 = -1$$, the two lines are perpendicular to each other.

Hence proved.