Find the equation of the line passing through the point (– 1, 3) and

1.	Find the equation of the line passing through the point (– 1, 3) and perpendicular to the line 3x – 4y – 16 = 0.
Answer» The given equation is 3x – 4y – 16 = 0 ⇒ y = \(\frac{3}{4}\) x – 4 ∴ Slope of the line, m₁ = \(\frac{3}{4}\) ∴ Slope of the perpendicular line, m₂ = – \(\frac{4}{3}\) . Since the line passes through (– 1, 3), so the equation of the line is y – y₀ = m(x – x₀) ⇒ y – 3 = m₂ (x+ 1) ⇒ y – 3 = –\(\frac{4}{3}\)(x + 1) ⇒ 3y – 9 = – 4x – 4 ⇒ 4x + 3y – 5 = 0

Find the equation of the line passing through the point (– 1, 3) and perpendicular to the line 3x – 4y – 16 = 0.

Answer»

The given equation is

3x – 4y – 16 = 0

⇒ y = \(\frac{3}{4}\) x – 4

∴ Slope of the line, m₁ = \(\frac{3}{4}\)

∴ Slope of the perpendicular line, m₂ = – \(\frac{4}{3}\) .

Since the line passes through (– 1, 3), so the equation of the line is

y – y₀ = m(x – x₀)

⇒ y – 3 = m₂ (x+ 1)

⇒ y – 3 = –\(\frac{4}{3}\)(x + 1)

⇒ 3y – 9 = – 4x – 4

⇒ 4x + 3y – 5 = 0