Find the perpendicular distance of the line 3y = 4x + 5 from (2, 1)

Question

Question

Answer 1

Concept:

The distance of a point (x₁, y₁) from a line ax + by + c = 0

D = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

Calculation:

Given line 3y = 4x + 5

⇒ 4x - 3y + 5 = 0

(x1, y1) = (2, 1)

∴ D = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

⇒ D = \(\rm \left|4\times 2 + (-3)\times 1+5\over\sqrt{4^2+(-3)^2}\right|\)

⇒ D = \(\rm \left|10\over5\right|\) = 2