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

1.	Find the perpendicular distance of the line 3y = 4x + 5 from (2, 1)1. 12. 23. 34. 4
Answer» Correct Answer - Option 2 : 2 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

Find the perpendicular distance of the line 3y = 4x + 5 from (2, 1)1. 12. 23. 34. 4

Answer» Correct Answer - Option 2 : 2

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