Find the perpendicular distance from the origin of the perpendicular f

1.	Find the perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line x - √3y + 4 = 0.
Answer» Given: Line x - √3y + 4 = 0 To find: The perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line x - √3y + 4 = 0 . Concept Used: Distance of a point from a line. Explanation: The equation of the line perpendicular to x - √3y + 4 = 0 is x - √3y + λ = 0 This line passes through (1,2) ∴ \(\sqrt{-3}\) + 2 + λ = 0 ⇒ λ = \(\sqrt{-3}\) - 2 Substituting the value of λ, we get \(\sqrt{3}x+y-\sqrt{3}-2=0\) Let d be the perpendicular distance from the origin to the line \(\sqrt{3}x+y-\sqrt{3}-2=0\) d = \(\frac{0-0-\sqrt{3}-2}{\sqrt{1+3}}=\frac{\sqrt{3}+2}{2}\) Hence, the required perpendicular distance is \(\frac{\sqrt{3}+2}{2}\)

Find the perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line x - √3y + 4 = 0.

Answer»

Given:

Line x - √3y + 4 = 0

To find:

The perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line x - √3y + 4 = 0 .

Concept Used:

Distance of a point from a line.

Explanation:

The equation of the line perpendicular to x - √3y + 4 = 0 is x - √3y + λ = 0 This line passes through (1,2)

∴ \(\sqrt{-3}\) + 2 + λ = 0

⇒ λ = \(\sqrt{-3}\) - 2

Substituting the value of λ, we get \(\sqrt{3}x+y-\sqrt{3}-2=0\)

Let d be the perpendicular distance from the origin to the line \(\sqrt{3}x+y-\sqrt{3}-2=0\)

d = \(\frac{0-0-\sqrt{3}-2}{\sqrt{1+3}}=\frac{\sqrt{3}+2}{2}\)

Hence, the required perpendicular distance is \(\frac{\sqrt{3}+2}{2}\)