A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.

The given line equation is

⇒ x = 3y + 2 ……….(1)

⇒ 3y = x - 2

⇒ y = $\dfrac{1}{3}x - \dfrac{2}{3}$

Comparing above equation with y = mx + c we get,

Slope (m) = $\dfrac{1}{3}$

Let slope of perpendicular line be m₁.

⇒ m₁ × m = -1 (As product of slope of perpendicular lines = -1).

⇒ $m_1 \times \dfrac{1}{3}$ = -1

⇒ m₁ = -3.

Equation of the line passing through origin and with slope = -3, by point-slope form is:

⇒ y – y₁ = m(x – x₁)

⇒ y – 0 = -3(x – 0)

⇒ y = -3x

⇒ 3x + y = 0 ………..(2)

Next,

Point X is the intersection of the lines (1) and (2).

Substituting value of x from (1) in (2),

⇒ 3(3y + 2) + y = 0

⇒ 9y + 6 + y = 0

⇒ 10y = -6

⇒ y = $-\dfrac{6}{10} = -\dfrac{3}{5}$.

⇒ x = 3y + 2 = $3 \times -\dfrac{3}{5} + 2 = \dfrac{-9 + 10}{5} = \dfrac{1}{5}$.

Hence, the co-ordinates of the point X are $\Big(\dfrac{1}{5}, -\dfrac{3}{5}\Big)$.