A line through origin meets the line 2x = 3y + 13 at right angles at point Q. Find the co-ordinates of Q.

Let the line passing through the origin O(0, 0) be L1 and slope m₁, and the given line be L2 : 2x = 3y + 13 and slope be m₂.

Given equation of line,

⇒ 2x = 3y + 13

⇒ 3y = 2x - 13

⇒ y = $\dfrac{2}{3}x - \dfrac{13}{3}$

Comparing above equation with y = mx + c we get, Slope (m₂) = $\dfrac{2}{3}$.

Since L1 is perpendicular to L2, the product of their slopes is -1.

⇒ m₁ × m₂ = -1

⇒ m₁ × $\dfrac{2}{3}$ = -1

⇒ m₁ = $-\dfrac{3}{2}$.

The equation of the line L1 having slope m₁ and passing through the origin can be given by point-slope form i.e.,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{3}{2}$(x - 0)

⇒ y = $-\dfrac{3x}{2}$

⇒ 2y = -3x

⇒ 2y + 3x = 0.

For finding the coordinates of the foot of the perpendicular which is the point of intersection of the lines.

-3y + 2x - 13 = 0 …….(i)

2y + 3x = 0 ……….(ii)

On multiplying equation (i) by 2, we get :

-6y + 4x - 26 = 0 ……….(iii)

On multiplying equation (ii) by 3 we get,

6y + 9x = 0 ………(iv)

Adding (iii) and (iv) we get,

⇒ -6y + 4x - 26 + 6y + 9x = 0

⇒ 13x - 26 = 0

⇒ 13x = 26

⇒ x = $\dfrac{26}{13}$

⇒ x = 2.

Substituting value of x in (ii), we get :

⇒ 2y + 3(2) = 0

⇒ 2y + 6 = 0

⇒ 2y = -6

⇒ y = $\dfrac{-6}{2}$

⇒ y = -3.

∴ Coordinates = (2, -3)

Hence, coordinates of Q = (2, -3).