Question

Find the equation of the line passing through the intersection of the lines x + 3y − 1 = 0 and x − 2y + 4 = 0 and perpendicular to the line 2x + 3y = 0

Answer 1

We have U₁ ≡ x + 3y − 1 = 0 and U₂ ≡ x − 2y + 4 = 0. Equation of the required line is

U₁ ≡ λU₂≡ (1 + λ)x + (3 - 2λ)y - 1 + 4λ = 0

Since this line is perpendicular to the line 2x + 3y = 0, we  have

Thus, the required line equation

Direct Method: 

Let U₁ ≡ x + 3y − 1 = 0 and U₂ ≡ x − 2y + 4 = 0. Therefore, the point of intersection of the lines U₁ = 0 and U₂ = 0 is (−2, 1).

Hence, the equation of the required line is 3(x + 2) - 2(y -1) = 0

3x - 2y + 8 = 0