Question

Find the joint equation of the line passing through (-1, 2) and perpendicular to the lines  x + 2y + 3 = 0 and 3x - 4y - 5 = 0

Answer 1

Let L₁ and L₂ be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x - 4y - 5 = 0 respectively.

Slopes of the lines x + 2y + 3 = 0 and 3x - 4y - 5 = 0 are `-1/2` and `- 3/-4 = 3/4`   respectively.

∴ slopes of the lines L₁ and L₂ are `2` and `(-4)/3` respectively.

Since the lines L₁ and L₂ pass through the point (-1, 2), their equations are

∴ (y - y₁) = m(x - x₁)

∴ (y - 2)= 2(x + 1)

⇒ y - 2 = 2x + 2

⇒ 2x - y + 4 = 0 and 

∴ (y - 2) = `((-4)/3)`(x + 1)

⇒ 3y - 6 = (- 4)(x + 1)

⇒ 3y - 6 = - 4x - 4

⇒ 4x + 3y - 6 + 4 = 0

⇒ 4x + 3y - 2 = 0

their combined equation is

∴ (2x - y + 4)(4x + 3y - 2) = 0

∴ 8x² + 6xy - 4x - 4xy - 3y² + 2y + 16x + 12y - 8 = 0

∴ 8x² + 2xy + 12x - 3y² + 14y - 8 = 0