Question

Find the joint equation of the line passing through the point (3, 2), one of which is parallel to the line x - 2y = 2, and other is perpendicular to the line y = 3.

Answer 1

Let L₁ be the line passes through (3, 2) and parallel to the line x - 2y = 2 whose slope is `(-1)/-2 = 1/2`

∴ slope of the line L₁ is `1/2`

∴ equation of the line L₂ is

y - 2 =`1/2`(x - 3)

∴ 2y - 4 = x - 3

∴ x - 2y + 1 = 0

Let L₂ be the line passes through (3, 2) and perpendicular to the line y = 3.

∴ equation of the line L₂ is of the form x = a. Since L₂ passes through (3, 2), 3 = a.

∴ equation of the line L₂ is x = 3, i.e. x - 3 = 0 

Hence, the equations of the required lines are

x - 2y + 1 = 0 and x - 3 = 0

∴ their joint equation is

(x - 2y + 1)(x - 3) = 0

∴ x² - 2xy + x - 3x + 6y - 3 = 0

∴ x² - 2xy - 2x + 6y - 3 = 0