Question

Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR.

Answer 1

Answer:

2x + y - 1 = 0

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

x - 2y + 4 = 0 ( subtract x + 4 from both sides )

- 2y = - x - 4 ( divide all terms by - 2 )

y = $\frac{1}{2}$ x + 2 ← in slope- intercept form

with slope m =  $\frac{1}{2}$

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{\frac{1}{2} }$ = - 2 , thus

y = - 2x + c ← is the partial equation of the perpendicular line

To find c substitute (1, - 1) into the partial equation

- 1 = - 2 + c ⇒ c = - 1 + 2 = 1

y = - 2x + 1 ← in slope intercept form

Subtract - 2x + 1 from both sides

2x + y - 1 = 0 ← in general form