Question

Find the equation of the line which passes

through the point (5,-6) and is perpendicular
to the line 2x - 4y + 3 =0​

Answer 1

y = - 2x + 4

Explanation:

The equation of a line in slope- intercept form is

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

Rearrange 2x - 4y + 3 = 0 into this form

Subtract 2x + 3 from both sides

- 4y = - 2x - 3 ( divide all terms by - 4 )

y =
`(1)/(2)` x +
`(3)/(4)` ← in slope- intercept form

with slope m =
`(1)/(2)`

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

`m_(perpendicular)` = -
`(1)/(m)` = -
`(1)/((1)/(2) )` = - 2, thus

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

To find c substitute (5, - 6) into the partial equation

- 6 = - 10 + c ⇒ c = - 6 + 10 = 4

y = - 2x + 4 ← equation of perpendicular line