# What is the equation of the line that passes through (-2, 5) and is perpendicular to the line whose equation is y = 1/2x + 5?

The equation would be set up as 5y+-0.5(-2)+b reciprocal of 1/2 is 0.5 it’s negative because it’s perpendicular multiply 0.5 by -2 which is 1 subtract 1 from the 5y you have 4.

let's keep in mind that perpendicular lines have negative reciprocal slopes, hmmmmm what's the slope of the line through (-6,1/2) and (-4,2/3) anyway?

so we're really looking for the equation of a line whose slope is -12 and runs through -2,5.

Step-by-step explanation:

The given lines are perpendicular and as AB = AC , Therefore △ ABC is art . angled isosceles . Hence the line BC through ( 1 , 2) will make an angles of ±45

∘

with the given lines . Its equations is y - 2 = m (x - 1) where m = 1 / 7 and -7 as in .Hence the possible equations are 7x + y - 9 = 0 and x - 7y + 13 = 0

Alt :

The two lines will be parallel to bisectors of angle between given lines and they pass through ( 1, 2)

∴ y - 2 = m ( x - 1)

where m is slope of any of bisectors given by

5

3x+4y−5

=±

5

4x−3y−15

or x - 7y + 13 = 0 or 7x + y - 20 = 0

∴ m = 1 / 7 or - 7

putting in (1) , the required lines are 7x + y - 9 = 0

and x - 7y + 13 = 0 as found above