What is the equation of the line perpendicular to #y=-2/7x # that passes through # (-2,5) #?

1 Answer
Dec 30, 2015

Answer:

#y-5 = 7/2(x+2)# Equation in point-slope form.
#y=7/2x+12# Equation of the line in slope-intercept form

Explanation:

To find the equation of the line perpendicular to the given line.

Step 1: Find the slope of the given line.

Step 2: Take the slope's negative reciprocal to find the slope of perpendicular.

Step 3: Use the given point and the slope use the Point-Slope form to find the equation of line.

Let us write our given line and go through the steps one by one.

#y=-2/7x#

Step 1: Finding the slope of #y=-2/7x#
This is of the form #y=mx+b# where #m# is the slope.

Slope of the given line is #-2/7#

Step 2: The slope of perpendicular is the negative reciprocal of the given slope.

#m= -1/(-2/7)#
#m=7/2#

Step 3: Use the slope #m=7/2# and the point #(-2,5) to find the equation of the line in the Point-Slope form.

Equation of line in Point-slope form when slope #m# and a point #(x_1,y_1)# is # y-y_1 = m(x-x_1)#

#y-5 = 7/2(x+2)# Solution in point-slope form.

Simplifying we can get
#y-5 = 7/2x+7 # using distributive propertly
#y = 7/2x +7+5# adding #5# both sides

#y=7/2x+12# Equation of the line in slope-intercept form