What is the equation of the line that passes through #(-3,1)# and is perpendicular to the line that passes through the following points: #(-2,4),(6,1) #?

1 Answer
Apr 28, 2017

#y=8/3x+9#

Explanation:

Begin by first finding the slope between #(-2,4)# and #(6,1)# by using the slope formula: #m=(y_2-y_1)/(x_2-x_1)#

If we let #(-2,4)-> (color(red)(x_1),color(blue)(y_1))# and #(6,1)-> (color(red)(x_2),color(blue)(y_2))# then,

#m=color(blue)((1-4))/color(red)((6-(-2)))= -3/8#

Since we are looking for the perpendicular slope we use the following: #m_|_=-1/m#

#m_|_=-1/(-3/8)=8/3#

Now that we have our slope #(8/3)# and given the point #(-3,1)# we can find the equation of the line by using the point-slope formula: #y-y_1=m(x-x_1)#

#y-1=8/3(x-(-3)) ->y-1=8/3(x+3)#

We can rewrite this into #y=mx+b# form if desirable

#y=8/3x+8/3*3/1+1 -> y=8/3x+24/3+3/3 -> y=8/3x+27/3#

Upon further simplification, our final answer is #y=8/3x+9#