How do you write an equation of a line passing through (5, 8), perpendicular to # y = 2/3 x - 3#?

1 Answer
May 17, 2017

Answer:

#y=-3/2x+31/2#

Explanation:

First find the slope of the perpendicular line:

The slope of a perpendicular line is defined as #m_|_=-1/m#

We know that the slope of the given line is #2/3# so the slope of the perpendicular line is then:

#-1/(2/3)# or #-1*3/2=-3/2#

Next, we can use the point slope formula: #y-y_1=m(x-x_1)# to find the equation of the line by substituting the perpendicular slope and the point #(5,8)# for #(x_1,y_1)# Thus:

#y-8=-3/2(x-5)#

We can express this in #y=mx+b# form is desired by simply solving for #y#:

#y-8=-3/2x+15/2#

#ycancel(-8+8)=-3/2x+15/2+8#

*Find the LCD for #15/2# and #8# which is #2#

#y=-3/2x+15/2+8/1(2/2)#

#y=-3/2x+15/2+16/2#

#y=-3/2x+31/2#

The graph of the two lines will then look like this: graph{(y-2/3x+3)(y+3/2x-31/2)=0 [-1.92, 18.08, -0.32, 9.68]}