What is the equation of the line that passes through the point (-2,3) and that is perpendicular to the line represented by 3x-2y= -2?

1 Answer
Jan 13, 2017

Answer:

#(y - 3) = -3/2(x + 2)#

Or

#y = -3/2x#

Explanation:

First, we need to convert the line into slope-intercept form to find the slope.

The slope-intercept form of a linear equation is:

#y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b# is the y-intercept value.

We can solve the equation in the problem for #y#:

#3x - 2y = -2#

#3x - color(red)(3x) - 2y = -2 - color(red)(3x)#

#0 - 2y = -3x - 2#

#-2y = -3x - 2#

#(-2y)/color(red)(-2) = (-3x - 2)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = (-3x)/color(red)(-2) - 2/color(red)(-2)#

#y = 3/2x + 1#

So for this equation the slope is #3/2#

A line perpendicular to this line will have a slope which is the negative inverse of our line or #-3/2#

We can now use the point-slope formula to write the equation for the perpendicular line:

The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #color(red)(((x_1, y_1)))# is a point the line passes through.

Substituting the point from problem and the slope we calculated gives:

#(y - color(red)(3)) = color(blue)(-3/2)(x - color(red)(-2))#

#(y - color(red)(3)) = color(blue)(-3/2)(x + color(red)(2))#

Or, we can put the equation in the more familiar slope-intercept form by solving for #y#:

#y - color(red)(3) = color(blue)(-3/2)x + (color(blue)(-3/2) xx color(red)(2))#

#y - color(red)(3) = -3/2x - 3#

#y - color(red)(3) + 3 = -3/2x - 3 + 3#

#y = -3/2x + 0#

#y = -3/2x#