How do you find the equation of the line that passes through (2,-1) and parallel to 2x-3y=5?

1 Answer
Nov 16, 2016

Answer:

#y=2/3x-1/3#

Explanation:

First, rewrite the equation in intercept form: #y=mx+b#

Where #m# is the slope and #b# is your #y# intercept

#2x-3y=5#

Begin to isolate #y# by subtracting #2x# on both sides of the equation

#-3y=-2x+5#

Divide both sides of the equation by #-3#

#y=2/3x-5/3#

The slope of this line is #2/3#

To find another equation that is parallel to this equation, you have to use that slope because parallel lines have the same slope.

To find the equation using the given slope and given set of points, plug the numbers into point slope form:

#y-y_1=m(x-x_1)#

Where #(2,-1)=(x_1,y_1)#

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

Distribute #2/3# throughout the set of parenthesis

#y+1=2/3x-4/3#

Subtract #1# on both sides of the equation

#y=2/3x-1/3#

Now, notice that the slope of this equation is the same as the other equation, so that means that the two equations create two lines that are parallel to eachother