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

2 Answers
Nov 15, 2016

Answer:

#y=2/3x+19/3#

Explanation:

Parallel lines have the same slopes. So your new line has slope #2/3#

Apply point-slope formula, #y-y_1=m(x-x_1)# where #(x_1,y_1)# is #(-5,3)#:
#y-3=2/3(x-(-5))#
#y-3=2/3(x+5)#
#y-3=2/3x+10/3#
#y=2/3x+19/3#
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Nov 15, 2016

Answer:

#y=2/3x+19/3#

Explanation:

We're going to use two things here, point-slope form and slope-intercept form. Slope intercept is #y=mx+b#, whereas point-slope is #y-y_1=m(x-x_1)#. The point that we know we go through can be written as #(x_1,y_1)# so #x_1=-5# and #y_1=3#. We know the slope from the equation given in the question by applying it to slope-intercept. The #m# is the slope, so the slope of the equation in question is #2/3#.

Just to organize this better:
#x_1=-5#
#y_1=3#
#m=2/3#

Now we plug this information into point-slope #y-y_1=m(x-x_1)#
#y-3=2/3(x-(-5)) -> y-3=2/3(x+5)#

This is a good answer but I'm just going to bring it to y intercept because that's what I'm used to personally.

Add #3# to both sides.

#y=2/3(x+5)+3#

Distribute #2/3#

#y=2/3x+10/3+3#

Add #10/3# and #3#

#y=2/3x+19/3#