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What is the equation of the line passing through #(-2,-3)# and #(0,2)#?

1 Answer

Mar 25, 2016

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

Explanation:

#x_1 = -2, x_2 = 0, y_1 = -3, y_2 = 2#
Plug into slope formula: #(y_2-y_1)/(x_2-x_1)#:

#(2-(-3))/(0-(-2)) = 5/2 = m#

Using one of the points, plug into point slope from, I used #(0,2)#:
#y-2 = (5/2)(x-0)#
Simplify:
#y=(5/2)x+2#

graph{y=(5/2)x+2 [-6.18, 6.24, -3.277, 3.19]}
Both points #(-2,-3)# and #(0,2)# are on the graph

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