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Find the equation of the line passing through the points (3,-2) and (3, 4).The answer is x = 3. I'm just wondering how my textbook got to this solution.My work:y-y1=m(x-x1). m=y2-y1 / x2-x1. y=mx+bm=4--2 / 3-3 = 6/0 = 0. m=0.y--2=0(x-3) = y=0-2 y=-2 <<<

User Hydra
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1 Answer

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Given two points. we can find the equation of a line passing through the points

The formula to be used is:


(y_2-y_1)/(x_2-x_!)=(y-y_1)/(x-x_!)

where


x_1=3,y_!=-2,x_2=3,y_2=4
(4-(-2))/(3-3)=(y-(-2))/(x-3)

=>


(6)/(0)=(y+2)/(x-3)

The next step is to cross multiply


6(x-3)=0(y+2)

=>


6(x-3)=0

Divide both sides by 6 and make x the subject

x=3

User Stefan Rogin
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