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Find the equation of the line with a slope of 3/4 that goes through the point (8,10)

Sagot :

the answer is ......... y=3/4+10
Olemakpaduid
This is good.

Using the general format:   y -y1 = m(x-x1), format of a slope and a point.

(x1, y1) = (8,10)  

y -10 = (3/4)*(x-8)
y -10 = 3x/4 -6     Multiply through by 4.
4y -40 = 3x -24
4y      =   3x -24 + 40
4y = 3x + 16.

So equation of the line is:  4y = 3x + 16.

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