Question

Find the equation in standard form of the line passing through the points (6, 3) and (1,5)

Answer 1

2x+5y=27

Explanation:

m=(y2-y1)/(x2-x1)

m=(5-3)/(1-6)

m=2/-5

m=-2/5

y-y1=m(x-x1)

y-3=-2/5(x-6)

y-3=-2/5x+12/5

y=-2/5x+12/5+3

y=-2/5x+12/5+15/5

y=-2/5x+27/5

y-(-2/5x)=27/5

y+2/5x=27/5

5(y+2/5x)=5(27/5)

5y+2x=27

2x+5y=27