Question

find the equation of a straight line that has a slope of m=2/3 and is passing thru the point of -2,-4

Answer 1

(y-y1) = m(x-x1)

where your (-2,-4) = (x1,y1) and m=2\3

so that becomes,

(y-(-4)) = 2\3(x-(-2))

(y+4) = 2\3(x+2)

3(y+4) = 2(x+2)

distribute 3 and 2,

3y + 12 = 2x + 4

3y - 2x +12 - 4 = 0


ans..
3y - 2x +8 = 0 ^_^