equation of the line parallel to 2x-y-3=0 that passes the point (-2,1)

Question

find the equation of the line which passes through the point (-2,1) and is parallel to the straight line with the equation 2x-y-3=0

Answer 1

Given equaion 2x-y-3 = 0

Add 3 to each side.

2x-y-3+3 = 0+3

2x-y = 3

Subtract 2x from each side.

2x-y-2x = -2x+3

-y = -2x+3

Multiple to each side by negitive one.

y = 2x-3

Slope of the given line say m1 = 2.

We know that parallel lines have equal slope.

Parallel line slope say m2 = 2, point (x1,y1) = (-2,1)

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

y-1 = 2(x+2)

y-1 = 2x+4

Add 1 to each side.

y-1+1 = 2x+4+1

Required line equation is y = 2x+5.