ALGEBRA 2 find the equation in standard form for the line that passes through (3,-2) and parallel to 4x-y=5?

Question

please......help

Answer 1

Find the equation in standard form for the line that passes through (3,-2) and parallel to 4x - y = 5.

Since the line is parallel to the given line, the only thing that will change in the equation for the desired line is the constant.

Plug in the point to solve for the new constant.

4*3 - 1*(-2) = 12 + 2 = 14

The equation of the desired line is:

4x - y = 14

Answer 2

in this equation you would take 4x-y=5 and solve for y to find the slope(. y=mx+b form ) (because two lines that are parallel have the same slope)

after the equation is solved for y you get. y=4x-5 (after you divide by (-1) it cancels out the -4 and makes the 5 negative).

Then, put the equation in point-slope form using the given points (3,-2) y+2=4(x-3)

and you get the answer: y=4x-14

Hope this helps!

Answer 3

first graph 3,-2 then 4x-y=5 turns into -y=-4x+5 work this part out then you do rise over run on (3,-2)