Find the value of \(B-A\) if the graph of \(Ax + By =3\) passes through the point \((-7, 2)\) and is parallel to the graph of \(x+3y=-5\).
I thought it was sort of easy at first, but then I got stuck on a step. Can you provide a detailed explanation? Thanks!
So the first step would be to plot the information given:
Then you want to figure out the equation of the line that passes through your point - Which in this case is \(x+3y=-1\)
Because the line needs to be parellel you dont change the A or B - That is the(1)x and 3y.
Therefore A(3) - B(1) = 2
I hope that all made sense
From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B. Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.
Then the line is y = -x/3 - 19/3. Then 3y = -x - 19, so 3y + x = -19. We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3. Therefore, B - A = -19/3 - 3 = -28/3.