+231 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 $x + 3y = -5.$
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, A + B = -9/19 - 3/19 = -12/19.
\(y=mx+c\) slope-intercept form
"Parallel to the graph of \(x+3y=-5\)" means that \(Ax+By=3\) have the same slope as it.
Both are written in standard form.
\(x+3y=-5\) Divide both sides by 3
\(\frac{x}{3}+y=-\frac{5}{3}\) Rearrange
\(y=-\frac{1}{3}x-\frac{5}{3}\) Thus slope is \(-\frac{1}{3}\)
\(Ax+By=3\)
Rearrange to solve for y
\(\frac{A}{B}x+y=\frac{3}{B}\)
\(y=-\frac{A}{B}x+\frac{3}{B}\)
Notice \(\frac{A}{B}=\frac{1}{3}\) (Since it is the same slope as the previous line)
Thus we have
\(y=-\frac{1}{3}x+\frac{3}{B}\)
Now, as given, \(Ax+By=3\) passes through the point (-7,2)
Substitute this into \(y=-\frac{1}{3}x+\frac{3}{B}\)
\(2=-\frac{1}{3}*(-7)+\frac{3}{B}\) , Solve for B
\(2-\frac{7}{3}=\frac{3}{B}\) , \(B=-9\)
So the equation now becomes \(y=-\frac{1}{3}x-\frac{1}{3}\)
However, we want to find the value of A in order to answer the question "What is B-A"
\(Ax+By=3\), use (-7,2) (Also knowing the value of B= -9 to solve for A)
\(-7A+(-9)*(2)=3\)
\(A=-3\)
\(B-A=-9-(-3)=-9+3=-6\)
