+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
x3+y=−53 Rearrange
y=−13x−53 Thus slope is −13
Ax+By=3
Rearrange to solve for y
ABx+y=3B
y=−ABx+3B
Notice AB=13 (Since it is the same slope as the previous line)
Thus we have
y=−13x+3B
Now, as given, Ax+By=3 passes through the point (-7,2)
Substitute this into y=−13x+3B
2=−13∗(−7)+3B , Solve for B
2−73=3B , B=−9
So the equation now becomes y=−13x−13
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
