+0

# help meee!

0
373
3

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 $x + 3y = -5.$

May 7, 2020

#1
-1

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.

May 7, 2020
#2
0

Why did you do A+B instead of B-A?

peepeepuupuu  May 7, 2020
#3
0

$$y=mx+c$$ 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)

$$-7A+(-9)*(2)=3$$

$$A=-3$$

$$B-A=-9-(-3)=-9+3=-6$$

May 8, 2020