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Find an equation for the given line in the form $ax +by =​c$, where​ a, b, and c are integers with no factor common to all three and a ≥0.

Through $(-40, 35)$; parallel to $7x + 8y = 13$

I put that in slope intercept form. Then use the Point Slope Formula.

Every time I do a question likes this I always get $c>0$.

So my answer would end up being $7x + 8y = 75$. But the answer seems to always be

$7x + 8y = 0$.

If they gave me $4x + 3y = 14$ (I made this up)

I can put in $4x + 3y = 0$ and get a correct answer.

However when I work this out I NEVER get $ax + by = 0$

I always get $ax + by > 0 $

And that is has never been the correct answer. I've spent 1 1/2 hours now trying to figure this out and I'm sure it's something simple.

I'm going to work through a problem on here.

9x + 4y = 13 (-8, 18)

y = -9/4x + 13

y - 18 = -9/4x + (x -- 8)

y - 18 = -9/4x - 36

4y + 9x = 36 But for my answer I put 9x + 4y = 0 and it was correct.

Ty for the responses, I need some time to digest what was just given.

Jude
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    As of right now, your question is very difficult to understand. – zz20s May 28 '16 at 16:26
  • The line $7x+8y=75$ has the right slope, but it clearly does not go through $(-40,35)$ because $-7\cdot40+8\cdot35=0\ne75$. So you did something wrong. Could you show your working? – almagest May 28 '16 at 16:30
  • When you need to find the general form for a line known to pass through a given point that is parallel to a given line $ \ ax \ + \ by \ = \ c \ $ , you can insert the coordinates into $ \ ax \ + \ by \ $ to compute the new value for the constant. So $ \ 7(-40) \ + \ 8(35) \ = \ ? \ $ . (Working with slope-intercept form will also work, but what did you use for the slope?) – colormegone May 28 '16 at 16:31

2 Answers2

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Given the equation $ax+by=c$ ( with $b \ne 0$) the slope of the straight line is $$ m=-\frac{a}{b} $$ so the slope of $7x+8y=13$ is $m=-\frac{7}{8}$.

A parallel line have the same slope $m$ so its equation has the form: $$ y=-\frac{7}{8}x+q \qquad (1) $$ and, using the coordinates of the point $P=(-40,35)$ you can find $q$:

$$ 35=-\frac{7}{8}(-40)+q \quad \Rightarrow \quad q=0 $$

if you want write the equation in the form $ax+by=c$ you can reorder $(1)$ and multiply for any number, e.g. $8$ to find $7x+8y=0$

Emilio Novati
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Obviously $c$ cannot be always $0,$ it depends on the point it needs to pass through. If the point is $(-40,35)$ and the line is parallel to $7x+8y=13,$ and $a,b,c$ has no common factor then it has to be $7x+8y=0.$

Think of this as follows. The slope of the line is $-7/8.$ Then any line parallel to it is of the form $y=-\frac{7}{8}x+r,$ or $7x+8y=r.$ This is already of the form $ax+by=c$ where $a,b$ has no common factor, and its not possible to divide them further since they has to be integers. Then for the line to pass through $(-40,35)$ only option is $c=0$

A. Ray
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