How do you show that if $a+b=0$ , then the slope of $x/a+y/b+c=0$ is $1$ ?

Question

1468 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-19T00:04:14

Solve the linear equation for $y$ to obtain the slope-intercept form, and use the conditions on $a$ and $b$ to show that the slope is $1$ .

If we solve a linear equation for $y$ we get the slope-intercept form $y=mx + b$ where $m$ is the slope of the graph and $b$ is the
$y$ -intercept.

Doing so in the given case, we have

$x/a + y/b + c = 0$

$=> y/b = -1/ax - c$

$=> y = -b/ax - c$

Thus the slope $m$ is $(-b)/a$

But from $a + b = 0$ we can subtract $b$ from both sides to obtain $a = -b$ . Substituting this into the above slope, we get

$m = (-b)/a = (-b)/(-b) = 1$

How do you show that if a+b=0a+b=0, then the slope of x/a+y/b+c=0x/a+y/b+c=0 is 11?