Question

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

Answer 1

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`.

Explanation:

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`