# Intersection Point Between Two Lines With a General Form

#### EvanJ

It looks like y = -x+a and y = bx where a and b are constants intersect when y = ab/(b+1). Is there a way to prove that?

#### Greens

Lines intersect when their $y$-coordinate and $x$-coordinate are equal.

We know $y= -x+a$ and $y=bx$, so we set the right hand sides equal to each other to receive the $x$-coordinate for the intersection.

$\displaystyle -x+a = bx$

$\displaystyle a = bx+x$

$\displaystyle a = x(b+1)$

$\displaystyle x = \frac{a}{b+1}$

This is the value of $x$ at the intersection, so we may put this value into the equations for $y$ to find the $y$-coordinate at the intersection.

$\displaystyle y= bx$

$\displaystyle y=b \left( \frac{a}{b+1} \right)$

$\displaystyle y=\frac{ab}{b+1}$

Putting $x= \frac{a}{b+1}$ into $y=-x+a$ also yields $y=\frac{ab}{b+1}$, but I'll leave it to you to work out.

• topsquark and EvanJ

#### skipjack

Forum Staff
It looks like y = -x+a and y = bx where a and b are constants intersect when y = ab/(b+1).
What happens when a = 0 and b = -1?

#### EvanJ

My equation produces the undefined fraction 0/0 in that case. However, my equation isn't necessary because both lines become y = -x. They're the same line, so there's not one point of intersection. If you find a time my equation doesn't work for lines that aren't the same, I'm interested.

#### skipjack

Forum Staff
You originally asked about a way to prove a general method (one without exceptions mentioned), but your method isn't general, as it fails when there are no common points (distinct parallel lines) and when there are common points that you don't call intersection points (because the lines are identical).