Intersection Point Between Two Lines With a General Form

Oct 2013
717
91
New York, USA
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?
 
Oct 2018
129
96
USA
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.
 
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Oct 2013
717
91
New York, USA
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
Dec 2006
21,482
2,472
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).
 
Oct 2013
717
91
New York, USA
So it's a general method if the lines intersect at one point, which is how systems of equations normally work.