Algebra or Geometry?
On an $x,y$ axis draw a line from $0,40$ diagonally down, to the right, to $10,0$. Now draw a line from $0,27$ to $10,27$ intersecting the diagonal.
Question: How do I determine the $x$ coordinate where the line intersects the diagonal? (I'm pretty sure that if I drew the line from $0,20$ to $10,20$ it would intersect at $x = 5$)
On
I guess when you write $0,40$ you actually mean the point in the plane with coordinates $(0,40)$. Well, these are two lines intersecting each other. Any line in the plane can be written as $ax+by+c=0$ for some coefficients $a,b,c$. Imposing that $(0,40)$ and $(10,0)$ are points on this line you get: $$0+40b+c=0,$$ $$10a+0+c=0,$$ that is $c= -40b = -10a$. So your line is given by the points $(x,y)$ in the plane satisfying $4bx+by-40b = 0$. Notice that if $b=0$, $ax+by+c=0$ implies $ax+c=0$, which means that all the points on your line have the same first coordinate. This is false, as $(10,0)$ and $(0,40)$ are two points on the line with different $x$-coordinates. So you can assume $b \neq 0$. Then your actual line is $4x+y-40 = 0$, i.e. $y = -4x+40$. Similarly, you can find that the equation of the other line is $y=27$. The points in the intersection of these two lines satisfy both equations. So they have $y$-coordinate equal to $27$, and then $x$-coordinate given by $$27 = -4x+40 \rightarrow x = \frac{13}{4}.$$ So the intersection is one point, namely $(13/4,27)$.
You have two lines. Find the equation corresponding to each line and solve them simultaneously. For the red one you have two points on the line, so the two point form will work nicely. That will give the coordinates of the point of intersection.