I've been looking for an equation to find where the X value collides with a slanted line, and I have been rather unsuccessful and would appreciate some help.
That was probably pretty unclear but this picture should make it more clear.
I'm pretty new to calculating with slanted lines/hypotenuses so please try to keep it understandable.
On
First of all, we want the equation of the blue line. It passes through points $(-1, -2)$ and $(2,2)$ so therefore its slope is $$ k = \frac{2-(-2)}{2-(-1)} = \frac{4}{3} $$ The line is of the form $y=kx + b$, so we need $2=\frac{4}{3}\times 2 +b \Rightarrow b = \frac{-2}{3}$. Therefore, the blue line has the equation $$ y = \frac{4x-2}{3} $$ Now we want to solve $y=0$, or $$ \frac{4x-2}{3} = 0 \qquad \Rightarrow \qquad 4x-2 = 0 \qquad \Rightarrow \qquad x=\frac{1}{2} $$
Considering the two points $(2,2)$ and $(-1,-2)$ you can find the straight line equation :
$$\frac{y-2}{2-(-2)} = \frac{x-2}{2-(-1)}$$
The equation of the line is : $$y=\frac{4}{3}x - \frac{2}{3}$$
Now the intersection of this line with the green line which is $y=0$ can be found out by putting $y=0$ in the straight line equation.
You will get the point as $(\frac{1}{2},0)$
Hope this helps ...