I know the coordinates of the $3$ vertices of a triangle and the $y$ coordinate of a fourth point on one side. How do I find the $x$ coordinate of the fourth point? I feel I'm missing something obvious.
$$A(0, 0), B(844, 0), C(844, 2000)$$
The mystery point is $D(??, 950)$.
Thank you for any and all help!
On
The point $D$ lies on the line segment $\overline{AC}$. Therefore, the slopes of $\overline{AD}$ and $\overline{AC}$ must be equal. Mathematically, $$\frac{950-0}{x-0}=\frac{2000-0}{844-0}.$$
On
If you drop a perpendicular form $D$ to the line $\overline{AB}$, you can then use properties of similar triangles. In particular, if we label that point $E$, we will have $\triangle AED \sim \triangle ABC$. Since one is just a scaled copy of the other, we have $$ \frac{AE}{DE} = \frac{AB}{CB} \implies AE = DE\frac{AB}{CB} = 950 \frac{844}{2000} = \frac{4009}{10}. $$ But note that $x$ is equal to the distance from $A$ to $E$, and so $x = 400.9$.
On
Given $A(A_x,A_y)$, $C(C_x,C_y)$, $D(D_x,D_y)\in AC$ with known $A,C,D_y$ and unknown $D_x$, location of the point $B$ is irrelevant, since the point $D$ is on the linear Bézier segment with endpoints $A,C$, $t\in[0,1]$:
\begin{align} D&=A(1-t)+C(t). \tag{1}\label{1} \end{align}
\eqref{1} is in fact two equations
\begin{align} D_x&=A_x(1-t)+C_x(t) ,\\ D_y&=A_y(1-t)+C_y(t), \end{align}
This gives
\begin{align} t&=\frac{D_y-A_y}{C_y-A_y} ,\\ D_x&=\frac{A_x\,(C_y-D_y)+C_x\,(D_y-A_y)}{C_y-A_y} . \end{align}
The solution suggests that $A_y\ne C_y$, otherwise $D_y=A_y=C_y$ and there would be not enough data to determine $D_x$.
It's a direct proportion:
$$ \frac{950}{x}=\frac{2000}{844}$$
which you should be able to solve for $x$ with no difficulty.