[Math] Finding a point on the line perpendicular to a line from another point

coordinate systemsgeometric-constructiongeometry

I'm sorry if this is kind of basic, it's been a while since I took geometry. I did find this answer, but it requires 4 points and I only have 3.

I have three points $A$, $B$ and $C$ that form a non-right triangle. I know the Cartesian coordinates of these three points. Suppose I draw a perpendicular from $B$ through $AC$ where $D$ is the point where the perpendicular intersects $AC$. How can I find the Cartesian coordinates of a point $T$ along $\overrightarrow {BD}$ that is $n$ beyond $D$?

I drew a diagram to demonstrate what I'm trying to find.

Diagram
enter image description here
Any help would be appreciated, thanks!

Best Answer

I'm assuming you can find the equation of a line i) given two points and ii) given a point and slope and that you can do some other basic algebra which will show up. Comment if you don't understand anything.

  1. Determine the equation of the line AC. In particular, we need its slope for the next step.

  2. The slope of BD is the negative reciprocal of the slope of AC (so if for instance AC has slope equal to $\frac{2}{3}$ then BD has slope $-\frac{3}{2}$). This follows because BD is perpendicular to AC.

  3. You know the slope of BD and one point (B) so you can find its equation. Do this.

  4. Solve for D by finding the point of intersection of AC and BD. This can be done by equating their y-values.

  5. Use the length formula $L=\sqrt{(\Delta x)^2 + (\Delta y)^2}$ as well as $\Delta y = m \Delta x$ (m is slope) to create a quadratic which you can solve for $\Delta x = x_t - x_d$. You can use this to find $x_t$.

  6. Solve for $y_t$ using the equation for BD.

Comment if you get stuck following this.

Related Question