Find Third point in right triangle given two points and a length

geometrytrigonometry

I am trying to find the third point in a right triangle. I am given Points A and B, and the distance from B to C, but I don't know how to find the coordinates of point C.

An Image of the problem described (Excuse my sub-Par GeoGebra skills)

Best Answer

Outline:

1) We know the coordinates of points $A=(u,v)$ and $B=(w,z)$.

So Q1) What is the slope, $m$, of the line $\overline{AB}$?

Answer:

If $A = (u,v)$ and $B= (w,z)$ then the slope is $m = \frac {z-v}{w-u}$.

2) We know the $\overline{BC}$ is a right angle to $\overline{AB}$.

So Q2) what is the slope, $m_2$, of the line $\overline{BC}$?

Answer:

If the slope of $\overline{AB}=m$. Then the slope of $\overline {BC} = -\frac 1m$. Call it $m_2 = -\frac 1m = -\frac {w-u}{z-v}$.

3) Given $m_2$ is the slope of $\overline{BC}$ and that $B = (w,z)$

So Q3) What is the formula for the line $\overline{BC}$ in terms of variables $x$ and $y$?

The slope of $\overline{BC}$ is $m_2=\frac {rise}{run}$ is if $x$ is $x-u$ away from $u$ then if $y$ is $y-v$ away from $y$ then $m_2 =\frac {y-v}{x-u}$ or in other words $y-v = m_2(x-u)$. Or in slope intercept form $y = m_2x + (v-m_2u)$. But we don't need slope intercept form.

4) So if $B= (u,v)$ and $C$ is an unknown $(x,y)$ then

Q4) what is formula for the distance $d$ between $(u,v)$ and $(x,y)$.

Answer:

$d = \sqrt{(x-u)^2 +(y-v)}$

5) So if $d$ is a known value for the distance between $B=(u,v)$ and $C$ the unknown $(x,y)$ and we have the line formula for the line $\overline{BC}$,

Q5) How do we solve for $x$ and how do we solve for $y$

Answer for $x$:

$d = \sqrt{(x-u)^2 + (y-v)^2}$ and $y-v= m_2(x-u)$. So $d =\sqrt{(x-u)^2 + m_2^2(x-u)^2}=\sqrt{(1+m_2^2)(x-u)^2}=|x-u|\sqrt{1+m_2^2}$. If we assume $u > x$ (from the diagram) we have $d= (x-u)\sqrt{1+m_2^2}$. So $x-u = \frac d{\sqrt{1+m_2^2}}$ and $x =\frac d{\sqrt{1+m_2^2}}+u$.

Answer for $y$:

$x-u = \frac d{\sqrt{1+m_2^2}}$ and $y-v = m_2(x-u)$ so $y-v=\frac {m_2d}{\sqrt{1+m_2^2}}$ and $y = \frac {m_2d}{\sqrt{1+m_2^2}}+v$.

6) So what is $C=(x,y)$?

$C = (x,y) = (\frac d{\sqrt{1+m_2^2}}+u,\frac {m_2d}{\sqrt{1+m_2^2}}+v)=$

.

$(\frac d{\sqrt{1 + (-\frac 1{m})^2}} +u, -\frac dm\frac 1{\sqrt{1+(-\frac 1m)^2}} + v) =$

.

$(\frac d{\sqrt{1 + (\frac{z-v}{w-u})^2}}+ u, -\frac {d(w-u)}{(z-v)\sqrt{1 + (\frac{z-v}{w-u})^2}}+v)$

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