3 coordinate of a right triangle

Question

My math and geometry skills are extremely rusty at this point in my life. Could someone walk me through a formula to find the 3 coordiate point's (XY) of this right triangle? I know the distance of two sides and coordinate points of x1,y1 and x2,y2. I need to solve for x3,y3. THanks]1

Answer 1

What is $x_1 - x_2$. Then $y_3 = y_2 + (x_1 - x_2)*\frac {10} 3$.

What is $y_1 - y_2$. Then $x_3 = x_2 + (y_2 - y_1)*\frac {10} 3$.

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Every line has a slope (which may be infinite). To find the slope take two points, say $(x_i, y_i)$ and $(x_k, y_k)$ and calculate $\frac {y_k - y_i}{x_k - x_i}$.

If two lines are perpendicular, and the slope of one is $m$, the slope of the other is $- \frac{1}{m}$.

So the slope of the line from $(x_2, y_2)$ to $(x_3,y_3)$ is $- \frac {x_2 - x_1}{y_2 - y_1}$.

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Every line has a formula $y = mx + b$ where $m$ is the slope (unless the slope $m$ is infinite; then the formula is $x = constant$).

So the formula for the line from $(x_2, y_2)$ to $(x_3,y_3)$ is $y = - \frac {x_2 - x_1}{y_2 - y_1}x + b$.

We can solve for $b$ by plugging in the point $(x_2, y_2)$.

$y_2 = - \frac {x_2 - x_1}{y_2 - y_1}x_2 + b$ so

$b = y_2 + \frac {x_2 - x_1}{y_2 - y_1}x_2$.

So the formula for the line is:

$y = mx + b$ where $m = - \frac {x_2 - x_1}{y_2 - y_1}$ and $b = y_2 + \frac {x_2 - x_1}{y_2 - y_1}x_2$.

So $y_3 = m*x_3 + b$. There are an infinite number of solutions for this.

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Now the distance between $(x_i,y_i)$ and $(x_j, y_j)$ is $D^2 = (y_i -y_j)^2 + (x_i - x_j)^2$. (This is because of the pythogorean theorem).

So the distance between $(x_2, y_2)$ and $(x_3,y_3)$ is:

$10^2 = (y_3 - y_2)^2 + (x_3 - x_2)^2$.

There are an infinite number of solutions.

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BUT there is are only two solutions to both:

$y_3 = m*x_3 + b$ and $10^2 = (y_3 - y_2)^2 + (x_3 - x_2)^2$

Find those two solutions and you have the only two points possible.