Find third coordinate for a right triangle with 45degree angles

Question

I have a right triangle with two 45degree angles.

I know the points for the two coordinates opposite the right angle. I need to calculate the missing point.

I have seen similar questions here, but I have not understood the answers. There were lots of symbols and fancy words that made no sense to me. I tried to do some calculations, but my results are not quite right.

I'm hoping someone can provide me with a solution using layman's terms and a straight-forward formula.

Thanks very much

Answer 1

The two given points are opposite corners of a square. The third point of the triangle will be one of the other two corners of this square. Since the diagonals of a square are the same length and perpendicular to each other, you can find these points by computing the midpoint of the two given points and then going out the same distance perpendicularly to either side of it.

Let the two given points be $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$. Their midpoint is at $$P_0=P_1+\frac12(P_2-P_1)=\frac12(P_1+P_2)=\left({x_1+x_2\over2},{y_1+y_2\over2}\right).$$ Now we need a vector that's perpendicular to the line between $P_1$ and $P_2$ and has length equal to half the distance between them. $\frac12(P_2-P_1)=\left({x_2-x_1\over2},{y_2-y_1\over2}\right)$ has the right length, but is 90 degrees off, so we rotate that:$$\left(-{y_2-y_1\over2},{x_2-x_1\over2}\right).$$ Adding to and subtracting from $P_0$ gives you the points you’re looking for:$$\left({x_1+x_2\over2},{y_1+y_2\over2}\right)\pm\left(-{y_2-y_1\over2},{x_2-x_1\over2}\right).$$

Answer 2

Let $A=(x_A,y_A)$ and $B=(x_B,y_B)$ be the given points. Let their midpoint be $M=((x_A+x_B)/2,(y_A+y_B)/2)$ and $X=(x,y)$ the point to be found.

Then you have that $MX$ is perpendicular to $AB$ and that $MX=AM$. This gives you two equations, to find $x$ and $y$.

Answer 3

You have ($W$ is auxiliary to write in a line the second equation) $$W=(X-x_1)^2+(Y-y_1)^2=(X-x_2)^2+(Y-y_2)^2\qquad (1)\\(X-\frac{x_1+x_2} {2})^2+(Y-\frac{y_1+y_2}{2})^2+(\frac {x_1+y_1}{2}-x_1)^2+(\frac {y_1+y_2}{2}-y_1)^2=W\qquad (2)$$ This simplifies to linear system $$aX+bY=c\qquad (1')\\dX+eY=f\qquad (2')$$ Try to determine the values of the constants $a,b,c,d,e,f$ and solve the linear system.