What are the coordinates of vertex C

Question

Triangle $ABC$ has 2 given vertices, $A(1,1)$ and $B(5,3)$. Also, AC=BC and $\angle ACB = \,^{\circ}\mathrm{90}$.

The triangle is in the first quadrant entirely. What are the coordinates of vertex C?

I could only figure out that AB = $2\sqrt{5}$ and that Line $ = 0,5x + 0,5 $

But I don't know what to do after that? Can anyone help? What also confuses me is $\angle ACB = \,^{\circ}\mathrm{90}$. What exactly is $\angle ACB$

Answer 1

As Arthur says, $\angle ACB$ is the angle between $AC$ and $CB$. If you have a right triangle with two equal sides, the other angles must be $45^\circ$. You are right that $AB=2\sqrt 5$ Then $AC^2+CB^2=20=2AC^2$. So $AC=\sqrt{10}$. Maybe you already knew that an isoceles right triangle has sides in the ratio $1:-1:\sqrt 2$, but we have confirmed that.

The easiest way to find $C$ is to use the point of Hagen von Eitzen: find the midpoint of $AB$, which is the center of the square. The slope of the other diagonal is $-2$ (the negative of the inverse of the slope of $AB$ as they are perpendicular) and $C$ is $\sqrt 5$ away along this line.

Answer 2

A simpler approach than my other answer is this. Let the vector from $A$ to $C$ be $(a,b)$. Then if the triangle is below $AB$ you need to make a left turn to get the vector from $C$ to $B$, so it will be $(-b,a)$. Adding those two vectors to $A$ needs to get us to $B$, so $(1+a-b,1+b+a)=(5,3)$. We can solve this to get $a=3, b=-1$, so the corner is $(1+a,1+b)=(4,0)$. Similarly if the corner is above $AB$ we need to make a right turn. If the vector from $A$ to $C$ is $(c,d)$, the one from $C$ to $B$ is $(d,-c)$, giving $(1+c+d,1+d-c)=(5,3), d=3, c=1, B=(2,4)$. If your quadrant definition does not include the axes, only the second solution is in the first quadrant.