# What are the coordinates of vertex C

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 $\angle ACB$

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By the conditions, the triangle is half of a square with $AB$ as diagonal. – Hagen von Eitzen Oct 6 '12 at 15:48
$\angle ACB$ is the angle between the two lines $AC$ and $CB$. – Arthur Oct 6 '12 at 15:50

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.

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Ok, the formula of the bisection of AB is $y=-2x+8$, I got that, but why is $C$ $\sqrt{5}$ along this line and how could I give the coordinates of C, $\sqrt{5}$ isn't exactly an integer or even a number with 1 or 2 decimals (I believe). – JohnPhteven Oct 6 '12 at 16:22
You are right for the bisector. As the side of the square is $\sqrt {10}$, the diagonal is $\sqrt {20}=2\sqrt 5$ and half the diagonal (the segment from the center to a corner) is $\sqrt 5$. You are also right that $\sqrt 5$ is not an integer or a number with a few decimals. It is irrational-it goes on forever without repeating. For math problems I usually leave it as $\sqrt 5$ – Ross Millikan Oct 6 '12 at 22:03
Thank you for your help. – JohnPhteven Oct 7 '12 at 7:49
@ZafarS: To get a distance of $\sqrt 5$ along this line, you go 2 units in the $y$ direction and $-1$ in $x$ (or the other way). The length is then $\sqrt {2^2+1^2}=\sqrt 5$. As the center is $(3,2)$, this leads to the other corners being $(4,0)$ and $(2,4)$ – Ross Millikan Oct 7 '12 at 14:37
Wow. how could I not have thought about that! Thank you a TON! I assume they want $(2,4)$ as an answer since that's entirely in the first quadrant and I'm not too sure about $(4,0)$ – JohnPhteven Oct 7 '12 at 14:51

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.

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