In the diagram, $A(2,2)$ and $C(8,4)$ are two of the vertices of an isosceles right-angled triangle $A B C$. If the vertex $B$ is located on the $x$-axis and $\angle A B C=90^{\circ}$, the $x$-coordinate of $B$ is
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-100.jpg?height=344&width=372&top_left_y=229&top_left_x=1294)
Solution 1

Since $A B=B C$, then $B$ lies on the perpendicular bisector of $A C$.

Since $A$ has coordinates $(2,2)$ and $C$ has coordinates $(8,4)$, then the midpoint of $A C$ is $\left(\frac{1}{2}(2+8), \frac{1}{2}(4+2)\right)=(5,3)$ and the slope of $A C$ is $\frac{4-2}{8-2}=\frac{1}{3}$.

Therefore, the slope of the perpendicular bisector is -3 (the negative reciprocal of $\frac{1}{3}$ ) and it passes through $(5,3)$, so has equation $y-3=-3(x-5)$ or $y=-3 x+18$.

The $x$-intercept of this line comes when $y$ is set to 0 ; here, we obtain $x=6$.

Therefore, since $B$ is the point where the perpendicular bisector of $A C$ crosses the $x$-axis, then the $x$-coordinate of $B$ is 6 .

(We can check that indeed if $B$ has coordinates $(6,0)$, then $A B$ and $B C$ are perpendicular.)

## Solution 2

Since $\triangle A B C$ is an isosceles right-angled triangle, then $\angle A B C=90^{\circ}$, and so $A B$ is perpendicular to $B C$.

Suppose $B$ has coordinates $(b, 0)$.

The slope of $A B$ is $\frac{2-0}{2-b}$ and the slope of $B C$ is $\frac{4-0}{8-b}$.

Since $A B$ and $B C$ are perpendicular, their slopes are negative reciprocals, so

$$
\begin{aligned}
\frac{2}{2-b} & =-\frac{8-b}{4} \\
-8 & =(2-b)(8-b) \\
-8 & =b^{2}-10 b+16 \\
b^{2}-10 b+24 & =0 \\
(b-4)(b-6) & =0
\end{aligned}
$$

and so $b=4$ or $b=6$.

We must determine which value of $b$ gives $A B=B C$ (since we have already used the perpendicularity).

If $b=4$, then $A B=\sqrt{(4-2)^{2}+(0-2)^{2}}=\sqrt{8}$ and $B C=\sqrt{(8-4)^{2}+(4-0)^{2}}=\sqrt{32}$ and so $A B \neq B C$.

Therefore, the $x$-coordinate of $B$ must be 6 .

(We can check that, in this case, $A B$ does equal $B C$.)

## Solution 3

To go from $A$ to $C$, we go 6 units right and 2 units up.

Suppose that to go from $A$ to $B$, we go $p$ units right and $q$ units down, were $p, q>0$.

Since $B C$ is equal and perpendicular to $A B$, then to go from $B$ to $C$, we must go $q$ units right and $p$ units up.

(We can see this by looking at the slopes of segments $A B$ and $B C$.)

Therefore, to get from $A$ to $C$ through $B$, we go $p+q$ units right and $q-p$ units up, so $p+q=6$ and $q-p=2$, as the result is the same as from going directly to $C$ from $A$.

Since $p+q=6$ and $q-p=2$, then $2 q=8$ (adding the equations), so $q=4$, and so $p=2$.

Since $A$ has coordinates $(2,2)$, then $B$ has coordinates $(6,0)$ which lies on the $x$-axis as required.

ANSWER: (D)