3. Let the point $T(a+\sqrt{2}, \sqrt{2}-a)$ be equidistant from the points $A(-\sqrt{2}, \sqrt{2})$ and $B(\sqrt{2},-\sqrt{2})$.

a) Determine the value of the parameter $a$.

b) Calculate the area of triangle $A B T$.
3. If point $T$ is equally distant from points $A$ and $B$, then it must hold that $\sqrt{(a+\sqrt{2}+\sqrt{2})^{2}+(\sqrt{2}-a-\sqrt{2})^{2}}=$ $\sqrt{(a+\sqrt{2}-\sqrt{2})^{2}+(\sqrt{2}-a+\sqrt{2})^{2}}$, from which we get $a^{2}+4 a \sqrt{2}+8+a^{2}=a^{2}+8-4 a \sqrt{2}+a^{2}$ or $8 a \sqrt{2}=0$ and finally $a=0$.

Points $A, B$ and $T$ are the vertices of an isosceles right triangle with legs of length $2 \sqrt{2}$. The area of this triangle is $\frac{2 \sqrt{2} \cdot 2 \sqrt{2}}{2}=4$.

![](https://cdn.mathpix.com/cropped/2024_06_07_06886e410b39c0216094g-08.jpg?height=491&width=491&top_left_y=220&top_left_x=1342)

![](https://cdn.mathpix.com/cropped/2024_06_07_06886e410b39c0216094g-08.jpg?height=63&width=1617&top_left_y=774&top_left_x=228)

Writing $\sqrt{(a+\sqrt{2}+\sqrt{2})^{2}+(\sqrt{2}-a-\sqrt{2})^{2}}=$

$\sqrt{(a+\sqrt{2}-\sqrt{2})^{2}+(\sqrt{2}-a+\sqrt{2})^{2}}$

.1 point

Simplifying the equation to the form $8 a \sqrt{2}=0$ .1 point

Calculating $a=0$

.1 point

(b) Writing the formula for calculating the area of the triangle .1 point Calculated area of the triangle: 4 .1 point