## Task A-1.6.

In a coordinate system in the plane, two lines intersect at a right angle at the point $A(6,8)$. The intersections $P$ and $Q$ of these lines with the $y$-axis are symmetric with respect to the origin. Determine the area of the triangle $A P Q$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ce71b02a7f068198719dg-06.jpg?height=653&width=597&top_left_y=276&top_left_x=655)

Points $P$ and $Q$ are on the $y$-axis and are symmetric with respect to the origin, which means their distances from the origin $O$ are equal, i.e., $O$ is the midpoint of segment $\overline{P Q}$.

Triangle $P A Q$ is a right triangle with hypotenuse $\overline{P Q}$ because $\varangle P A Q=90^{\circ}$. Since in any right triangle, the midpoint of the hypotenuse is also the center of the circumscribed circle, we conclude that $|A O|=|P O|=|Q O|$.

We can easily calculate $|A O|=\sqrt{6^{2}+8^{2}}=10$.

Therefore,

$$
|P Q|=|P O|+|O Q|=2|A O|=20
$$

The length of the altitude to the hypotenuse is equal to the $x$-coordinate of point $A$.

It follows that the area of triangle $A P Q$ is

$$
P(A P Q)=\frac{20 \cdot 6}{2}=60
$$