Let $t$ be TNYWR.

What is the area of the triangle enclosed by the $x$-axis, the $y$-axis, and the line with equation $y=t x+6$ ?

## The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2022 Canadian Team Mathematics Contest 
The vertices of the triangle are the origin and the points at which the line intersects the $x$ - and $y$-axes.

When $y=0$, we get $0=t x+6$ so $x=\frac{-6}{t}$.

When $x=0$, we get $y=t(0)+6$ so $y=6$.

The triangle is right-angled at the origin, its height is 6 , and its base the distance from $\frac{-6}{t}$ to the origin.

If $t$ is negative, then the base is $\frac{-6}{t}$, and if $t$ is positive, then the base is $\frac{6}{t}$.

Therefore, if $t$ is negative, then the area of the triangle is

$$
\frac{1}{2} \times 6 \times \frac{-6}{t}=\frac{-18}{t}
$$

and if $t$ is positive, then the area of the triangle is

$$
\frac{1}{2} \times 6 \times \frac{6}{t}=\frac{18}{t}
$$

Since $t=-4$ which is negative, the area of the triangle is $\frac{-18}{-4}=\frac{9}{2}$.

ANswer: $\left(7,-4, \frac{9}{2}\right)$