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: $\left(7,-4, \frac{9}{2}\right)$
Let $t$ be TNYWR.

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

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