, 2022

The line with equation $y = 2x -
6$has$y$-intercept$-6$.

Also, the $x$-intercept of $y = 2x - 6$ occurs when $y = 0$, which gives $0 = 2x - 6$ or $2x = 6$ which gives $x = 3$.

Therefore, the triangle bounded by the $x$-axis, the $y$-axis, and the line with equation $y = 2x - 6$ has base of length 3 and
height of length 6, and so has area $\frac{1}{2} \times 3 \times 6 = 9$.

We want the area of the triangle bounded by the $x$-axis, the vertical line with equation
$x=d$, and the line with equation
$y = 2x - 6$ to be 4 times this area,
or 36.

This means that $x=d$ is to the right
of the point $(3,0)$, because the new
area is larger. In other words, $d>3$.

[[IMAGE0]]

The base of this triangle has length $d-3$, and its height is $2d-6$, since the height is measured along
the vertical line with equation $x =
d$.

Thus, we want $\$\frac{1}{2}(d-3)(2d-6)$ =
36$or$(d-3)(d-3) = 36$
which means $(d-3)^2 = 36$.

Since $d-3 > 0$, then $d-3 = 6$ which gives $d = 9$.

Alternatively, we could note that if similar triangles have areas in the
ratio $4:1$ then their corresponding
lengths are in the ratio $\sqrt{4}:1$
or $2:1$.

Since the two triangles in question are similar (both are right-angled
and they have equal angles at the point $(3,0)$), the larger triangle has base of
length $2 \times 3 = 6$ and so $d = 3 + 6 = 9$.