Alexandra draws a letter A which stands on the $x$-axis.

(a) The left side of the letter A lies along the line with equation $y=3 x+6$. What is the $x$-intercept of the line with equation $y=3 x+6$ ?

(b) The right side of the letter A lies along the line $L_{2}$ and the letter is symmetric about the $y$-axis. What is the equation of line $L_{2}$ ?

(c) Determine the area of the triangle formed by the $x$-axis and the left and right sides of the letter A.

![](https://cdn.mathpix.com/cropped/2024_04_17_d3d67b8c1418cbc5debag-3.jpg?height=417&width=442&top_left_y=819&top_left_x=1297)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_17_d3d67b8c1418cbc5debag-3.jpg?height=415&width=420&top_left_y=1316&top_left_x=1297)

Figure 2
(a) To find the $x$-intercept of the line with equation $y=3 x+6$, we set $y=0$ to obtain $3 x+6=0$ or $3 x=-6$ and so $x=-2$.

Therefore, the $x$-intercept is -2 .

(b) Solution 1

To find the equation of the line that is symmetric across the $y$-axis with the line with the equation $y=3 x+6$, we reverse the sign of the slope and keep the same $y$-intercept to obtain $y=-3 x+6$.

Solution 2

The line with equation $y=3 x+6$ has $x$-intercept -2 (from (a)) and $y$-intercept 6 (from the equation of the line).

Since the letter A is symmetric about the $y$-axis, the right side of the letter $A$ will lie along the line with $x$-intercept 2 (that is, $-(-2)$ ) and $y$-intercept 6 .

The equation of this line is thus of the form $y=m x+6$ for some $m$.

Since it passes through the point $(2,0)$, then $0=2 m+6$ and so $2 m=-6$ or $m=-3$.

Thus, the equation of this line is $y=-3 x+6$.

(c) Since the $x$-intercepts of the lines that form the sides are -2 and 2 , then the base of the triangle has length $2-(-2)=4$.

Since both lines pass through $(0,6)$, then the height of the triangle is 6 .

Therefore, the area of the triangle is $\frac{1}{2} \cdot 4 \cdot 6=12$.

(d) Solution 1

We note first that $c0$ and so $6-c=4$ which gives $c=2$.

## Solution 2

Since the base of the shaded triangle is parallel to the base of the larger triangle, then their base angles are equal and so the triangles are similar.

Since the area of the shaded triangle is $\frac{4}{9}$ of the area of the larger triangle, then the side lengths of the shaded triangle are $\sqrt{\frac{4}{9}}=\frac{2}{3}$ of the corresponding side lengths of the larger triangle.

Furthermore, the height of the shaded triangle is $\frac{2}{3}$ that of the larger triangle.

Since the height of the larger triangle is 6 , then the height of the shaded triangle is $\frac{2}{3} \cdot 6$ which equals 4.

Since the top vertex of the shaded triangle lies at $y=6$ and the base of the triangle lies along $y=c$, then the height of the shaded triangle is $6-c$.

Thus, $6-c=4$ and so $c=2$.