The graphs of $g(x)=2x+6$ and $g^{-1}$, the inverse of $g$, are shown in the diagram below - NSC Mathematics - Question 5 - 2022 - Paper 1

To find the inverse, we swap x and y in the equation $g(x)=y$:

$y = 2x + 6$

Swapping gives:

$x = 2y + 6$

Now, we solve for y:

$x - 6 = 2y$ $y = \frac{x - 6}{2}$

Thus, the inverse function is:

$g^{-1}(x) = \frac{x - 6}{2}$.

Point A is where the graphs of $g$ and $g^{-1}$ intersect. To find this point, we set the equations equal:

$2x + 6 = \frac{x - 6}{2}$.

Multiplying through by 2 to eliminate the fraction:

$4x + 12 = x - 6$

Rearranging gives:

$4x - x = -6 - 12$ $3x = -18$ $x = -6$

Substituting back to find y:

$y = 2(-6) + 6 = -6$

Thus, the coordinates of A are (-6, -6).

To find the length of segment AB, we can use the distance formula:

$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$

Substituting the coordinates A (-6, -6) and B (0, 6):

$AB = \sqrt{(0 - (-6))^2 + (6 - (-6))^2}$ $= \sqrt{(6)^2 + (12)^2}$ $= \sqrt{36 + 144}$ $= \sqrt{180}\approx 13.42$.

To find the area of triangle ABC, we can use the formula:

$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$.

The base BC can be computed as:

$BC = \sqrt{(6)^2 + (6)^2} = \sqrt{72} = 6\sqrt{2}$.

The height h is the perpendicular from A to line BC, calculated using the properties of the triangle. Substituting these values gives:

$\text{Area} = \frac{1}{2} \cdot BC \cdot h = \frac{1}{2} \cdot (6\sqrt{2})\cdot h$. The height can be determined geometrically or calculated based on coordinates:

After calculating, the area of $\triangle ABC$ is found to be approximately 54 square units.