Figure 2 shows the line with equation $y = 10 - x$ and the curve with equation $y = 10x - x^2 - 8$ - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 3

To find the coordinates of A and B, we need to set the equations of the line and the curve equal to each other:

$10 - x = 10x - x^2 - 8$

Rearranging gives us:

$x^2 - 11x + 18 = 0$

Now we can solve this quadratic equation using the quadratic formula, where $a = 1$, $b = -11$, and $c = 18$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} = \frac{11 \pm \sqrt{121 - 72}}{2} = \frac{11 \pm \sqrt{49}}{2} = \frac{11 \pm 7}{2}$

Thus, the solutions for x are:

$x_1 = \frac{18}{2} = 9$ $x_2 = \frac{4}{2} = 2$

Now substituting these x-values back into the line equation to find the corresponding y-values:

For $x = 9$:
$y = 10 - 9 = 1$
Thus, one coordinate is $A(9, 1)$.

For $x = 2$:
$y = 10 - 2 = 8$
Thus, the other coordinate is $B(2, 8)$.