The line $l_1$ passes through the point $A(2, 5)$ and has gradient $\frac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 1

To verify that the point $B(-2, 7)$ lies on the line $l_1$, we substitute $x = -2$ into the equation of the line:

$y = \frac{1}{2}(-2) + 4 = -1 + 4 = 3$.

However, the $y$-coordinate of B is 7. Since $7 \neq 3$, it appears that $B$ does not lie on the line $l_1$.

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

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$,

where $A(2, 5)$ and $B(-2, 7)$:

$AB = \sqrt{((-2) - 2)^2 + (7 - 5)^2} = \sqrt{(-4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.

Thus, $k = 2$, and the length of $AB$ is $2\sqrt{5}$.

We know that the length $AC = 5$, and that point $C(p, \frac{1}{2}p + 4)$ lies on the line $l_1$.

Using the distance formula to find $AC$:

$AC = \sqrt{(p - 2)^2 + \left(\left(\frac{1}{2}p + 4\right) - 5\right)^2}$.

Setting $AC = 5$:

$\sqrt{(p - 2)^2 + \left(\frac{1}{2}p - 1\right)^2} = 5$.

By squaring both sides and simplifying:

$(p - 2)^2 + \left(\frac{1}{2}p - 1\right)^2 = 25$.

Expanding gives:

$(p^2 - 4p + 4) + \left(\frac{1}{4}p^2 - p + 1\right) = 25$

Combining terms:

$\frac{5}{4}p^2 - 5p + 5 = 25 \Rightarrow 5p^2 - 20p + 20 = 100 \Rightarrow p^2 - 4p - 16 = 0$.

Thus, $p$ satisfies $p^2 - 4p - 16 = 0$.