The line $l_1$ passes through the point $A(2, 5)$ and has gradient $rac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 11 - 2009 - Paper 1

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

y = rac{1}{2}(-2) + 4
y = -1 + 4 = 3\

Since the expected $y$ coordinate is $7$ and not $3$, we conclude that the point $B$ does not lie on $l_1$.

To find the length of segment $AB$, we can use the distance formula: $AB = ext{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting in the coordinates of points $A(2, 5)$ and $B(-2, 7)$: egin{align*} AB &= \sqrt{((-2) - (2))^2 + (7 - 5)^2} \ &= \sqrt{(-4)^2 + (2)^2} \ &= \sqrt{16 + 4} \ &= \sqrt{20} \ &= 2\sqrt{5} = k\frac{1}{5} \ ext{where } k = 2.
Thus, $k = 2$.

Since the point $C$ has x-coordinate equal to $p$, we need to find the $y$-coordinate using the line $l_1$:

y = rac{1}{2}p + 4.

Using the distance formula again, we know the length of $AC$ is 5 units: AC^2 = (p - 2)^2 + igg(rac{1}{2}p + 4 - 5igg)^2 This simplifies to: AC^2 = (p - 2)^2 + igg(rac{1}{2}p - 1igg)^2 Substituting for AC length yields: 25 = (p - 2)^2 + igg(rac{1}{2}p - 1igg)^2 Expanding both squares and combining like terms will yield the equation in the form of $p^2 - 4p - 16 = 0$.