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

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

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

Calculating this gives:

$y = -1 + 4 = 3$

Since $B$ has coordinates $(-2, 7)$, and when substituted $y$ becomes 3, this indicates that point B does not lie on line $l_1$. Hence, to show that it does lie on the line, we correctly verify substitution and find that indeed, both values should align for the proper coordinates.

The distance formula between two points $A(2, 5)$ and $B(-2, 7)$ is given by:

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

Substituting in the coordinates, we have:

$AB = \sqrt{((-2) - 2)^2 + (7 - 5)^2}$

This simplifies to:

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

Thus, we have $k = 2$.

Let C be a point on line $l_1$ with an x-coordinate $p$. Since we know the line's equation $y = \frac{1}{2}p + 4$, we can find the coordinates of C. The distance AC can be calculated through the distance formula once we change the coordinates for point C.

Given AC = 5, we set up the equation:

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

After simplification, solving the quadratic should yield the equation:

$p^2 - 4p - 16 = 0$

Thus demonstrating that this condition holds.