The line l_1 passes through the point A (2, 5) and has gradient \frac{-1}{2}. The point C lies on l_1 and has x-coordinate equal to p. The length of AC is 5 units. (i) Show that p satisfies p^2 - 4p - 16

Step 1: find the line equation or y-coordinate
Skill 1: Substitute either values slope(m) and any one point into the
equation of a straight line.
FORMULA: Equation of a straight line
y - y_1 = m(x - x_1). Where m = slope,
(x_1, y_1) = any point
EXAMPLE: I took (2, 5) as a point and m=\frac{-1}{2}
y-5=\frac{-1}{2}\left(x-2\right)
y = \frac{-x}{2} + 6
Step 2: Locate the coordinates of the end points.
EXAMPLE: The two endpoints M (2, 5) and N (x = p , y = \frac{-x}{2} + 6 ).
Therefore, (x_1, y_1) = (2, 5) and (x_2, y_2) = (x = p , y = \frac{-x}{2} + 6 ).
Step 3: Plug the corresponding coordinates into the Distance Formula. .
FORMULA:
d = \sqrt{(x_2 - x_2)^2 + (y_2 -y_1)^2}
EXAMPLE: Endpoints M (2, 1) and N (6, 4).
d = \sqrt{(6 - 2)^2 + (4 - 1)^2}
Step 4: Substitute the d(given distance) value in the equation got in step 3 and simplify the equation
NOTE: Simplification steps: Calculate the subtraction in parentheses.
Square the value in parentheses.
Add the numbers under the radical sign.
Step 5: Compare the salved equation with the given equation and verify they are satisfying or not