8. (a) Find an equation of the line joining A(7, 4) and B(2, 0), giving your answer in the form ax+by+c=0, where a, b and c are integers - Edexcel - A-Level Maths Pure - Question 10 - 2010 - Paper 1

The length of the line segment AB can be calculated using the distance formula:

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

Substituting the coordinates of A(7, 4) and B(2, 0):

$AB = \sqrt{(2 - 7)^2 + (0 - 4)^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}$

Thus, the length of AB in surd form is:

$AB = \sqrt{41}$

Given that the length AC is equal to AB, we have:

$AC = \sqrt{(2 - 7)^2 + (t - 4)^2} = \sqrt{41}$

This leads to:

$(2 - 7)^2 + (t - 4)^2 = 41$

Calculating the left side:

$(-5)^2 + (t - 4)^2 = 25 + (t - 4)^2 = 41$

Thus:

$(t - 4)^2 = 16$

Taking the square root,

$t - 4 = 4 \quad \text{or} \quad t - 4 = -4$

Solving these gives:

1. $t = 8$
2. $t = 0$ (not acceptable as t > 0)

Therefore, the value of t is:

$t = 8$

The area of triangle ABC can be calculated using the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Here, we can consider AB as the base, and AC as the height.

The base AB is already calculated as:

$AB = \sqrt{41}$

For the height, since point C (2, 8) is directly above the x-coordinate of A and B, the height is the vertical distance from point C to the line AB.

Using the coordinates of points A(7, 4), B(2, 0), and C(2, 8), the area is:

$\text{Area} = \frac{1}{2} \times (7 - 2) \times (8 - 0) = \frac{1}{2} \times 5 \times 8 = 20$

So the area of triangle ABC is:

$\text{Area} = 20$