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

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

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

Using the coordinates 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 is \sqrt{41}.

Since AC = AB, we know:

$AC = \sqrt{(2 - 7)^2 + (t - 4)^2} = \sqrt{(-5)^2 + (t - 4)^2}$

Setting this equal to the length of AB:

$\sqrt{41} = \sqrt{25 + (t - 4)^2}$

Squaring both sides gives:

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

Solving for (t - 4)^2:

$(t - 4)^2 = 16$

Taking the square root:

$t - 4 = 4 ext{ or } t - 4 = -4$

Thus:

$t = 8 ext{ or } t = 0$

Since t > 0, we have t = 8.

To find the area of triangle ABC, we can use the formula:

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

The base can be taken as AB, with length \sqrt{41}, and the height is the y-coordinate of point C, which is t = 8. Thus the area is:

$\text{Area} = \frac{1}{2} \times \sqrt{41} \times 8 = 4\sqrt{41}$

Therefore, the area of triangle ABC = 4\sqrt{41}.