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 8 - 2010 - Paper 1

To calculate the length of segment AB, we use the distance formula:

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

Substituting points A(7, 4) and B(2, 0):

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

This simplifies to:

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

Since AC = AB and point C has coordinates (2, t), we can apply the distance formula again:

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

Setting AC equal to AB:

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

Squaring both sides gives:

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

This simplifies to $25 + (t - 4)^2 = 41$. Then, $(t - 4)^2 = 16$

Taking the square root:

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

Thus, $t = 8 ext{ or } t = 0$. Given that t > 0, we conclude: t = 8.

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

$\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$

Substituting the coordinates A(7, 4), B(2, 0), and C(2, 8):

$\text{Area} = \frac{1}{2} \left| 7(0-8) + 2(8-4) + 2(4-0) \right|$

Calculating:

$= \frac{1}{2} \left| 7(-8) + 2(4) + 2(4) \right|$ $= \frac{1}{2} \left| -56 + 8 + 8 \right|$ $= \frac{1}{2} \left| -40 \right|$ $= \frac{40}{2} = 20.$