Find $|AB|$, the length of $[AB]$ - Leaving Cert Mathematics - Question 2 - 2020

To find the midpoint D of the segment $[AB]$, we use the midpoint formula:

$D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Substituting the coordinates of A(4, 6) and B(-2, 2):

$D = \left( \frac{4 + (-2)}{2}, \frac{6 + 2}{2} \right) = \left( \frac{2}{2}, \frac{8}{2} \right) = (1, 4)$

Therefore, the coordinates of D are (1, 4).

Since E(7, 3) is the midpoint of $[AC]$, we first find the coordinates of point C(10, 0) and point A(4, 6).

To show that $DE$ is parallel to $BC$, we need to compare their slopes.

The slope of segment $DE$ is given by:

$m_{DE} = \frac{y_E - y_D}{x_E - x_D} = \frac{3 - 4}{7 - 1} = \frac{-1}{6}$

Next, we find the slope of segment $BC$:

$m_{BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{0 - 2}{10 - (-2)} = \frac{-2}{12} = \frac{-1}{6}$

Since both slopes $m_{DE}$ and $m_{BC}$ are equal, $DE$ is parallel to $BC$.

To find the area of triangle $ABC$ using vertices A(4, 6), B(-2, 2), and C(10, 0), we 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 points:

$\text{Area} = \frac{1}{2} \left| 4(2 - 0) + (-2)(0 - 6) + 10(6 - 2) \right|$

Calculating:

• $4(2) = 8$
• $(-2)(-6) = 12$
• $10(4) = 40$

Thus,

$\text{Area} = \frac{1}{2} \left| 8 + 12 + 40 \right| = \frac{1}{2} \left| 60 \right| = 30 \text{ units}^2$

Therefore, the area of triangle $ABC$ is 30 square units.