The points A(−9, 3), B(−4, 3) and C(4, 10) are the vertices of the triangle ABC, as shown - Leaving Cert Mathematics - Question a) - 2014

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

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

Using coordinates A(−9, 3), B(−4, 3), and C(4, 10):

• $x_1 = -9$, $y_1 = 3$
• $x_2 = -4$, $y_2 = 3$
• $x_3 = 4$, $y_3 = 10$

Substituting these values:

$ext{Area} = \frac{1}{2} \times |-9(3 - 10) + (-4)(10 - 3) + 4(3 - 3)|$ $= \frac{1}{2} \times |-9(-7) - 4(7) + 4(0)|$ $= \frac{1}{2} \times |63 - 28|$ $= \frac{1}{2} \times 35$ $= 17.5$

Thus, the area of the triangle ABC is 17.5.

To draw triangle XYZ congruent to triangle ABC, we can use the same lengths for sides AB, BC, and CA. The vertices will correspond such that:

• $|AB| = |XY|$
• $|BC| = |YZ|$
• $|CA| = |XZ|$

Place point X at coordinate (-5, 1) and use the calculated dimensions to plot the remaining points Y and Z appropriately.

The coordinates of point Z can be determined to lie at $ext{Z}(-5, -4)$ or any point that maintains the congruence conditions.

Reason: The triangles XYZ and ABC are congruent by the Side-Side-Side (SSS) criterion since all corresponding sides are equal in length.