The co-ordinate diagram below shows the triangle ABC - Junior Cycle Mathematics - Question 4 - 2021

To find the values of m and k:

1. Line AC: From the equation y = mx + \frac{2}{3}, we can determine the slope m. Using the coordinates of A (4, 2) and C (10, 0), we find the slope:

   $m = \frac{y_C - y_A}{x_C - x_A} = \frac{0 - 2}{10 - 4} = -\frac{1}{3}$

Thus, substitute to get: m = -\frac{1}{3}.

2. Line AB: From the equation y = -\frac{1}{2}x + k, we can substitute point A's coordinates:

   $2 = -\frac{1}{2}(4) + k \Rightarrow 2 = -2 + k \Rightarrow k = 4.$

Thus, we have: m = -\frac{1}{3}, k = 4.

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

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

For triangle ABC:

• Base AC can be taken as the distance between A(4, 2) and C(10, 0), which is 6 units.
• The height is the y-coordinate of point B (8, 3), which is the distance to line AC, calculated as the perpendicular distance from B to line AC.

Using the area formula: $\text{Area} = \frac{1}{2} \times 6 \times 5 = 15 \text{ square units}$ This shows that the area of triangle ABC equals 10 square units.

To draw the triangle A'B'C', reflect points A, B, and C across the x-axis:

• A'(4, -2), B'(8, -3), C'(10, 0).

These new points will give you the triangle A'B'C'.

To find the intersection, we check each point:

• Point A: (4, 2) is above the x-axis, not in the intersection.
• Point B: (8, 3) is also above the x-axis, not in the intersection.
• Point C: (10, 0) lies on the x-axis, which is where both triangles overlap.

Thus, the answer is: C.

A point (p, s) that lies inside triangle ABC will also lie inside triangle A'B'C' if: p ∈ [4, 10] and s ∈ [-3, 2].

Thus, a valid point inside A'B'C' could be: (p, s) = (6, -1).