ABC is a triangle where the co-ordinates of A and C are (0, 6) and (4, 2) respectively - Leaving Cert Mathematics - Question 3 - 2017

To find the co-ordinates of B, we need to use the fact that G is located at the centroid:

$G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right)$

Given G(2/3, 4/3), and knowing A(0, 6) and C(4, 2), let B be (x, y). Thus we set up two equations:

1. For x-coordinates:
   $\frac{0 + x + 4}{3} = \frac{2}{3}$
   Therefore, $$$$

ightarrow x = -2 $$

2. For y-coordinates: $\frac{6 + y + 2}{3} = \frac{4}{3}$
   This gives us: $$$$

ightarrow y = -4 $$

Thus, the co-ordinates of B are $B(-2, -4)$.

To prove that C is the orthocentre of triangle ABC, we need to demonstrate that the altitudes of the triangle intersect at C.

1. Calculate slopes:
   • Slope AC: $m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{2 - 6}{4 - 0} = -1$
   • Slope BC: $m_{BC} = \frac{y_B - y_C}{x_B - x_C} = \frac{-4 - 2}{-2 - 4} = 1$

Since the product of the slopes (m_{AC} * m_{BC}) = -1, lines AC and BC are perpendicular. Therefore:

The altitude from C to AB must be perpendicular to AB. Thus:

• Use points A and B to find slope AB: $m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{-4 - 6}{-2 - 0} = -5$

Thus the line perpendicular will have slope = 1/5. Using point-slope form for altitude from C: $y - 2 = \frac{1}{5}(x - 4)$

2. Similarly, for the altitude from B: This should intersect at C, confirming C is the orthocentre.

Using both equations, find intersection point, confirming that altitudes meet at C. Therefore C is indeed the orthocentre.