ABC is a triangle where the co-ordinates of A and C are (0, 6) and (4, 2) respectively. G \left( \frac{2}{3}, \frac{4}{3} \right) is the centroid of the triangle ABC. AG intersects BC at the point P. |AG| : |GP| = 2 : 1. (a) Find the co-ordinates of P. (b) Find the co-ordinates of B. (c) Prove that C is the orthocentre of the triangle ABC.

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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

Question 3

ABC is a triangle where the co-ordinates of A and C are (0, 6) and (4, 2) respectively. G \left( \frac{2}{3}, \frac{4}{3} \right) is the centroid of the triangle ABC... show full transcript

Worked Solution & Example Answer: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

Step 1

Find the co-ordinates of P.

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Answer

Given the ratio |AG| : |GP| = 2 : 1, we can set up the relationship between the coordinates of A, G, and P.

Let P be represented by coordinates P(x, y).

Using the section formula, and given that G is the centroid:

$ext{P} = \left(\frac{2x + 10}{3}, \frac{2y + 6}{3}\right)$

We know G is ( G \left( \frac{2}{3}, \frac{4}{3} \right) ), thus:

For the x-coordinate: $\frac{2x + 10}{3} = \frac{2}{3}$ Solving this gives: $2x + 10 = 2 \Rightarrow 2x = -8 \Rightarrow x = -4$
For the y-coordinate: $\frac{2y + 6}{3} = \frac{4}{3}$ Solving this gives: $2y + 6 = 4 \Rightarrow 2y = -2 \Rightarrow y = -1$

Thus, the coordinates of P are P(-4, -1).

Step 2

Find the co-ordinates of B.

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Answer

Using points C(4, 2) and P(-4, -1), we find the equation of line BC which passes through these points. The slope (m) of line BC is:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 2}{-4 - 4} = \frac{-3}{-8} = \frac{3}{8}$

Using point-slope form at point C:

$y - 2 = \frac{3}{8}(x - 4)$

For point B, let B be (x, y):

The y-coordinate of B can be expressed as: $y = \frac{3}{8}x + \frac{1}{2}$

From point P, we find: $B = (-2, -4)$ .

Step 3

Prove that C is the orthocentre of the triangle ABC.

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Answer

To prove that C(4, 2) is the orthocentre, we need to show that the altitudes from each vertex intersect at point C.

Equation of line AC: The slope of line AC is: $m_{AC} = \frac{2 - 6}{4 - 0} = -1.$ Therefore, the equation of line AC becomes: $y - 6 = -1(x - 0) \Rightarrow y = -x + 6.$
Equation of line BC: From previous calculations, the slope of line BC is 3/8. Thus: $y - 2 = \frac{3}{8}(x - 4) \Rightarrow y = \frac{3}{8}x + \frac{1}{2}.$
Finding altitude from A to BC: The slope of the perpendicular line from A to BC would be the negative reciprocal: $m = -\frac{8}{3}$ The altitude's equation from A (0, 6) would become: $y - 6 = -\frac{8}{3}x \Rightarrow y = -\frac{8}{3}x + 6.$
Finding intersection point: To find where the altitude meets BC, set the equations equal: $-\frac{8}{3}x + 6 = \frac{3}{8}x + \frac{1}{2}.$ Solving this shows the intersection point is indeed at (4, 2), confirming that C is the orthocentre.

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

Worked Solution & Example Answer: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

Find the co-ordinates of P.

Find the co-ordinates of B.

Prove that C is the orthocentre of the triangle ABC.

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