Question

Find the coordinates of the incentre and centroid of the triangle whose sides have the equations 3x− 4y = 0, 12y + 5x = 0 and y − 15 = 0.

Answer 1

The given lines are as follows:
3x − 4y = 0                     ... (1)
12y + 5x = 0                   ... (2)
y − 15 = 0                       ... (3)
In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = 0, y = 0
Thus, AB and BC intersect at B (0, 0).
Solving (1) and (3):
x = 20 , y = 15
Thus, AB and CA intersect at A (20, 15).
Solving (2) and (3):
x = −36 , y = 15
Thus, BC and CA intersect at C (−36, 15).
Let us find the lengths of sides AB, BC and CA.

\[AB = \sqrt{\left( 20 - 0 \right)^2 + \left( 15 - 0 \right)^2} = 25\]

\[BC = \sqrt{\left( 0 + 36 \right)^2 + \left( 0 - 15 \right)^2} = 39\]

\[AC = \sqrt{\left( 20 + 36 \right)^2 + \left( 15 - 15 \right)^2} = 56\]

Here, a = BC = 39, b = CA = 56 and c = AB = 25
Also,

\[\left( x_1 , y_1 \right)\] = A (20, 15), \[\left( x_2 , y_2 \right)\] = B (0, 0) and \[\left( x_3 , y_3 \right)\] = C (−36, 15)

\[\therefore\text{ Centroid }= \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]

       \[ = \left( \frac{20 + 0 - 36}{3}, \frac{15 + 0 + 15}{3} \right) = \left( \frac{- 16}{3}, 10 \right)\]

\[\text { And, incentre } = \left( \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c} \right)\]

\[ = \left( \frac{39 \times 20 + 56 \times 0 - 25 \times 36}{39 + 56 + 25}, \frac{39 \times 15 + 56 \times 0 + 25 \times 15}{39 + 56 + 25} \right)\]

\[ = \left( \frac{- 120}{120}, \frac{120 \times 8}{120} \right) = \left( - 1, 8 \right)\]