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.

$$\begin{gathered} AB=\sqrt{{\left(20-0\right)}^2+{\left(15-0\right)}^2}=25 \\ BC=\sqrt{{\left(0+36\right)}^2+{\left(0-15\right)}^2}=39 \\ AC=\sqrt{{\left(20+36\right)}^2+{\left(15-15\right)}^2}=56 \end{gathered}$$

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)

$$\begin{gathered} \therefore \mathrm{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right) \\ =\left(\frac{20+0-36}{3},\frac{15+0+15}{3}\right)=\left(\frac{-16}{3},10\right) \end{gathered}$$

$$\begin{gathered} \mathrm{And},\mathrm{incentre}=\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right) \\ =\left(\frac{39\times 20+56\times 0-25\times 36}{39+56+25},\frac{39\times 15+56\times 0+25\times 15}{39+56+25}\right) \\ =\left(\frac{-120}{120},\frac{120\times 8}{120}\right)=\left(-1,8\right) \end{gathered}$$