Coordinate System and Coordinates
11.0 Incentre of a triangle
11.0 Incentre of a triangle
The point of intersection of internal angle bisectors of a triangle is called the incenter of the triangle.
If $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$ are the vertices of a $\Delta ABC$, whose sides $BC$, $CA$ and $AB$ are of lengths $a$, $b$ and $c$ respectively, then co-ordinates of incentre of $\Delta ABC$ is $$I \equiv (\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}})$$
Note:
- If $\Delta ABC$ is equilateral triangle, then $a=b=c$. Therefore, $$I \equiv (\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}) \equiv (\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}) \equiv G$$ i.e., incentre and centroid coincide in equilateral triangle.
- Incentre divides the angle bisectors in the ratio $b + c:a$, $c + a:b$ and $a + b:c$.
Question 3. Find the co-ordinates of $(i)$ centroid $(ii)$ in-centre of the triangle whose vertices are $(0,6)$, $(8,12)$ and $(8,0)$.
Solution: $(i)$ As we know that the co-ordinates of centroid of a triangle whose angular points $({x_1},{y_1})$, $({x_2},{y_2})$ and $({x_3},{y_3})$ is $$G \equiv (\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3})$$ Therefore, the centroid is $$\begin{equation} \begin{aligned} (\frac{{0 + 8 + 8}}{3},\frac{{6 + 12 + 0}}{3}) \\ (\frac{{16}}{3},6) \\\end{aligned} \end{equation} $$
$(ii)$ Let $A(0,6)$, $B(8,12)$ and $C(8,0)$ are the vertices of $\Delta ABC$.
Then, $$\begin{equation} \begin{aligned} c = AB = \sqrt {{{(0 - 8)}^2} + {{(6 - 12)}^2}} = 10 \\ b = CA = \sqrt {{{(0 - 8)}^2} + {{(6 - 0)}^2}} = 10 \\ a = BC = \sqrt {{{(8 - 8)}^2} + {{(12 - 0)}^2}} = 12 \\\end{aligned} \end{equation} $$
The co-ordinates of in-centre are $$I \equiv (\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}})$$$$\begin{equation} \begin{aligned} I \equiv (\frac{{(12 \times 0) + (10 \times 8) + (10 \times 8)}}{{12 + 10 + 10}},\frac{{(12 \times 6) + (10 \times 12) + (10 \times 0)}}{{12 + 10 + 10}}) \\ I \equiv (\frac{{160}}{{32}},\frac{{192}}{{32}}) \\ I \equiv (5,6) \\\end{aligned} \end{equation} $$
