Question

Find the incentre of the triangle with vertices $$\left(7,1\right),\left(-1,5\right)$$and $$\left(3+2\sqrt{3},3+4\sqrt{3}\right)$$
1) $$\left(3+{\displaystyle \frac{2}{\sqrt{3}}},3+{\displaystyle \frac{4}{\sqrt{3}}}\right)$$
2) $$\left(1+{\displaystyle \frac{2}{3\sqrt{3}}},1+{\displaystyle \frac{4}{3\sqrt{3}}}\right)$$
3) (7, 1)
4) None of these

Answer 1

Hint: Using the vertices first find out the length of the sides of the triangle and proceed. Then we have to determine the type of the triangle by observing the sides of the triangle. Now using the incentre formula for that particular type of triangle the coordinates are found.

Complete step-by-step answer:
Lest us consider the three vertices to be $$A\left(7,1\right),B\left(-1,5\right)$$, $$C\left(3+2\sqrt{3},3+4\sqrt{3}\right)$$ which forms the triangle $$\mathrm{\Delta }ABC$$ .
So, to find the length of the sides of the triangle distance formula is used and solve it
$$d=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}$$
So, the length of the side AB we get,
$$\Rightarrow AB=\sqrt{{\left(\left(-1\right)-7\right)}^2-{\left(5-1\right)}^2}$$
$$\Rightarrow AB=\sqrt{8^2+4^2}$$
$$\Rightarrow AB=\sqrt{80}$$
$$\Rightarrow AB=4\sqrt{5}$$
Now, the length of the side BC we get,
$$\Rightarrow BC=\sqrt{{\left(\left(3+2\sqrt{3}\right)-\left(-1\right)\right)}^2-{\left(\left(3+4\sqrt{3}\right)-5\right)}^2}$$
$$\Rightarrow Bc=4\sqrt{5}$$
Now, the length of the side CA we get,
$$\Rightarrow CA=\sqrt{{\left(\left(3+2\sqrt{3}\right)-7\right)}^2-{\left(\left(3+4\sqrt{3}\right)-5\right)}^2}$$
$$\Rightarrow CA=4\sqrt{5}$$

Since AB = BC = CA = $$4\sqrt{5}$$
Hence, we can say that it is an equilateral triangle. Since $$\mathrm{\Delta }ABC$$ is an equilateral triangle. The incentre is nothing but equal to the centroid of $$\mathrm{\Delta }ABC$$ .
So, the coordinates of the incentre = coordinates of centroid
Now, coordinates of centroid can be written using the formula
$$\Rightarrow \left({\displaystyle \frac{x_1+x_2+x_3}{3}},{\displaystyle \frac{y_1+y_2+y_3}{3}}\right)$$
$$\Rightarrow \left({\displaystyle \frac{7+\left(-1\right)+\left(3+2\sqrt{3}\right)}{3}},{\displaystyle \frac{1+5+\left(3+4\sqrt{3}\right)}{3}}\right)$$
$$\Rightarrow \left({\displaystyle \frac{9+2\sqrt{3}}{3}},{\displaystyle \frac{9+4\sqrt{3}}{3}}\right)$$
$$\Rightarrow \left(3+{\displaystyle \frac{2}{\sqrt{3}}},3+{\displaystyle \frac{4}{\sqrt{3}}}\right)$$
Thus, coordinates of incentre are: $$\Rightarrow \left(3+{\displaystyle \frac{2}{\sqrt{3}}},3+{\displaystyle \frac{4}{\sqrt{3}}}\right)$$
Hence option (1) is the correct answer to this question.
So, the correct answer is “Option 1”.

Note: When we have to solve this type question, first calculate the distance of the side of the triangle and apply the formula of incentre with respect to the triangle that formed. Alternate way to find the incentre is using the formula-
$$\left({\displaystyle \frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c}},{\displaystyle \frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}}\right)$$
Where,
Length of side AB is a
Length of side BC is b
Length of side CA is c