Question

find the incentre of the triangle formed by the points (1,2) (3,4) (2,3).

Answer 1

Step-by-step explanation:

here is the ans

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Answer 2

The incenter of the triangle formed by the points (1, 2), (3, 4), and (2, 3) will be (2, 3).

Given,

Coordinates of 3 points of a triangle: (1, 2), (3, 4), and (2, 3).

To find,

Incenter of the trinagle.

Solution,

The incenter, or the coordinates of the incenter of a triangle ABC is determined by the formula given below.

$(x,y)=(\frac{ax_1+bx_2+cx_3}{a+b+c} ,\frac{ay_1+by_2+cy_3}{a+b+c}) \hfill ...(1)$

where,

a, b, and c are the sides of the triangle such that

BC = a,

AC = b, and

AB = c.

Let the given triangle be ABC such that coordinates are,

A (1, 2),

B (2, 3), and,

C (3, 4)

Firstly, we need to determine the side lengths of the triangle. These can be found using the distance formula, which is,

$d = \sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2} } \hfill ...(2)$

So,

$a=BC= \sqrt{(3-2)^{2} +(4-3)^{2} } = \sqrt{(1)^{2} +(1)^{2} } =\sqrt{2}$

$b=AC=\sqrt{(3-1)^{2} +(4-2)^{2} } = \sqrt{(2)^{2} +(2)^{2} } =\sqrt{8}=2\sqrt{2}$

$c=AB= \sqrt{(2-1)^{2} +(3-2)^{2} } = \sqrt{(1)^{2} +(1)^{2} } =\sqrt{2}$

Here, we can see that,

$\sqrt{2} +\sqrt{2} =2\sqrt{2}$

that is,

AB + BC = AC,

It means ABC is not a triangle, but the 3 points A, B, and C are collinear, such that a line AC is having B as the midpoint.

However, using (1), we can find the incenter, which will be the mid point of the line AB as follows.

$(x,y)=(\frac{\sqrt{2} (1)+2\sqrt{2} (2)+\sqrt{2} (3)}{\sqrt{2} +2\sqrt{2} +\sqrt{2} } ,\frac{\sqrt{2} (2)+2\sqrt{2}(3)+\sqrt{2} (4)}{\sqrt{2} +2\sqrt{2} +\sqrt{2} })$

$\implies (x,y)=(\frac{\sqrt{2}+4\sqrt{2} + 3\sqrt{2}}{\sqrt{2} +2\sqrt{2} +\sqrt{2} } ,\frac{2\sqrt{2}+6\sqrt{2}+4\sqrt{2}}{\sqrt{2} +2\sqrt{2} +\sqrt{2} })$

$\implies (x,y)=(\frac{8\sqrt{2}}{4\sqrt{2}} ,\frac{12\sqrt{2}}{4\sqrt{2}})$

$\implies (x,y)=(2,3).$

Therefore, the incenter of the triangle formed by the points (1, 2), (3, 4), and (2, 3) will be (2, 3).

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