Question

the vertices of a triangle are (0,0),(root3,3),(-root3,3) then the incenter is​

Answer 1

Answer:

see the attachment

hope it helps you

Answer 2

Answer:

The coordinates of the incenter are (0, 2).

Step-by-step explanation:

Given:

A = (0, 0), B = (√3, 3), C = (-√3, 3)

To find:

Incenter of the triangle (I)

Formula:

$I = (\frac{ax_{1}+bx_{2}+cx_{3} }{a+b+c}, \frac{ay_{1}+by_{2}+cy_{3} }{a+b+c})$

Step 1:

Consider the triangle ΔABC with vertices A(0, 0), B(√3, 3), and C(-√3, 3).

We know that the distance between two points is given as

$D = \sqrt{(x_{1}-x_{2} )^{2} + (y_{1}-y_{2} )^{2} }$

Using this formula to find the lengths of side AB, side BC, and side AC.

$AB = \sqrt{(\sqrt{3} -0 )^{2} + (3-0 )^{2} }$

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

$AB = \sqrt{3+9 }$

$AB = \sqrt{12 }$

$AB = 2\sqrt{3}$

AB = 2√3 units

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

$BC = \sqrt{(-2\sqrt{3} )^{2} +0 }$

$BC = \sqrt{4*3} }$

$BC = \sqrt{12 }$

$BC = 2\sqrt{3}$

BC =  2√3 units

$AC = \sqrt{(-\sqrt{3} -0)^{2} + (3-0 )^{2} }$

$AC = \sqrt{(-\sqrt{3} )^{2} + (3)^{2} }$

$AC = \sqrt{3 + 9 }$

$AC = \sqrt{12}$

$AC=2\sqrt{3}$

AC = 2√3 units

Step 2:

Since, AB = BC = AC, ΔABC is an equilateral triangle.

Therefore, the incenter is given as

$I = (\frac{ax_{1}+bx_{2}+cx_{3} }{a+b+c}, \frac{ay_{1}+by_{2}+cy_{3} }{a+b+c})$, where a = side BC, b = side AC, and c = side AB.

Substituting the values, we get

$I = (\frac{(2\sqrt{3})(0)+(2\sqrt{3})(\sqrt{3}) +(2\sqrt{3})(-\sqrt{3} )}{2\sqrt{3} +2\sqrt{3}+2\sqrt{3}}, \frac{(2\sqrt{3})(0)+(2\sqrt{3})(3)+(2\sqrt{3})(3) }{2\sqrt{3} +2\sqrt{3}+2\sqrt{3}})$

$I=(\frac{0+6-6}{6\sqrt{3} } , \frac{0+6\sqrt{3}+6\sqrt{3} }{6\sqrt{3\\} } )$

$I=( 0 , \frac{12\sqrt{3} }{6\sqrt{3} } )$

$I = (0,2)$

Therefore, the incenter of the given triangle is at (0, 2).

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