Question

find the incentre of the triangle formed by the points A(7,9),B(3,-7),C(-3,3)​

Answer 1

$\tt\blue{\underline{Answer:}}$

$\green{\therefore I= \bigg(\frac{ 7\sqrt{136} + 3 \sqrt{136} - 3 \sqrt{272} }{ \sqrt{136 } + \sqrt{136} + \sqrt{262} } \: \: , \: \: \frac{ 9\sqrt{136} - 7\sqrt{136} + 3 \sqrt{272} }{ \sqrt{136 } + \sqrt{136} + \sqrt{272} } \bigg)}$

$\tt\orange{\underline{Step-by-step\:\:explanation:}}$

$\sf \green{\underline{Given :}} \\ : \tt\implies Coordinate \: \: a(7,9) \\ \\ : \tt\implies Coordinate \: \: b(3,- 7) \\ \\ : \tt\implies Coordinate \: \: c( - 3,3) \\ \\ \sf \red{\underline{To \: Find :}} \\ \tt: \implies Coordinate \: of \: incentre \: of \: triangle = ?$

• According to given question :

$\tt for \: Incentre \: we \: have \: to \: first \: find \: side \: of \: triangle \\ \\ \sf \orange{Using \: distance \: formula : } \\ \tt : \implies ab = \sqrt{ (x_{2} - x_{1} )^{2} + (y_{2} - y_{1})^{2} } \\ \\ \tt : \implies ab = \sqrt{(3 - 7)^{2} +( - 7 - 9)^{2} } \\ \\ \tt : \implies ab = \sqrt{16 +256} \\ \\ \tt : \implies ab = \sqrt{272} \: units \\ \\ \sf \orange{Similarly:} \\ \\ \tt : \implies bc = \sqrt{136} \\ \\ \tt : \implies ca = \sqrt{136} \\ \\ \tt {As \: we \: know \: that} \\ \tt : \implies I= \bigg(\frac{ax_{1} + b x_{2} + cx_{3} }{a + b + c} \: \: , \: \frac{ay_{1} + b y_{2} + cy_{3} }{a + b + c} \bigg) \\ \\ \tt : \implies I= \bigg(\frac{ 7\sqrt{136} + 3 \sqrt{136} - 3 \sqrt{272} }{ \sqrt{136 } + \sqrt{136} + \sqrt{262} } \: \: ,\: \: \frac{ 9\sqrt{136} - 7\sqrt{136} + 3 \sqrt{272} }{ \sqrt{136 } + \sqrt{136} + \sqrt{272} } \bigg)$

Answer 2

$\huge\bold \green{Given:-}$

• coordinate a ( 7 , 9 )
• coordinate b ( 3 , -7 )
• coordinate c ( -3 , 3 )

$\huge\bold\pink{To \: Find:-}$

• coordinate of incentre of the triangle.

$\huge\bold\red{Solution:-}$

PLEASE SEE THE ATTACHED FILE .