TS ICET 24 May 2018 Shift 1 Paper

Question : 140 of 200

Marks: +1, -0

The in-centre o f the triangle formed by A(2. 3), B(-2. -5) and C(-4, 6) is

Solution:

Given

A(2, 3) B(–2, –5) C(–4, 6)

a=B C=√{2^{2}+11^{2}}=√{125}=5 √{5}

b=C A=√{6^{2}+3^{2}}=√{45}=3 √{5}

c=A B=√{4^{2}+8^{2}=} √{80}=4 √{5}

In centre

=(\;{a x_{1}+b x_{2}+c x_{3}}/{a+b+c}, \;{a y_{1}+b y_{2}+c y_{3}}/{a+b+c})

=(\;{(5 √{5})(2)+(3 √{5})(-2)+(4 √{5})(-4)}/{5 √{5}+3 √{5}+4 √{5}}, {(5 √{5})(3)+(3 √{5})(-5)+(4 √{5}) 6}/{5 √{5}+3 √{5}+4 √{5}})

=(\;{-12 √{5}}/{12 √{5}}, \;{24 √{5}}/{12 √{5}})=(-1,2)

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