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EAMCET Engineering 2001 Solved Paper

Section: Mathematics

Question : 116 of 200

Marks: +1, -0

x+y=1, x=1, y=1

Solution:

From the figure,

A (x_1, y_1) = (1,0)

B (x_2, y_2) \; = (1,1)

\text{ and } C (x_3, y_3) \; = (0,1)

a = BC \; = \sqrt{(1-0)^2+(1-1)^2} = 1

b = CA \; = \sqrt{(1-0)^2+(0-1)^2} = \sqrt{1+1} = \sqrt{2}

c = AB \; = \sqrt{(1-1)^2+(1-0)^2} = 1

Incentre of triangle

\; = \left( \frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c} \right)

\; = \left( \frac{1+\sqrt{2}}{1+1+\sqrt{2}}, \frac{\sqrt{2}+1}{1+1+\sqrt{2}} \right) = \left( \frac{1+\sqrt{2}}{2+\sqrt{2}}, \frac{1+\sqrt{2}}{2+\sqrt{2}} \right)

\; = \left( \frac{1+\sqrt{2}}{\sqrt{2}(1+\sqrt{2})}, \frac{1+\sqrt{2}}{\sqrt{2}(1+\sqrt{2})} \right) = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)

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