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  • Given vertices <img alt="\mathrm{A(1, 1), B(4, -2) \; and \; C(5, 5)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%281%2C%201%29%2C%20B%284%2C%20-2%29%20%5C%3B%20and%20%5C%3B

Given vertices \mathrm{A(1, 1), B(4, -2) \; and \; C(5, 5)} are of a triangle. Find the equation of the perpendicular from C to the interior bisector of the angle A

Option: 1

\mathrm{x=4}


Option: 2

\mathrm{x=5}


Option: 3

\mathrm{x=2}


Option: 4

\mathrm{x=1}


Answers (1)

best_answer

For triangle ABC, \mathrm{a=\sqrt{(4-5)^2+(-2-5)^2}=5 \sqrt{2}}

\mathrm{b=\sqrt{(1-5)^2+(1-5)^2}=4 \sqrt{2}}

\mathrm{c=\sqrt{(1-4)^2+(1+2)^2}=3 \sqrt{2}}  if incentre of this triangle be (h,k)

then

\mathrm{h=\frac{5 \sqrt{2} \cdot 1+4 \sqrt{2} \cdot 4+3 \sqrt{2} \cdot 5}{5 \sqrt{2}+4 \sqrt{2}+3 \sqrt{2}}=3}

and 

\mathrm{k=\frac{5 \sqrt{2} \cdot 1+4 \sqrt{2} \cdot(-2)+3 \sqrt{2} \cdot 5}{5 \sqrt{2}+4 \sqrt{2}+3 \sqrt{2}}=1}

so incentre is (3, 1) 

equation of interior angle bisector through A is y = 1 

So line perpendicular to it and passing through C(5, 5) will be x = 5.

Posted by

Anam Khan

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