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  • For a given triangle whose vertices are (-12,0),(0,12) & (-14,14), which of the following is true?Option: 1 centre of incircle is (-9,9)</div

For a given triangle whose vertices are (-12,0),(0,12) & (-14,14), which of the following is true?

Option: 1

centre of incircle is (-9,9)


Option: 2

\mathrm{\text { radius of incircle is } 3 \sqrt{2}}


Option: 3

\mathrm{\text { Equation of incircle is }(x+9)^2+(y-9)^2=(3 \sqrt{2})^2}


Option: 4

All of the above 


Answers (1)

best_answer

\mathrm{\text { Let } A(-12,0), B(0,12), C(-14,14)}

\mathrm{\begin{aligned} \therefore \quad & A B=\sqrt{144+144}=\sqrt{2 \cdot 144}=12 \sqrt{2} \\ & B C=\sqrt{196+4}=10 \sqrt{2}=A C \end{aligned}}

∴ ?ABC is an isosceles triangle.
Therefore the median through C is bisector of ∠C and the equation of angle bisector of
C passes through the mid point of AB where mid point of AB is M(–6, 6).
Now, equation of angle bisector of C is OC or CM given by y – 6 = –1(x + 6) ⇒ y = –x, let any point on this line be P(–α, α) where α > 0.

Again, equation of AB is x – y + 12 = 0 and equation of AC is 7x + y + 84 = 0
Now, the ⊥ distance from (–α, α) to the lines AB & AC is the radius and (–α, α) be the centre of incircle. Also, the ⊥ distance from (–α, α) to the lines AB & AC are equal.

\mathrm{\begin{aligned} & \therefore\left|\frac{-7 \alpha+\alpha+84}{\sqrt{50}}\right|=\left|\frac{-\alpha-\alpha+12}{\sqrt{2}}\right| \\ & \Rightarrow \frac{-6 \alpha+84}{5}= \pm(-2 \alpha+12) \end{aligned}}

⇒ –6α + 84 = –5(–2α + 12) and –6α + 84 = 5(–2α + 12)
⇒ α = 9 and α = –6 (rejected as α > 0)
∴ Centre of incircle is (–9, 9) and

\mathrm{\text { radius }=\left|\frac{-2 \alpha+12}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2}}

\mathrm{\text { Thus, equation of incircle is }(x+9)^2+(y-9)^2=(3 \sqrt{2})^2}

 

Posted by

Ritika Jonwal

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