Question

The vertices of a point are A(1,2);B(-3,-4);&C(5,-6) .Find the CircumCentre.​

Answer 1

$\huge\bf{\red{\overbrace{\underbrace{\purple{Given:}}}}}$

★Vertices of a ∆ are

•A(1, 2)

•B(-3, -4) &

•C(5, -6)

$\huge\bf{\red{\overbrace{\underbrace{\purple{To\:\:Find:}}}}}$

★The coordinates of the CircumCentre of the ∆ .

$\huge\bf{\red {\overbrace{\underbrace{\purple{Concept\:\:Used:}}}}}$

★We will use 'Distance formula '.

$\huge\bf{\red {\overbrace{\underbrace{\purple{Answer:}}}}}$

We have,

•A(1, 2)

•B(-3, -4) &

•C(5, -6)

Let the point S(x, y) be the CircumCentre of the ∆.

So, SA = SB = SC

. °. $SC^{2}=SB^{2}=SC^{2}$

Now, taking SA=SB,

$\implies SA=SB$

$\implies SA^{2}=SB^{2}$

$\implies(x-1)^{2}+(y-2) ^{2}=(x-3) ^{2}+(y+4) ^{2}$

$\implies x^{2}-2x+1+y^{2}+4-4y=x^{2}-6x+9+y^{2}+8y+16$

$\large\blue{\boxed{\red{(a+b) ^{2}=a^{2}+b^{2}+2ab}}}$

$\large\blue{\boxed{\red{(a-b) ^{2}=a^{2}+b^{2}-2ab}}}$

$\implies\cancel{x^{2}}-2x+1+\cancel{y^{2}}+4-4y=\cancel{x^{2}}-6x+9+\cancel{y^{2}}+8y+16$

$\implies -2x-4y+5=-6x+8y+25$

$\implies 6x-2x-8y-4y=20$

$\implies 4x-12y=20$

$\implies 4(x-3y) =20$

$\implies x-3y=\dfrac{20}{4}$

$\implies x-3y=\dfrac{\cancel{20}^{5}}{\cancel{4}}$

${\underline{\boxed{\red{.\degree. (x-3y) =5}}}}$...........(1)

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Now, again taking SB=SC,

$\implies SA=SC$

$\implies SA^{2}=SB^{2}$

$\implies (x-3) ^{2}+(y+4) ^{2}=(x-5) ^{2}+(y-6) ^{2}$

$\implies x^{2}+9-6x+y^{2}+16+8y=x^{2}+25-10x+y^{2}+36+12y$

$\footnotesize{\implies \cancel{x^{2}}+9-6x+\cancel{y^{2}}+16+8y=\cancel{x^{2}}+25-10x+\cancel{y^{2}}+36+12y}$

$\implies 25+8y-6x=25+12y-10x+36$

$\implies \cancel{25}+8y-6x=\cancel{25}+12y-10x+36$

$\implies 10x-6x+8y-12y=36$

$\implies 4x-4y=36$

$\implies 4(x-y) =36$

$\implies (x-y) =\dfrac{36}{4}$

$\implies (x-y) =\dfrac{\cancel{36}^{9}}{\cancel{4}}$

${\underline{\boxed{\purple{.\degree. x-y=9}}}}$.........(2)

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Now (2) -(1), we have,

$\implies x-y-(x-3y)=9-5$

$\implies x-y-x+3y=4$

$\implies 2y=4$

$\implies y =\dfrac{4}{2}$

$\implies y =\dfrac{\cancel{4}^{2}}{\cancel{2}}$

${\underline{\boxed{\orange{.\degree. y=2}}}}$

Putting it in (2)

$\implies x -2=9$

$\implies x=9+2$

${\underline{\boxed{\blue{.\degree. x =11}}}}$

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$\Large{Answer}$$\begin{cases} \longrightarrow x=11\\\longrightarrow y=2\end{cases}$

Therefore the coordinate of point of CircumCentre is (11, 2).

$<marquee scrollamount="1300">♥Answer by Rishabh♥</marquee>$