Question

The circumcentre of the triangle formed by (1, - 2), (2, 4) and (3,-6) is



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Answer 1

Let (x, y) be the circumcenter of the triangle having vertices (1, -2), (2, 4) and (3, -6).

The circumcenter is equidistant from all the vertices of the triangle. Therefore, if d is the distance of the circumcenter from any of the vertex,

$\longrightarrow d^2=(x-1)^2+(y+2)^2\quad\quad\dots(1)$

$\longrightarrow d^2=(x-2)^2+(y-4)^2\quad\quad\dots(2)$

$\longrightarrow d^2=(x-3)^2+(y+6)^2\quad\quad\dots(3)$

Equating (1) and (2),

$\longrightarrow(x-1)^2+(y+2)^2=(x-2)^2+(y-4)^2$

$\longrightarrow x^2+y^2-2x+4y+5=x^2+y^2-4x-8y+20$

$\longrightarrow x=\dfrac{15}{2}-6y\quad\quad\dots(4)$

Equating (2) and (3),

$\longrightarrow (x-2)^2+(y-4)^2=(x-3)^2+(y+6)^2$

$\longrightarrow x^2+y^2-4x-8y+20=x^2+y^2-6x+12y+45$

$\longrightarrow 2x-20y-25=0$

Putting value of $x$ from (4),

$\longrightarrow 2\left(\dfrac{15}{2}-6y\right)-20y-25=0$

$\longrightarrow y=-\dfrac{5}{16}$

And from (4),

$\longrightarrow x=\dfrac{15}{2}-6\left(-\dfrac{5}{16}\right)$

$\longrightarrow x=\dfrac{75}{8}$

Hence the circumcenter is $\bf{\left(\dfrac{75}{8},\ -\dfrac{5}{16}\right)}.$