Question

Find the cirumcentre of the (5-2)(-1;2)and(2;-1)

Answer 1

★ Rationale : -

We are given with three points of a triangle and are asked to find the circumcentre of the triangle.

Let's name the points as A(5 -2) , B(-1,2) and C(2,-1)

$\bf \: Mid \: point \: = ( \dfrac{x1+ x2}{2} , \dfrac{y1+ y2}{2} )$

$\bf \: Mid \: point \: of AB \: = ( \dfrac{4}{2},0)$

$\implies \: Mid \: point \: of AB \: =(2,0)$

$\bf \: Slope = \dfrac{y2 - y1}{x2 - x1}$

$Slope \: of \: AB \: = \dfrac{ 4}{ - 6}$

$\implies \bf Slope \: of \: AB \: = \dfrac{ - 2}{3}$

$\bf \: Slope \: of \: perpendicular \: bisector = \dfrac{ - 3}{2}$

$Equation \: of \: AB \: with \: respect \: to \: slope \: = \dfrac{ - 3}{2} and \: coordinate \: (2,0)$

$\bf \: y - y1 = m(x - x1)$

$y = \dfrac{ - 3}{2} (x - 2)$

$2y = - 3x + 6$

$3x + 2y - 6 = 0 \: \: \: \: \: - (1)$

$\bf \: Midpoint \: of \: AC = ( \dfrac{ 7}{2} , \dfrac{ - 3}{2} )$

$\bf \: Slope \: of \: AC = \dfrac{ - 1 + 2}{2 - 5} = \dfrac{ - 1}{3}$

$\bf \: Slope \: of \: perpendicular \: bisector \: = - 3$

$\bf \: equation \: of \: AC\: with \: respect \: to \: slope \: and \: coordinate ( \dfrac{7}{2} , \dfrac{ - 3}{2} )$

$y + \dfrac{3}{2} = - 3(x - \dfrac{7}{2} )$

$\dfrac{2y + 3}{2} = - 3( \dfrac{2x - 7}{2} )$

$2y + 3 = - 6x + 21$

$6x + 2y - 18= 0 \: \: \: \: \: \: - (2)$

By solving equation (1) and (2) ,

The circumcenter of triangle = ( 4 , - 3 )

____________________.

★ Rough calculations of two equations :

3x + 2y = 6 ----- ( 1 )

6x + 2y = 18 ------( 2 )

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-3x = - 12

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» Therefore x = 4

By Substituting the value of x in equation we get y = - 3

Therefore the point is ( 4 , - 3 )