Find The Circumcentre Of The Triangle Formed By The Points (2, 3), (1, -5) And (-1, 4).

Given three points are:

A=(2,3)

B=(1,-5)

C=(-1,4)

Mid point of AB = [latex] \left ( \frac{2+1}{2} , \frac{3-5}{2} \right ) = \left ( \frac{3}{2}, -1\right )[/latex]

Slope of AB=[latex] \frac{y2-y1}{x2-x1} = \frac{-5-3}{1-2} = \frac{-8}{-1} = 8[/latex]

Slope of perpendicular bisector = [latex] \frac{-1}{8}[/latex]

Equation of AB with respect to slope = [latex] \frac{-1}{8} and coordinate \left ( \frac{3}{2}, -1 \right)[/latex]

y – y1 = m (x – x1)

= y + 1 =[latex] \frac{-1}{8} (x\frac{-3}{2})[/latex]

y + 1 = [latex] \frac{-1}{16} (2x – 3) [/latex] [latex] \Rightarrow [/latex]16y + 16 = -2x + 3

2x + 16y + 13 = 0—-[1]

Slope of =[latex] \left ( \frac{2 – 1}{2}, \frac{3 + 4}{5}\right ) [/latex] [latex] \Rightarrow \left ( \frac{1}{2},\frac{7}{2} \right )[/latex]

Slope of = [latex] \frac{y2 – y1}{x2 – x1}[/latex] [latex] \Rightarrow \frac{4-3}{-1-2}[/latex] [latex] \Rightarrow \frac{-1}{3}[/latex]

The slope of perpendicular bisector = 3

Equation of AC with respect to slope and 3 coordinate =

[latex] \left ( \frac{1}{2},\frac{7}{2} \right )[/latex] [latex] \Rightarrow y – y1 = m(x – x1)[/latex] [latex] \Rightarrow y – \frac{7}{2} = 3 (x – \frac{1}{2})[/latex] [latex] \Rightarrow 2y – 7 = 6x – 3[/latex] [latex] \Rightarrow 6x – 2y + 4 = 0[/latex] [latex] \Rightarrow 3x – y + 2 = 0—-[2][/latex]

Now by multiplying equation 1 with 3 and equation 2 with 2, we get:

(2x + 16y + 13 = 0) * 3

[latex] \Rightarrow 6x + 48y + 39 = 0[/latex] [latex] \Rightarrow 6x + 48y = -39[/latex]

(3x – y + 2 = 0) * 2

[latex] \Rightarrow 6x + 2y + 4 =0[/latex] [latex] \Rightarrow 6x + 2y = -4 [/latex] [latex] \frac{6x + 48y = -39-6x + 2y = -4}{50y = -35}[/latex] [latex] \Rightarrow x = \frac{-9}{10}[/latex] [latex] \Rightarrow y = \frac{-35}{50} \Rightarrow \frac{-7}{10}[/latex]

Therefore, The circumcenter of a triangle is

[latex] \left ( \frac{-9}{10},\frac{-7}{10} \right)[/latex]

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