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Find the coordinates of the circumcenter 0,0 4,0 4,-3

Sagot :

The coordinates of the circumcenter of triangle having coordinates of (0,0),(4,0),(4,-3) are (1,-17/6).

Given The coordinates of triangle are (0,0),(4,0),(4,-3).

Let the triangle be ABC.

Let the coordinates of the circumcenter be D(x, y).

We know that the length of circumcenter from the corner points are equal to each other.

In this way AD=BD=DC

AD=[tex]\sqrt{(x-0)^{2} +(y-0)^{2} }[/tex]

=[tex]\sqrt{x^{2} +y^{2} }[/tex]

DC=[tex]\sqrt{(4-x)^{2} +(-3-y)^{2} }[/tex]

BD=[tex]\sqrt{(4-x)^{2} +(0-y)^{2} }[/tex]

=[tex]\sqrt{(4-x)^{2} +y^{2} }[/tex]

AD=DC

[tex]\sqrt{x^{2} +y^{2} }[/tex]==[tex]\sqrt{(4-x)^{2} +(-3-y)^{2} }[/tex]

Squaring both sides we get

[tex]x^{2} +y^{2}=16+x^{2} -8x+9+3y+3y+y^{2}[/tex]

8x-6y=25--------------------1

DC=BD

[tex]\sqrt{(4-x)^{2} +(-3-y)^{2} } =\sqrt{(4-x)^{2} +y^{2} }[/tex]

8x+6y=-9---------------------2

Solve equation 1 and 2

Add both equations

8x-6y+8x+6y=25-9

16x=16

x=1

put the value of x in 1

8x-6y=25

8*1-6y=25

-6y=25-8

y=-17/6

Hence the coordinates of circumcenter is (1,-17/6).

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