Question

the vertices of triangle is (6,6) (0,6) (6,0) find the distance between circumcenter and centroid

Answer 1

Given :-

• Vertices of triangle = (6,6) , (0,6) , (6,0)

To find :-

• Distance between circumcentre and centroid

SOLUTION :-

• Firstly we find the circumcentre and centroid of the triangle

Finding centroid of the triangle :-

$G = \bigg(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \bigg)$

Substituting the values,

$G = \bigg( \dfrac{6 + 0 + 6}{3} , \dfrac{6 + 6 + 0}{3} \bigg)$

$G = \bigg( \cfrac{12}{3} , \dfrac{12}{3} \bigg)$

$G = (4 ,\: 4)$

So, centroid of the triangle is (4,4)

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Finding circumcentre of triangle:-

》 Before finding circumcentre of triangle firstly we need to know the vertices are the vertices of right angle triangle or acute angle triangle or obtuse angle triangle.

Conditions required:-

》 If the vertices are right angle triangle circumcentre is mid point of hypotenuse .

》 If the vertices are acute angle triangle circumcentre lies inside the triangle.

》 If the vertices are equilateral triangle centroid , circumcentre, orthocentre coincides.

So, We find distance between them points

$\sqrt{(x_1 - x_2) {}^{2} + (y_1 - y_2) {}^{2} }$

Once refer the attachment

$AB = \sqrt{(6 - 0) {}^{2} + (6 - 6) {}^{2} }$

$AB = \sqrt{(6) {}^{2} + (0) {}^{2} }$

$AB = \sqrt{36}$

$AB = 6$

$BC = \sqrt{(0 - 6) {}^{2} + (6 - 0) {}^{2} }$

$BC = \sqrt{( - 6) {}^{2} + (6) {}^{2} }$

$BC = \sqrt{36 + 36}$

$BC = \sqrt{72}$

$AC = \sqrt{(6 - 6) {}^{2} + (6 - 0) {}^{2} }$

$AC = \sqrt{(0) {}^{2} + (6) {}^{2} }$

$AC = \sqrt{(6) {}^{2} }$

$AC = 6$

By observing these distances ,

We will check the Pythagoras theorem

AB² + AC² = BC²

$(6) {}^{2} + (6) {}^{2} = ( \sqrt{72} ) {}^{2}$

$36 + 36 = 72$

$72 = 72$

Hence , it satisfies Pythagoras theorem

So, the hypotenuse at BC

As per the above conditions,

It is a right angle triangle So, the circumcentre is midpoint of hypotenuse .

Refer the attachment

$Midpoint = \bigg( \dfrac{x_1 + x_2}{2} ,\cfrac{y_1 + y_2}{2} \bigg)$

$= \bigg( \cfrac{ 0 + 6}{2} ,\dfrac{6 + 0}{2} \bigg)$

$= \bigg( \cfrac{6}{2} , \dfrac{6}{2} \bigg)$

$= (3 ,\: 3)$

So, the circumcentre of triangle is (3,3)

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Now ,finding distance between circumcentre and centroid

Centroid = (4,4)

Circumcentre = (3,3)

• x1 = 3
• y1 = 3
• x2 = 4
• y2 = 4

By using distance formula,

$\sqrt{(x_1 - x_2) {}^{2} + (y_1 - y_2) {}^{2} }$

$= \sqrt{(3 - 4) {}^{2} + (3 - 4) {}^{2} }$

$= \sqrt{( - 1) {}^{2} + ( - 1) {}^{2} }$

$= \sqrt{1 + 1}$

$= \sqrt{2}$

So, the distance between circumcentre and centroid is $= \sqrt{2}$ units.