Question

The vertices of a triangle are (6,0) ,(0,6) and (6,6) . find the distance between the circumcentre and centroid of the triangle

Answer 1

Final Answer : √2 units

Steps:
1) Let there be triangle ABC, where
A = (6,0)
B = (0,6)
C = (6,6)
Clearly,
We observe that ABC is right -angled triangle.

2) In right - angled triangle, circumcentre is mid-point of hypotenuse.
Here, BA is hypotenuse .
So, Mid point of AB
D = ((0+6)/2 ,(6+0)/2 )
= (3,3)

Hence, Circumcentre is D(3,3).

3) And, Centroid
E = (6+0+6)/3 , (0+6+6)/3
E = (4,4)

Therefore, Distance between Centroid and Circumvented is ED
$= \sqrt{ {(4 - 3)}^{2} + {(4 - 3)}^{2} } \\ = \sqrt{ {1}^{2} + {1}^{2} } = \sqrt{2} \: units \:$

Hence, Distance between Circumcentre and Centroid of triangle is
$\boxed { \sqrt{2} \: units \: }$

Answer 2

Given vertices are A(6,0), B(0,6), C(6,6).

Circumcentre:

We need to find the midpoint of AB.

A = (6,0) = (x1,y1)

B = (0,6) = (x2,y2)

Mid-point of AB = (x1 + x2)/2, (y1 + y2)/2

= > (6 + 0)/2, (0 + 6)/2

= > 6/2, 6/2

= > 3, 3

Therefore the circumcentre of the triangle is (3,3).


Centroid:

A(x1,y1), B(x2,y2), C(x3,y3).

The coordinates of the centroid of a triangle can be determined by;

= > (x1 + x2 + x3)/3, (y1 + y2 + y3)/3

= > (6 + 0 + 6)/3,(0 + 6 + 6)/3

= > (12/3),(12/3)

= > (4,4)


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


Now,

The distance between the circumference and centroid of the triangle:

$= \ \textgreater \ D = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

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

$= \ \textgreater \ \sqrt{1 + 1}$

$= \ \textgreater \ \sqrt{2}$



Hope this helps!