Question

Find the co-ordinates of the circumcentre of the triangle formed by the
points whose vertices are (1, 1, 0), (1, 2, 1) and (-2, 2, -1).​

Answer 1

Answer:

Step-by-step explanation:

Let say co-ordinates of circum center are ( x , y , z)

circumcenter will be equidistant from each vertex

so (x -1)² + (y - 1)² + z² = (x -1)² + (y - 2)² + (z - 1)² = (x +2)² + (y -2)² + (z + 1)²

(x -1)² + (y - 1)² + z² = (x -1)² + (y - 2)² + (z - 1)²

=> (y - 1)² + z² = (y - 2)² + (z - 1)²

=> y² + 1 - 2y + z² =  y² + 4 - 4y + z² + 1 - 2z

=> 2y + 2z = 4

=> y + z = 2

(x -1)² + (y - 2)² + (z - 1)² = (x +2)² + (y -2)² + (z + 1)²

=> (x -1)²+ (z - 1)² = (x +2)² + (z + 1)²

=> x² + 1 - 2x + z² + 1 - 2z  = x² + 4 + 4x + z² + 1 + 2z

=> 6x + 4z = - 3

=> 6x = -3 - 4z

(x -1)² + (y - 1)² + z² = (x +2)² + (y -2)² + (z + 1)²

=> x² + 1 - 2x + y² +  1 - 2y + z² = x² + 4 + 4x + y² + 4 - 4y + z² + 1 + 2z

=> 6x -2y + 2z = -7

putting y = (2-z)

=>6x - 2(2 -z) + 2z = -7

=>6x - 4 + 2z + 2z = -7

=> 6x = -3 -4z  ( same as eealier equation)

so

y + z = 2 & 6x = -3 - 4z

will give coordinates of circumcentre of the triangle

Answer 2

Answer:

Let's the co-ordinates of circumcentre be ( x , y , z).

circumcentre will be always at the equal distance from each vertex.

So,

(x -1)² + (y - 1)² + z² = (x -1)² + (y - 2)² + (z - 1)² = (x +2)² + (y -2)² + (z + 1)²(x -1)² + (y - 1)² + z² = (x -1)² + (y - 2)² + (z - 1)²

• (y - 1)² + z² = (y - 2)² + (z - 1)²
• y² + 1 - 2y + z² =  y² + 4 - 4y + z² + 1 - 2z
• 2y + 2z = 4
• y + z = 2

(x -1)² + (y - 2)² + (z - 1)² = (x +2)² + (y -2)² + (z + 1)²

• (x -1)²+ (z - 1)² = (x +2)² + (z + 1)²
• x² + 1 - 2x + z² + 1 - 2z  = x² + 4 + 4x + z² + 1 + 2z
• 6x + 4z = - 3
• 6x = -3 - 4z

(x -1)² + (y - 1)² + z² = (x +2)² + (y -2)² + (z + 1)²

• x² + 1 - 2x + y² +  1 - 2y + z² = x² + 4 + 4x + y² + 4 - 4y + z² + 1 + 2z
• 6x -2y + 2z = -7

Putting y = (2-z)

• 6x - 2(2 -z) + 2z = -7
• 6x - 4 + 2z + 2z = -7
• 6x = -3 -4z

So,

 y + z = 2

6x = -3 - 4z