Question

find the circuncentre of a triangle whose vertices are (-2,-3),(-1,0),(7,-6)

Please don't write anappropriate answer.

Answer 1

Pls forgive me for my handwriting...:))

Answer 2

GIVEN: A triangle ABC, A(7,-6), B(-2,-3), C(-1,0) Circumcentre P ( x, y)
We calculate the coordinates x1,y1 of point M ( mid point of BC). Then apply Pythagoras law in triangle PBM.
Coordinates are found by section formula
X1 = (m1x2 + m2x1)/2, y1 = (m1y1 + m2y2) /2
And the lengthof sides PB, BM, PM are calculated by distance formula
√{( x1-x2)² + ( y1-y2)²}
So, here, x1= -3/2 & y1 = -3/2
PB² = BM² + PM²
=> (x+2)² + (y+3)² =( -2+ 3/2)² + (-3 + 3/2)² + ( x+3/2)² + (y + 3/2)²
=> x² + 4 + 4x+ y² +9 + 6y= 1/4. + 9/4. + +x² + 9/4. + 3x y² + 9/4. + 3y
=> 4x + 6y + 13 = 7 + 3x + 3y
=> x + 3y = -6 …………………..(1)
Similarly, the midpoint AB is N, the coordinates of which x2 = (-2+7)/2 = 5/2 &
y2 = (-3 + -6) /2 = -9/2
Then apply Pythagoras law in triangle PAN
PA² = AN² = + PN²
=> (x-7)² + (y+6)² = (7- 5/2)² + (-6+9/2)² + (x-5/2)² + (y+9/2)²
=> x² + 49 - 14x +y² +36 + 12y = 81/4 + 9/4 + x² + 25/4 - 5x + y² + 9y + 81/4
=> 85 -14x + 12y = 49 -5x + 9y
=> -9x + 3y = -36
=> -3x + y = -12 …………..(2)
Now, by solving eq ( 1) & eq (2)
x+ 3y =-6 ………….(1)
-3x + y = -12 ……….(2)
x = 3
y = -3
So, coordinates of circumcentre = (3, -3)