Question

The Coordinates of the circumcentre of the right triangle ABC having vertices A (0,-4) B (0,0) , and C (6,0)
a) (6,4)
b) (6,-4)
c) (2,-4)
d) (3,-2)
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Answer 1

Step-by-step explanation:

Let A(4,6), B(0,4), C(6,2) be the vertices of the given △ABC.

Let P(x,y) be the circumcentre of △ABC. Then,

PA=PB=PC⟹PA2=PB2=PC2

Now, PA2=PB2

(x−4)2+(y−6)2=(x−0)2+(y−4)2

x2+y2−8x−12y+52=x2+y2−8y+16

8x+4y=36

2x+y=9         

Again, PB2=PC2

(x−0)2+(y−4)2=(x−6)2+(y−2)2

x2+y2−8y+16=x2+y2−12x−4y+40

12x−4y=24

3x−y