1.

The coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, –2) and (2, –2).(a) (5, 2) (b) (–2, –5) (c) (0, 0) (d) (5, 0)

Answer»

(a) (5, 2)

Let A(8, 6) , B(8, –2) and C(2, –2) be the vertices of the given triangle and P(x, y) be the circum-centre of this triangle. 

Then, PA2 = PB2 = PC

Now, PA2 = PB2 ⇒ (x – 8)2 + (y – 6)2 = (x – 8) + (y + 2)2 

⇒ x2 – 16x + 64 + y2 – 12y + 36 = x2 – 16x + 64 + y2 + 4y + 4 

⇒ 16y = 32 ⇒ y = 2. 

Now, PB2 = PC2 ⇒ (x – 8)2 + (y + 2)2 = (x – 2)2 + (y + 2)

⇒ x2 – 16x + 64 + y2 + 4y + 4 = x2 – 4x + 4 + y2 + 4y + 4 

⇒ 12x = 60 ⇒ x = 5. 

∴ Co-ordinates of the circumcentre are (5, 2).



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