Test Index

KEAM 2014 Math Paper

Question : 105 of 120

Marks: +1, -0

(2, -1 )
(1,-2)
(5,2)
(2,5)
(4, 5)

Solution:

Let the vertices of a triangle are

A(8,6)

B(8,-2)

and

C(2,-2)

Now,

\;\; AB=\sqrt{(8-8)^{2}+(6+2)^{2}}

=\sqrt{0+8^{2}}=8

BC=\sqrt{(2-8)^{2}+(-2+2)^{2}}=\sqrt{36+0}=6

and

CA=\sqrt{(8-2)^{2}+(6+2)^{2}}

=\sqrt{6^{2}+8^{2}}

=\sqrt{36+64}=\sqrt{100}=10

Now,

\;\; AB^{2}+BC^{2}=(8)^{2}+(6)^{2}

=64+36=100=AC^{2}

So,

ABC

is a right angled triangle and right angled at

B

We know that, in a right angled triangle, the circumcenter is the mid-point of hypotenuse.

\therefore

Mid-point of

AC=\left(\;\frac{8+2}{2},\;\frac{6-2}{2}\right)=(5,2)

Hence, the required circumcenter is (5,2)

Go to Question:

Prev Question

Next Question