Question

. In the given figure, a right triangle BOA is given and C is the
mid-point of the hypotenuse AB. Show that is equidistant from the vertices 0.
A and B.
(CBSE 2010)​

Answer 1

Given a right triangle BOA with vertices B(0,2b), O(0,0), A(2a,0) Since, C is the midpoint of AB Therefore, coordinates of C are

$( \frac{2a + 0}{2} )$

$( \frac{0 + 2b}{2})$

= (a,b)

Now,

$co = \sqrt{(a - 0)^{2} + (b - 0) ^{2} } = \sqrt{ {a}^{2} + {b}^{2} }$

$ca = \sqrt{( {2a - a)}^{2} + {(0 - b)}^{2} } = \sqrt{ {a}^{2} + {b}^{2} }$

$cb = \sqrt{ {(a - 0)}^{2} + {(b - 2b)}^{2} } = \sqrt{ {a}^{2} + {b}^{2} }$

Since, CO=CA =CB Therefore, C is equidistant from O, A and B.

Answer 2

Answer:

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