Q18 of 48 Page 102

Prove that the mid–point C of the hypotenuse in a right angled triangle AOB is situated at equal distances from the vertices O, A and B of the triangle.

Consider a right angled ∆AOB, such that C is the mid–point of hypotenuse AB.



We have O → (0, 0)


Since A lies on y-axis, A→ (0, y)


and B lies on x-axis, B→ (x, 0)


Using mid–point formula, coordinates of C are


Using distance formula,


AC =


=


=


=


=


BC =


=


=


=


=


We know that distance of a point P (x,y) from origin O (0, 0) is given as OP =


∴ OC =


=


=


We can observe that OA = OB = OC.


∴ C (mid–point of hypotenuse AB) is equidistant from all the three vertices of the right angled ∆AOB.


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