Q. 184.7( 7 Votes )

# 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.

Answer :

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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If the coordinates of P and Q are (a cos θ, b sin θ) and (–a sin θ, b cos θ) respectively, then prove that OP^{2} + OQ^{2} = a^{2} + b^{2}, where O is the origin.

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.

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