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In the figure, a right triangle BOA is given. $C$ is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices $O, A$ and $B$."

Academic Mathematics NCERT Class 10

Given:

A right triangle BOA is given. $C$ is the mid-point of the hypotenuse AB.
To do:

We have to show that $C$ is equidistant from the vertices $O, A$ and $B$.

Solution:

In $\triangle OAB$,

Coordinates of $O$ are $(0, 0)$, $A$ are $(2a, 0)$ and $B$ are $(0, 2b)$.
$C$ is the mid-point of $AB$.

Using mid-point formula, we get,
The coordinates of $C$ are \( \left(\frac{2 a+0}{2}, \frac{0+2 b}{2}\right) \)

\( =(a, b) \)

\( \mathrm{CO}=\sqrt{(a+0)^{2}+(b+0)^{2}} \)

\( =\sqrt{a^{2}+b^{2}} \)

Using distance formula, we get,

\( \mathrm{CA}=\sqrt{(2 a-a)^{2}+(0-b)^{2}} \)

\( =\sqrt{(a)^{2}+(b)^{2}} \)

\( =\sqrt{a^{2}+b^{2}} \)

\( \mathrm{CB}=\sqrt{(0-a)^{2}+(2 b-b)^{2}} \)

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

\( =\sqrt{a^{2}+b^{2}} \)
Here,

$CO = CA = CB$
This implies, $C$ is equidistant from the vertices $O, A$ and $B$.
Hence proved.

Updated on: 10-Oct-2022

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