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Prove that the points $( 3,\ 0) ,\ ( 6,\ 4)$ and $( -1,\ 3)$ are the vertices of a right angled isosceles triangle.
Given: The points $( 3,\ 0) ,\ ( 6,\ 4)$ and $( -1,\ 3)$ are given.
To do: To prove are the vertices of a right angled isosceles triangle.
Solution:
Let $A( 3,\ 0) ,\ B( 6,\ 4)$ and $C( –1,\ 3)$ be the given points.
We know if there two points $(x_{1} ,\ y_{1} )$ and $(x_{2} ,\ y_{2} )$,
Distance between the two points,$=\sqrt{( x_{2} -x_{1})^{2} +( y_{2} -y_{1})^{2}}$
Using the distancee formula,
$AB=\sqrt{( 3-6)^{2} +( 0-4)^{2}}$
$\Rightarrow AB=\sqrt{( -3)^{2} +( -4)^{2}}$
$\Rightarrow AB=\sqrt{9+16}$
$\Rightarrow AB=\sqrt{25}$
$\Rightarrow AB=5\ unit$
Similarly $BC=\sqrt{( -1-6)^{2} +( 3-4)^{2}}$
$\Rightarrow BC=\sqrt{( -7)^{2} +( -1)^{2}}$
$\Rightarrow BC=\sqrt{49+1}$
$\Rightarrow BC=\sqrt{50}$
$\Rightarrow BC=5\sqrt{2} \ unit$
And $CA=\sqrt{( -1-3)^{2} +( 3-0)^{2}}$
$\Rightarrow CA=\sqrt{( -4)^{2} +( 3)^{2}}$
$\Rightarrow CA=\sqrt{16+9}$
$\Rightarrow CA=\sqrt{25}$
$\Rightarrow CA=5\ unit$
here we find that,
$BC^{2} =AB^{2} +CA^{2}$ which satisfy pythagoras theorem,
and $AB=CA$
Thus The given points are the vertices of an issocele right triangle.
Updated on: 10-Oct-2022
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