Show that the points$(  -2,\ 3) ,\ (  8,\ 3)$ and $( 6,\ 7)$ are the vertices of a right triangle.

Given: Points (-2, 3), (8, 3) and (6, 7).

To do: To prove that the given points are the vertices of a triangle.

Solution: 

The given points are A(-2, 3), B(8, 3) and C(6, 7) 

AB, BC and CA are the sides of triangle.

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 above distance formula, We have

$AB=\sqrt{( -2-8)^{2} +( 3-3)^{2}}$

$=\sqrt{( -10)^{2}}$

$=10$

$BC=\ \sqrt{( 8-6)^{2} +( 3-7)^{2}}$ 

$=\sqrt{( 2)^{2} +( -4)^{2}}$

$=\sqrt{4+16}$

$=2\sqrt{5}$

$CA=\sqrt{( -2-6)^{2} +( 3-7)^{2}}$ 

$=\sqrt{( -8)^{2} +( -4)^{2}}$

$=\sqrt{64+16}$

$=\sqrt{80}$

$=4\sqrt{5}$

Now sum of the squares of BC and CA,

$BC^{2} +CA^{2} =( 2\sqrt{5})^{2} +( 4\sqrt{5})^{2}$

$=20+80$

$=100$

$=10^{2}$

$=AB^{2}$

Thus we get $AB^{2} =BC^{2} +CA^{2}$

Which satisfy the pythagoras theorem.

$\vartriangle ABC$ is a right triangle.

Hence proved.

 

Updated on: 10-Oct-2022

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