Question

Show that the points A(3,-1) B(5,-1) and C(3,-3) are vertices of a right angle triangle

Answer 1

To Prove :

• We have to prove that A(3,-1) B(5,-1) and C(3,-3) are the vertices of a right angled triangle.

Have a look at the attachment for the diagram!

Using distance formula, Let's find three sides.

The first side,

$\sf{AB = \sqrt{(x_{2} - x_{1})^{2} + {(y_{2} - y_{1})}^{2} } }$

$\sf{AB = \sqrt{ {(3 - 5)}^{2} + {( - 1 + 1)}^{2} } }$

$\sf{AB = 2 \: units}$

The second side,

$\sf{BC = \sqrt{(x_{2} - x_{1})^{2} + {(y_{2} - y_{1})}^{2} } }$

$\sf{BC = \sqrt{ {(5 - 3)}^{2} + {( - 1 + 3)}^{2} } }$

$\sf{BC = \sqrt{4 + 4} }$

$\sf{BC = 2 \sqrt{2} \: units}$

The third and the final side,

$\sf{CA = \sqrt{(x_{2} - x_{1})^{2} + {(y_{2} - y_{1})}^{2} } }$

$\sf{CA = \sqrt{ {(3 - 3)}^{2} + {(3 - 1)}^{2} } }$

$\sf{CA = 2 \: units}$

_________________

According to Pythagoras Theorem,

$\sf{ {BC}^{2} = {(2 \sqrt{2}) }^{2} } = 8$

$\sf{ {AB}^{2} + {CA}^{2} = {2}^{2} + {2}^{2} = 8}$

Hence, the vertices of the given triangle form a right angled triangle.

Answer 2

Given ,

The points A(3,-1) B(5,-1) and C(3,-3) are vertices of a right angle triangle

We know that ,

$\boxed{ \sf{Distance = \sqrt{ {( x_{2} - x_{1} )}^{2} + {(y_{2} - y_{1})}^{2} }}}$

Thus , the distance between A(3,-1) and B(5,-1)

AB = √{(5 - 3)² + (-1 + 1)²}

AB = √{4}

AB = 2 units

Similarly , the distance between B(5,-1) and C(3,-3)

BC = √{(3 - 5)² + (-3 + 1)²}

BC = √{4 + 4}

BC = √{8}

BC = 2√2 units

And the distance between A(3,-1) and C(3,-3)

AC = √{(3 - 3)² + (-3 + 1)²}

AC = √{4}

AC = 2 units

It is observed that ,

(BC)² = (AC)² + (AB)²

Therefore ,

• The given points are the vertices of right angled triangle