# Find out whether the points $\left( {3,2} \right),\left( { - 2, - 3} \right)$and $\left( {2,3} \right)$form a triangle or not. If yes, then name the type of triangle.

Last updated date: 18th Mar 2023

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Answer

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Hint: Use distance formula for finding the length of sides and then apply the condition for triangle.

The given points in the question are $\left( {3,2} \right),\left( { - 2, - 3} \right)$and$\left( {2,3} \right)$.

Let $\left( {3,2} \right)$is denoted as point$A$, $\left( { - 2, - 3} \right)$as point $B$and $\left( {2,3} \right)$as point $C$.

Now, we can use the distance formula for finding out the distance between two points $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$, it is:

$ \Rightarrow D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Applying this formula, we can find out the distance between points. We’ll get:

$

\Rightarrow AB = \sqrt {{{\left( {3 + 2} \right)}^2} + {{\left( {2 + 3} \right)}^2}} = \sqrt {50} = 5\sqrt 2 , \\

\Rightarrow BC = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {52} = 2\sqrt {13} , \\

\Rightarrow AC = \sqrt {{{\left( {3 - 2} \right)}^2} + {{\left( {2 - 3} \right)}^2}} = \sqrt 2 , \\

$

From this, we can see that:

$

\Rightarrow A{B^2} = {\left( {5\sqrt 2 } \right)^2} = 50, \\

\Rightarrow B{C^2} = {\left( {2\sqrt {13} } \right)^2} = 52, \\

\Rightarrow A{C^2} = {\left( {\sqrt 2 } \right)^2} = 2. \\

\Rightarrow B{C^2} = A{B^2} + A{C^2} \\

$

The sides are satisfying Pythagoras theorem. $AB$ is perpendicular to $AC$.

Therefore, the above points are forming a triangle and it is a right angled triangle.

Note: If one of the angles of a triangle is ${90^ \circ }$,then the triangle is called a right angled triangle.

If one of the angles of a triangle is greater than ${90^ \circ }$,then the triangle is called an obtuse angled triangle. And if all the angles of a triangle are less than ${90^ \circ }$,then the triangle is called acute angled triangle.

The given points in the question are $\left( {3,2} \right),\left( { - 2, - 3} \right)$and$\left( {2,3} \right)$.

Let $\left( {3,2} \right)$is denoted as point$A$, $\left( { - 2, - 3} \right)$as point $B$and $\left( {2,3} \right)$as point $C$.

Now, we can use the distance formula for finding out the distance between two points $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$, it is:

$ \Rightarrow D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Applying this formula, we can find out the distance between points. We’ll get:

$

\Rightarrow AB = \sqrt {{{\left( {3 + 2} \right)}^2} + {{\left( {2 + 3} \right)}^2}} = \sqrt {50} = 5\sqrt 2 , \\

\Rightarrow BC = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {52} = 2\sqrt {13} , \\

\Rightarrow AC = \sqrt {{{\left( {3 - 2} \right)}^2} + {{\left( {2 - 3} \right)}^2}} = \sqrt 2 , \\

$

From this, we can see that:

$

\Rightarrow A{B^2} = {\left( {5\sqrt 2 } \right)^2} = 50, \\

\Rightarrow B{C^2} = {\left( {2\sqrt {13} } \right)^2} = 52, \\

\Rightarrow A{C^2} = {\left( {\sqrt 2 } \right)^2} = 2. \\

\Rightarrow B{C^2} = A{B^2} + A{C^2} \\

$

The sides are satisfying Pythagoras theorem. $AB$ is perpendicular to $AC$.

Therefore, the above points are forming a triangle and it is a right angled triangle.

Note: If one of the angles of a triangle is ${90^ \circ }$,then the triangle is called a right angled triangle.

If one of the angles of a triangle is greater than ${90^ \circ }$,then the triangle is called an obtuse angled triangle. And if all the angles of a triangle are less than ${90^ \circ }$,then the triangle is called acute angled triangle.

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