Question

Show that $(3,2),\,\,(2, - 3),\,\,(0,0)$ are the vertices of a right- angled triangle using the distance formula.

Answer 1

Hint: We have to show that $(3,2),\,\,(2, - 3),\,\,(0,0)$ are the vertices of a right- angled triangle using the distance formula. A right- angled triangle is a triangle in which one angle is of $90$ degrees and according to the Pythagoras theorem the square of the longest side is equal to the sum of squares of other two. We will use the distance formula between two points $({x_1},{y_1})$ and $({x_2},{y_{2)}}$ which can be expressed as $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ where $d$ denotes the distance between the two points.

Complete step by step answer:
A right- angled triangle is a triangle in which one angle is of $90$ degrees and according to the Pythagoras theorem the square of the longest side is equal to the sum of squares of other two.
Now, we have to show that $(3,2),\,\,(2, - 3),\,\,(0,0)$ are the vertices of a right- angled triangle using distance formula.
Let the given points be named as $A(3,2),\,\,B(2, - 3)$ and $C\,(0,0)$
Now we will find the length of $AB,\,BC$ and $CA$ using the distance formula.
We know that distance formula is given by $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Now, we will calculate the length of $A(3,2)$ and $B(2, - 3)$.so,
$ \Rightarrow AB = \sqrt {{{(2 - 3)}^2} + {{( - 3 - 2)}^2}} $
Solving the above equation. We get,
$ \Rightarrow AB = \sqrt {{{(1)}^2} + {{( - 5)}^2}} $
Solving squares of the numbers. We get,
$ \Rightarrow AB = \sqrt {1 + 25} $
$ \Rightarrow AB = \sqrt {26} \,units$
Now, we will calculate the length of $B(2, - 3)$and $C(0,0)$. So,
$ \Rightarrow BC = \sqrt {{{(0 - 2)}^2} + {{(0 + 3)}^2}} $
Solving the above equation. We get,
$ \Rightarrow BC = \sqrt {{{( - 2)}^2} + {{(3)}^2}} $
Solving squares of the numbers. We get,
$ \Rightarrow BC = \sqrt {4 + 9} $
$ \Rightarrow BC = \sqrt {13} \,units$
Now, we will calculate the length of $C(0,0)$and $A(3,2)$. So,
$ \Rightarrow CA = \sqrt {{{(3 - 0)}^2} + {{(2 - 0)}^2}} $
Solving the above equation. We get,
$ \Rightarrow CA = \sqrt {{3^2} + {2^2}} $
Solving squares of the numbers. We get,
$ \Rightarrow CA = \sqrt {9 + 4} $
$ \Rightarrow BC = \sqrt {13} \,units$
Now, for right angled triangles they must satisfy that the square of the longest side is equal to the sum of squares of the other two. So,
${\left( {\sqrt {13} } \right)^2} + {\left( {\sqrt {13} } \right)^2} = {\left( {\sqrt {26} } \right)^2}$
Solving the squares of the numbers. We get,
$ \Rightarrow 13 + 13 = 26$
Therefore, ${(BC)^2} + {(CA)^2} = {(AB)^2}$
Hence, the condition satisfied and the given points are the vertices of a triangle.

Note:
Before solving these types of problems we must be aware of all the properties of triangles so that we can prove these types of questions and then we can apply the distance formula carefully. Most of the students are confused between the subtraction sign and addition sign in the distance formula. If signs get wrong the answer gets wrong and we don’t get the required result. From the above data we can also conclude that the above triangle is an isosceles triangle as its two sides are equal and one is different from the other two.