Question

Prove that the points (-3, 0), (1, -3) and (4, 1) are the vertices of a right angled isosceles triangle.

Answer 1

Hint: We will first name the points A, B and C and then find lengths of AB, BC and CA. We will observe that two of them are equal and thus, it is an isosceles triangle if it is a triangle. Now, for the right angled triangle, we will show that it follows Pythagorean Theorem and thus is right angled.

Complete step-by-step answer:
Let A = (-3, 0), B = (1, -3) and C = (4, 1) be the points given to us.
Now, we know that the distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by:
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ units
We have A = (-3, 0), B = (1, -3) and C = (4, 1) with us
So, the length of AB will be:- $AB = \sqrt {{{(1 + 3)}^2} + {{( - 3 - 0)}^2}} $ units
$ \Rightarrow AB = \sqrt {{4^2} + {3^2}} = 5$ units …………..(1)
Now, the length of BC will be:- $BC = \sqrt {{{(4 - 1)}^2} + {{(1 + 3)}^2}} $ units
$ \Rightarrow BC = \sqrt {{3^2} + {4^2}} = 5$ units …………..(2)
Now, the length of AC will be:- $AC = \sqrt {{{(4 + 3)}^2} + {{(1 - 0)}^2}} $ units
$AC = \sqrt {{7^2} + {1^2}} = \sqrt {50} $ units …………..(3)
We can clearly see that AB = BC. Hence, if ABC is a triangle, then it must be isosceles.
Let us now see if $\vartriangle ABC $ follows Pythagorean Theorem or not, if it does then it is definitely an isosceles triangle.
Pythagorean Theorem: It states that in a right angled triangle, \[{H^2} = {P^2} + {B^2}\], where H is the hypotenuse, P and B are the other two sides of triangle.
We see that \[A{C^2} = 50\] and also \[A{B^2} + B{C^2} = 25 + 25 = 50\].
Hence, \[A{B^2} + B{C^2} = A{C^2}\].
Hence, we have proved that it is a right angled isosceles triangle.

Note: The students must note that after equation (3), we wrote that if it is a triangle. We said that because we are not even sure if it is a triangle or not. They may all be a bunch of collinear points with B as the midpoint of A and C. So, when we proved that it follows Pythagorean, only then we will get to know whether it is actually a triangle or not.
To prove that these points are non collinear, you may find the area formed by these points using the formula:
$A = \left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|$ , where \[({x_1},{y_1}),({x_2},{y_2})\] and \[({x_3},{y_3})\] are the vertices of the triangle.
If the area comes out to be zero, then the points are collinear, otherwise they form a triangle.