Question

Prove that the points (3,3),(3,7) and (9,3) are the vertices of a right triangle

Answer 1

The points (3,3), (3,7) and (9,3) do not form a right angled triangle

Solution:

Given vertices of triangle are (3, 3) , (3, 7) and (9, 3)

Let the triangle be ABC

Let the vertices be,

A = (3, 3)

B = (3, 7)

C = (9, 3)

For a triangle to be right angled triangle, it must obey Pythagoras theorem

Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle.

Therefore by above definition,

$AC^2 = AB^2 + BC^2$

Let us use the distance formula

$D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

The distance between A and C is given as:

$\text{ Here } x_1 = 3 ; y_1 = 3 ; x_2 = 9 ; y_2 = 3$

Therefore,

$\begin{array}{l}{D=\sqrt{(9-3)^{2}+(3-3)^{2}}} \\\\{D=\sqrt{6^{2}+0}=\sqrt{36}=6}\end{array}$

Thus length of AC is 6 units

The distance between A and B is given as:

$\text{ Here } x_1 =3 ; y_1 = 3 ; x_2 = 3; y_2 = 7$

$\begin{array}{l}{D=\sqrt{(3-3)^{2}+(7-3)^{2}}} \\\\{D=\sqrt{0+4^{2}}=\sqrt{16}=4}\end{array}$

The distance between B and C is given as:

$\text{ Here } x_1 = 3 ; y_1 = 7 ; x_2 = 9 ; y_2 = 3$

$\begin{aligned}&D=\sqrt{(9-3)^{2}+(3-7)^{2}}\\\\&D=\sqrt{6^{2}+(-4)^{2}}=\sqrt{36+16}=\sqrt{52}\end{aligned}$

Thus by Pythagorean theorem,

$AC^2 = AB^2 + BC^2$

Substituting the obtained values we get,

$6^2 = 4^2 + (\sqrt{52})^2\\\\ 36 = 16 + 52\\\\36 \neq 68$

Thus the vertices does not form a right angled triangle