Question

Show that the points $\left( {12,8} \right)$, $\left( { - 2,6} \right)$ and $\left( {6,0} \right)$ are the vertices of a right angled triangle.

Answer 1

Here we will first find the distance between the two points by using the distance formula. Then after finding all the sides we will substitute the value of sides in the formula of the Pythagoras theorem. If it satisfies the Pythagoras theorem then the points are the vertices of a right-angled triangle.

Formula Used:
We will use the following formulas:

1. The distance between the two points with coordinate $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ is given by the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$.
2. Pythagoras theorem given by ${\left( {{\rm{Hypotenuse}}} \right)^2} = {\left( {{\rm{Perpendicular}}} \right)^2} + {\left( {{\rm{Base}}} \right)^2}$

Complete step by step solution:
Given coordinates of the vertices are $\left( {12,8} \right)$, $\left( { - 2,6} \right)$ and $\left( {6,0} \right)$.
We will use the distance between two points formula to calculate the length of the sides of the triangle.
So we will find the distance between the points $\left( {12,8} \right),\left( { - 2,6} \right)\,$ using the distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$. Therefore, we get
${d_1} = \sqrt {{{\left( { - 2 - 12} \right)}^2} + {{\left( {6 - 8} \right)}^2}} = \sqrt {{{\left( { - 14} \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt {196 + 4} = \sqrt {200} = 10\sqrt 2$
Now we will find the distance between the points $\left( { - 2,6} \right),\left( {6,0} \right)$. Therefore, we get
${d_2} = \sqrt {{{\left( { - 6 - 2} \right)}^2} + {{\left( {0 - 6} \right)}^2}} = \sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 6} \right)}^2}} = \sqrt {64 + 36} = \sqrt {100} = 10$
Now we will find the distance between the points $\left( {12,8} \right),\left( {6,0} \right)$. Therefore, we get
${d_3} = \sqrt {{{\left( {6 - 12} \right)}^2} + {{\left( {0 - 8} \right)}^2}} = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 8} \right)}^2}} = \sqrt {36 + 64} = \sqrt {100} = 10$
Now we will apply the Pythagoras theorem to check whether the vertices are of the right angles triangle or not. Therefore we will apply Pythagoras theorem, we get
$\Rightarrow {d_2}^2 + {d_3}^2 = {10^2} + {10^2} = 200$
Also we know that
${d_1}^2 = {\left( {10\sqrt 2 } \right)^2} = 200$
Therefore we can clearly see that ${d_1}^2 = {d_2}^2 + {d_3}^2$.

Hence it satisfies the Pythagoras theorem. So, the vertices are the vertices of a right-angled triangle.

Note:
We should know that the Pythagoras theorem is applied only to right-angled triangles. Right-angled triangle is the triangle which has one of its angles equal to $90^\circ$. The hypotenuse is the longest side of the right angled triangle. We should know the basic condition of the Pythagoras theorem which states that the square of the hypotenuse is equal to the sum of the square of the other two sides of the triangle.