How would you show that a triangle with vertices \[\left( {13, - 2} \right)\], \[\left( {9, - 8} \right)\], \[\left( {5, - 2} \right)\] is isosceles?

Hint: Here we will use the distance formula to find the length of each side of the given triangle. Then, we will compare the lengths of all three sides. If the length of any two of the sides is found to be equal and the length of the third side will be different, then the given triangle will be proved to be isosceles.

Complete step-by-step solution:
Let us label the vertices of the given triangle as \[A\left( {13, - 2} \right);B\left( {9, - 8} \right);C\left( {5, - 2} \right)\].
Now, we know from the distance formula that the distance between the two points is given by
\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \].
The length of a side of the triangle will be equal to the distance between the vertices at the end of that side.
Therefore, the length of the side \[AB\] is given by
\[AB = \sqrt {{{\left( {9 - 13} \right)}^2} + {{\left( { - 8 - \left( { - 2} \right)} \right)}^2}} \]
Subtracting the terms, we get
\[ \Rightarrow AB = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2}} \]
Applying the exponent on the terms, we get
\[ \Rightarrow AB = \sqrt {16 + 36} \]
Adding the terms, we get
\[ \Rightarrow AB = \sqrt {52} = 2\sqrt {13} \] ………………………………..\[\left( 1 \right)\]
Similarly, the length of the side \[BC\] is given by
\[BC = \sqrt {{{\left( {5 - 9} \right)}^2} + {{\left( { - 2 - \left( { - 8} \right)} \right)}^2}} \]
Subtracting the terms, we get
\[ \Rightarrow BC = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} \]
Applying the exponent on the terms, we get
\[ \Rightarrow BC = \sqrt {16 + 36} \]
Adding the terms, we get
\[ \Rightarrow BC = \sqrt {52} = 2\sqrt {13} \] ………………………..\[\left( 2 \right)\]
Similarly, the length of the side \[CA\] is given by
\[CA = \sqrt {{{\left( {13 - 5} \right)}^2} + {{\left( { - 2 - \left( { - 2} \right)} \right)}^2}} \]
Subtracting the terms, we get
\[ \Rightarrow CA = \sqrt {{8^2} + {0^2}} \]
Applying the exponent on the terms and adding them, we get
\[ \Rightarrow CA = 8\]………………………\[\left( 3 \right)\]
From equation \[\left( 1 \right)\], \[\left( 2 \right)\], and \[\left( 3 \right)\] we can say that
\[AB = BC \ne CA\]
Thus, in the given triangle two of the sides have equal lengths whereas the third side is of different lengths.
We know that an isosceles triangle is a triangle whose two of the three sides are equal in length, while the third side has a different length.
Hence, the triangle with the given vertices is proved to be isosceles.

Note: We can make a mistake by checking the length of only two sides of a triangle and not the thirds side. If the two sides are the same we can assume that the triangle is isosceles but if we don’t check the third side we might be wrong. This is because if the third side is equal to the other two sides then it will be an equilateral triangle and not isosceles. So, to compare the lengths of all of the three sides we must find the length of the third side also.