Question

Prove that the points $ \left( {7,10} \right),\left( { - 2,5} \right) $ and $ \left( {3, - 4} \right) $ are the vertices of an isosceles right triangle.

Answer 1

Hint: In order to prove that the given coordinates of vertices are of an isosceles triangle , use the distance formula and prove that the any two sides of the triangle are equal. Which is the necessary and sufficient condition for a triangle to be isosceles type.
Distance Formula:
 $ D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Complete step-by-step answer:
We are given the coordinates of points of a triangle as $ \left( {7,10} \right),\left( { - 2,5} \right) $ and $ \left( {3, - 4} \right) $
Let the triangle be ABC with points $ A\left( {7,10} \right),B\left( { - 2,5} \right) $ and $ C\left( {3, - 4} \right) $ .
As per the question, we have proven that the triangle ABC is an isosceles triangle.
Now to prove any triangle as an isosceles triangle we must know what is an isosceles triangle . So Isosceles is a triangle having its two sides equal in length.
Therefore, we have to find all the lengths of the sides of triangle ABC and then see that any of its two sides are equal or not.
In order to find the length of sides of a triangle from the coordinates of its vertices, we will be using distance formula .
According to distance formula , the distance $ PQ $ between point $ P\left( {{x_1},{y_1}} \right)and\,Q\left( {{x_2},{y_2}} \right) $ is given by
 $ PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Using the formula above we will the distances AB,BC,AC
1. Length for AB
As we know $ A\left( {7,10} \right) $ and $ B\left( { - 2,5} \right) $ .
Assuming A $ \left( {{x_1},{y_1}} \right) $ and B $ \left( {{x_2},{y_2}} \right) $ Putting these coordinates into the formula we get the distance AB as
 \[
  AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \\
  AB = \sqrt {{{\left( { - 2 - 7} \right)}^2} + {{\left( {5 - 10} \right)}^2}} \\
  AB = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( { - 5} \right)}^2}} \\
  AB = \sqrt {81 + 25} \\
  AB = \sqrt {106} \;
 \]
2. Length of BC-
We have B $ \left( { - 2,5} \right) $ and C $ \left( {3, - 4} \right) $
Same as above assume B $ \left( {{x_1},{y_1}} \right) $ and C $ \left( {{x_2},{y_2}} \right) $ , we have BC as
 \[
  BC = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \\
  BC = \sqrt {{{\left( {3 - \left( { - 2} \right)} \right)}^2} + {{\left( { - 4 - 5} \right)}^2}} \\
  BC = \sqrt {{{\left( {3 + 2} \right)}^2} + {{\left( { - 9} \right)}^2}} \\
  BC = \sqrt {{{\left( 5 \right)}^2} + 81} \\
  BC = \sqrt {25 + 81} \\
  BC = \sqrt {102} \;
 \]
As we can clearly see the length of side AB and BC are equal i.e. $ AB = BC = \sqrt {102} $ .
Hence we can say that the triangle ABC is an isosceles triangle.
Therefore, the triangle with vertices $ \left( {7,10} \right),\left( { - 2,5} \right) $ and $ \left( {3, - 4} \right) $ is an isosceles triangle.

Note: While using the distance formula, use the notations for the points like B $ \left( {{x_1},{y_1}} \right) $ and C $ \left( {{x_2},{y_2}} \right) $ for B $ \left( { - 2,5} \right) $ and C $ \left( {3, - 4} \right) $ . This will help students and will create no confusion while substituting the values in the formula.