Question

Find the length of the hypotenuse of the right triangle whose vertices are given by the points $( - 2,1),(1,1)$and$(1,2)$
(A)$\sqrt {10} $
(B)$\sqrt 5 $
(C)$10$
(D)$2\sqrt 5 $

Answer 1

Hint: Let\[X \equiv ( - 2,1)\], $Y \equiv (1,1)$, and$Z \equiv (1,2)$. Compute the lengths of the sides XY, YZ and XZ using the distance formula:
Distance between the two points $P \equiv (a,b)$ and $Q \equiv (c,d)$ is given by
$ d(P,Q) \equiv \sqrt {{{(c - a)}^2} + {{(d - b)}^2}} $
Compare the lengths of the sides. The largest length is the required answer.

Complete step by step solution: Let’s look at what is given to us.
We have a right angled triangle.
We also have the coordinates of the vertices of this right triangle, namely, $( - 2,1),(1,1)$and$(1,2)$
We are asked to compute the length of the hypotenuse of the said right triangle.
The hypotenuse of a right triangle is the side of the triangle which is opposite to the right angle.
Suppose $\vartriangle XYZ$ is the given right triangle.
Also, let \[X \equiv ( - 2,1)\],$Y \equiv (1,1)$, and $Z \equiv (1,2)$
Here is a pictorial representation of a right angled triangle.

Since we do not know the vertex of $\vartriangle XYZ$ which forms a right angle, we will need to calculate the length of each side.

Since the coordinates of the vertices are given, we will use the distance formula:
Distance between the two points $P \equiv (a,b)$ and $Q \equiv (c,d)$ is given by
$d(P,Q) \equiv \sqrt {{{(c - a)}^2} + {{(d - b)}^2}} $
Therefore, in $\vartriangle XYZ$, let us consider the length of side XY.
$XY \equiv \sqrt {{{(1 - ( - 2))}^2} + {{(1 - 1)}^2}} = \sqrt {{{(1 + 2)}^2} + {{(0)}^2}} = \sqrt 9 = 3$
Similarly, we can find the lengths of sides YZ and XZ.
$YZ \equiv \sqrt {{{(1 - 1)}^2} + {{(2 - 1)}^2}} = \sqrt {{{(0)}^2} + {{(1)}^2}} = \sqrt 1 = 1$
$ZX\equiv \sqrt {{{(1 - ( - 2))}^2} + {{(2 - 1)}^2}} = \sqrt {{{(3)}^2} + {{(1)}^2}} = \sqrt {10} $
We know that the hypotenuse is the longest side of the right angled triangle.
So, let us compare the distances that we just calculated.
We know that $1 < 3$.
Also, we know that $9 < 10$.
Therefore, taking square roots on both the sides, we get $3 < \sqrt {10} $
Thus, we have $1 < 3 < \sqrt {10} $
This implies that the length of side XZ is greater than the lengths of the sides XY and YZ.
Hence XZ is the hypotenuse of the given triangle and its length is $\sqrt {10} $ units.

Note: While using the distance formula, one should be careful as some students tend to miss one of the steps.
The steps can be done one by one:
1) Substitute the values
2) Find the differences
3) Square the differences
4) Add the square values
5) Take the square root.