Chapter P, Problem 15RE

Solution

Show that the figure determined by the given points is right triangle

Yes, the given figure is a right triangle.

Given information:

Given points are $\left(-2,1\right),\left(3,11\right),\text{ and }\left(7,9\right)$

Formula used:

1. A triangle is right triangle if sum square of length of two sides of the triangle is equals to the square of third side, i.e. if length of the length of sides of the triangles are a, b, and c then it is right triangle if $a^2=b^2+c^2$
2. The distance between two points, $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$, is given by:

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

Calculation:

To proof that the triangle is right triangle use the property that a triangle is right triangle if sum square of length of two sides of the triangle is equals to the square of third side.

First, calculate the length of the sides of the triangle in given figure use the formula for the distance between two points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ that is

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

The length of the side with end points $\left(-2,1\right),\text{ and }\left(3,11\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(3-\left(-2\right)\right)}^2+{\left(11-1\right)}^2}=\sqrt{{\left(3+2\right)}^2+{\left(10\right)}^2}\\ =\sqrt{5^2+100}\\ =\sqrt{25+100}\\ =\sqrt{125}\\ =5\sqrt{5}\end{array}$

Similarly, the length of the side with end points $\left(-2,1\right),\text{ and }\left(7,9\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(7-\left(-2\right)\right)}^2+{\left(9-1\right)}^2}=\sqrt{{\left(7+2\right)}^2+8^2}\\ =\sqrt{9^2+64}\\ =\sqrt{81+64}\\ =\sqrt{145}\end{array}$

And, the length of the side with end points $\left(3,11\right),\text{ and }\left(7,9\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(7-3\right)}^2+{\left(9-11\right)}^2}=\sqrt{4^2+{\left(-2\right)}^2}\\ =\sqrt{16+4}\\ =\sqrt{20}\\ =2\sqrt{5}\end{array}$

Thus, the length of the sides of the triangle in given figure are $5\sqrt{5}$ , $\sqrt{145}$ , and $2\sqrt{5}$ .

Now note that

  ${\left(\sqrt{145}\right)}^2=145$ , and

  $\begin{array}{c}{\left(5\sqrt{5}\right)}^2=25\times 5\\ =125\end{array}$

And,

  $\begin{array}{c}{\left(2\sqrt{5}\right)}^2=4\times 5\\ =20\end{array}$

And,

  $\begin{array}{c}{\left(5\sqrt{5}\right)}^2+{\left(2\sqrt{5}\right)}^2=125+20\\ =145\end{array}$

Thus,

  ${\left(5\sqrt{5}\right)}^2+{\left(2\sqrt{5}\right)}^2={\left(\sqrt{145}\right)}^2$

Thus, it satisfies the property sum square of length of two sides of the triangle is equals to the square of third side.

Thus, given triangle is right triangle.