Chapter 9.5, Problem 3PSA

a.

To determine

Toshow: The triangle with vertices at $\left(8,4\right),\left(3,5\right)$ and $\left(4,10\right)$ is a right-angle triangle using distance formula.

Explanation of Solution

Given information:

The triangle with vertices at $\left(8,4\right),\left(3,5\right)$ , and $\left(4,10\right)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

D $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Proof:

Let A = $\left(8,4\right),B=\left(3,5\right)$ and C = $\left(4,10\right)$

  $\begin{array}{l}AB=\sqrt{{\left(8-3\right)}^2+{\left(4-5\right)}^2}\\ =\sqrt{25+1}\\ =\sqrt{26}\\ BC=\sqrt{{\left(4-3\right)}^2+{\left(10-5\right)}^2}\\ =\sqrt{1+25}\\ =\sqrt{26}\\ AC=\sqrt{{\left(8-4\right)}^2+{\left(4-10\right)}^2}\\ =\sqrt{16+36}\\ =\sqrt{52}\end{array}$

Side AC can be calculated using Pythagoras Theorem.

In right angle triangle ABC , we get

  $\begin{array}{l}{\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2\\ {\left(\sqrt{26}\right)}^2+{\left(\sqrt{26}\right)}^2={\left(AC\right)}^2\\ 26+26={\left(AC\right)}^2\\ \sqrt{52}=AC\\ \therefore AC=\sqrt{52}\end{array}$

The triangle with vertices at $\left(8,4\right),\left(3,5\right)$ and $\left(4,10\right)$ is a right angle triangle.

b.

To determine

To show: The triangle with vertices at $\left(8,4\right),\left(3,5\right)$ and $\left(4,10\right)$ is a right-angle triangle using slopes.

Explanation of Solution

Given information:

The triangle with vertices at $\left(8,4\right),\left(3,5\right)$ , and $\left(4,10\right)$ .

Formula used:

Slope of a line:

  $m=\frac{y_2-y_1}{x_2-x_1}$

  $m=$ slope.

  $(x_1,y_1)=$ coordinates of first point in the line.

  $(x_2,y_2)=$ coordinates of second point in the line.

Proof:

Slope of $\overline{AB}=\frac{5-4}{3-8}=-\frac{1}{5}$

Slope of $\overline{DA}=\frac{10-5}{4-3}=\frac{5}{1}=5$

The slopes are negative inverses, so the sides are perpendicular.

The triangle with vertices at $\left(8,4\right),\left(3,5\right)$ and $\left(4,10\right)$ is right triangle.