Q32. - Precalculus Enhanced with Graphing Utilities

Question

Show that the points $A=\left(-2,0\right) , B=\left(-4,4\right)$ and $C=\left(8,5\right)$

 are the vertices of a right triangle in two ways:

(a) By using the converse of the Pythagorean Theorem .

(b) By using the slopes of the lines joining the vertices .

Step-by-Step Solution

Answer

(a) A ,B ,C are the vertices of a right angle triangle because the sum of the square of two side is equal to the square of third side.

(b) The product of slopes of two line is equal to $-1$ therefore A,B,C are the vertices of a right angle triangle.

Step 1. Given information.

Consider the given information such that the A$\left(-2,0\right)$ ,B$\left(-4,4\right)$ and C$\left(8,5\right)$ are the vertices of right angle triangle.

Step 2. Using the pythagoras theorem to show that the points are vertices of right angle triangle by using distance formula. d = x 2 - x 1 2 + y 2 - y 1 2

$$\begin{gathered} d\left(AB\right)=\sqrt{{\left(-4+2\right)}^2+{\left(4-0\right)}^2} \\ d\left(AB=\sqrt{20}\right) \\ d\left(BC\right)=\sqrt{{\left(8+4\right)}^2+{\left(5-4\right)}^2} \end{gathered}$$

Further simplify

$$\begin{gathered} d\left(BC\right)=\sqrt{145} \\ d\left(CA\right)=\sqrt{{\left(-2-8\right)}^2+{\left(0-5\right)}^2} \\ d\left(CA\right)=\sqrt{125} \end{gathered}$$

For further simplification using pythagoras theorem.

$$\begin{gathered} \left(A{B}^2\right)+{\left(CA\right)}^2={\left(BC\right)}^2 \\ {\left(\sqrt{20}\right)}^2+{\left(\sqrt{125}\right)}^2={\left(\sqrt{145}\right)}^2 \\ 20+125=145 \\ 145=145 \end{gathered}$$

Step 3. Using product of slopes.

Slope of $\left(AB\right)$ that is $m_1=\frac{y_2-y_1}{x_2-x_1}$

Substitute the values of $x_{1 } ,x_2$ and $y_1 ,y_2$.

Therefore the value of $m_1=-2$ $m_2=\frac{1}{12}$and $m_3=\frac{1}{2}$.

Further simplify .

 $m_1{m}_3=-1$

Line segments $AB$and $CA$are perpendicular to each other therefore the given triangle is right triangle at $A\left(-2,0\right) ,B\left(-4,4\right)$ and $C\left(8,5\right)$.