Chapter 0.1, Problem 52E

To determine

To find: Find the equation of line passing through $\left(-2,-1\right)$ parallel to the line $y=5-3\left(x+4\right)$ .

Answer to Problem 52E

The side $AC$ is the hypotenuse.

Explanation of Solution

Given information:

It is given that the three vertices of triangle are $A(-3,10),B(1,3),C\left(15,11\right)$

Concept used: The condition for two lines are to be perpendicular is $m_1\cdot m_2=-1$

Calculation:

The slope of the all three lines of the triangle is to be found.

The slope of line that passes through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Let’s take the points $A\left(-3,10\right)$ and $B\left(1,3\right)$ .

The slope of line $AB$ will be:

  $\begin{array}{l}{m}_1=\frac{3-10}{1-\left(-3\right)}\\ m_1=\frac{-7}{4}\end{array}$

Let’s take the points $B\left(1,3\right)$ and $C\left(15,11\right)$

The slope of line $BC$ will be:

  $\begin{array}{l}{m}_2=\frac{11-3}{15-1}\\ m_2=\frac{8}{14}\\ m_2=\frac{4}{7}\end{array}$

Let’s take the points $A\left(-3,10\right)$ and $C\left(15,11\right)$

The slope of line $AC$ will be:

  $\begin{array}{l}{m}_3=\frac{11-10}{15-\left(-3\right)}\\ m_3=\frac{1}{18}\\ m_3=\frac{1}{18}\end{array}$

All of the three line if any of two lines fulfill the condition of perpendicularity then they will be perpendicular to each other and the triangle is right angle triangle.

Check the condition of perpendicularity between lines $AB$ and $BC$ .

Substitute the value of $m_1$ and $m_2$ in $m_1\cdot m_2=-1$

  $\begin{array}{l}\frac{-7}{4}\cdot \frac{4}{7}=-1\\ -1=-1\end{array}$

Hence these two lines are perpendicular. Therefore, the triangle is right angle triangle.

Now the length of the sides to be found.

Apply the formula of the distance between two points:

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

The distance between $A\left(-3,10\right)$ and $B\left(1,3\right)$ will be:

  $\begin{array}{l}AB=\sqrt{{\left(1-\left(-3\right)\right)}^2+{\left(3-10\right)}^2}\\ AB=\sqrt{{\left(1+3\right)}^2+{\left(-7\right)}^2}\\ AB=\sqrt{{\left(4\right)}^2+49}\\ AB=\sqrt{16+49}\\ AB=\sqrt{65}\\ AB=8.06\end{array}$

The distance between $B\left(1,3\right)$ and $C\left(15,11\right)$ will be:

  $\begin{array}{l}BC=\sqrt{{\left(15-1\right)}^2+{\left(11-3\right)}^2}\\ BC=\sqrt{{\left(14\right)}^2+8^2}\\ BC=\sqrt{196+64}\\ BC=\sqrt{260}\\ BC=16.12\end{array}$

The distance between $A\left(-3,10\right)$ and $C\left(15,11\right)$ is:

  $\begin{array}{l}AC=\sqrt{{\left(15-\left(-3\right)\right)}^2+{\left(11-10\right)}^2}\\ AC=\sqrt{{\left(15+3\right)}^2+1^2}\\ AC=\sqrt{{\left(18\right)}^2+1}\\ AC=\sqrt{324+1}\\ AC=\sqrt{325}\\ AC=18.025\end{array}$

The largest distance is of $AC$ hence this is the hypotenuse.