Chapter 13.2, Problem 8CE

To determine

To find: The reason for each step.

Answer to Problem 8CE

The reason for each step is given below.

Explanation of Solution

Given information:

We can use any two points on a line to determine the slope of the line.

Key steps of proof

  $\begin{array}{l}1.\angle B\cong \angle A\\ 2.\angle 1\cong \angle 2\\ 3.\Delta LBN\sim \Delta DAJ\\ 4.\frac{BN}{AJ}=\frac{LB}{DA},or\frac{BN}{LB}=\frac{AJ}{DA}\end{array}$

  $5.$ The slope of $\overline{LN}$ equals $\frac{BN}{LB}$ , and the slope of $\overline{DJ}$ equals $\frac{AJ}{DA}$ .

  $6.$ Slope of $\overline{LN}$ equals to slope of $\overline{DJ}$ .

  

Figure- $\left(1\right)$

Calculation:

Consider the Figure- $\left(1\right)$ ,

  $\overline{LN}$ and $\overline{DJ}$ are the two lines.

  $1.$

  $\angle B\cong \angle A$

In the given figure $\angle B=\angle A={90}^{\circ }$

Therefore, the reason is that both the angles are right angles.

  $2.$

  $\angle 1\cong \angle 2$

In Figure $\left(1\right)$ ,

  $\overline{LB}\parallel \overline{AD}$

So, the corresponding angles are equal.

Therefore, the line $\overline{LB}$ is parallel to the line $\overline{AD}$ , So the corresponding angles are equal.

  $3.$

  $\Delta LBN\sim \Delta DAJ$

In the $\Delta LBN$ and $\Delta DAJ$ ,

  $\begin{array}{l}\angle B\cong \angle A\\ \angle 1\cong \angle 2\end{array}$

Therefore, they are the similar triangles.

  $4.$

  $\frac{BN}{AJ}=\frac{LB}{DA},or\frac{BN}{LB}=\frac{AJ}{DA}$

In the figure $\Delta LBN\sim \Delta DAJ$ are the similar triangles. So the sides of the similar triangles are equal.

Therefore, The corresponding sides are proportional because $\text{ΔLBN}$ is similar to the $\text{ΔDAJ}$ .

  $5.$

We know that a line makes an angle ${\text{n}}^{\text{o}}$ with the horizontal

So,

  $\left(\text{m}\right)=\text{tan}{\text{n}}^{\text{o}}$

Therefore, tangent of the angle,

  $\text{tan}{\text{n}}^{\circ }=\frac{\text{perpendicular}}{\text{base}}$

Substitute the values,

Slope of $\overline{LN}$

  $\overline{LN}=\frac{BN}{LB}$

And,

Slope of $\overline{DJ}$

  $\overline{LN}=\frac{AJ}{DA}$

Therefore, slope of $\overline{DJ}$ is $\frac{BN}{LB}$ similar to the slope of $\overline{DJ}$ is $\frac{AJ}{DA}$ .

  $6.$

In the given figure

  $\angle 1\cong \angle 2$

So,

  $\tan \angle 1\cong \tan \angle 2$

The slope of the line is given by the tangent of the angle.

Slope of the line $\left(\text{m}\right)$ ,

  $\left(m\right)=\tan n^o$

Therefore, the slope of the $\overline{LN}$ is equal to the slope of the $\overline{DJ}$ .