Chapter 13.3, Problem 11CE

To determine

To find: The reason for each step.

Answer to Problem 11CE

The reason for each step is given below.

Explanation of Solution

Given information:

The two non vertical lines are perpendicular, then the product of the slopes is $-1$ .

  $l_1$ has slope $m_1$ , $l_2$ has slope $m_2$ , $l_1\perp l_2$ .

Key points of proof,

  $1.$ Draw the vertical segments shown.

  $2.$ $\frac{u}{v}=\frac{v}{w}$

  $3.$ $m_1=\frac{v}{u}$

  $4.$ $m_2=-\frac{v}{w}$

  $5.$ $m_{1.}{m}_2=\left(\frac{v}{u}\right).\left(-\frac{v}{w}\right)$

  $6.$ $m_{1.}{m}_2=\left(\frac{v}{u}\right).\left(-\frac{v}{w}\right)=-1$

  

figure $\left(1\right)$

Calculation:

  

Figure $\left(2\right)$

Consider the triangles, $\Delta BAD$ and $\Delta ABC$ ,

  $\angle BAD+\angle ABD={90}^{\text{o}}$ …. $\left(1\right)$

  $\angle ABD+\angle DBC={90}^{\text{o}}$ …. $\left(2\right)$

Solve $\left(1\right)$ and $\left(2\right)$ ,

  $\begin{array}{c}\angle BAD+\angle ABD=\angle ABD+\angle DBC\\ \angle BAD=\angle DBC\end{array}$

So,

  $\angle ADB=\angle BDC\left(\text{each}{90}^{\text{o}}\right)$

  $\Delta BAD\sim \Delta CBD$

Therefore, $\Delta BAD$ is similar to $\Delta CBD$ by $\text{AA}$ criterion.

Corresponding sides of similar triangles are proportional,

  $\frac{\overline{AD}}{\overline{BD}}=\frac{\overline{BD}}{\overline{CD}}$

Substitute the values,

  $\frac{u}{v}=\frac{v}{w}$ …. $\left(1\right)$

The slope of the line $l_1$ passing through $A$ and $B$

  $m_1=\frac{v}{u}$

The slope of the line $l_2$ passing through $B$ and $C$

  $m_2=-\frac{v}{w}$

The line $A$ and $B$ is perpendicular to the line through $B$ and $C$ .

So,

  $m_{1.}{m}_2=\left(\frac{v}{u}\right).\left(-\frac{v}{w}\right)$

Substitute the values from equation $\left(1\right)$

  $\begin{array}{c}{m}_{1.}{m}_2=\frac{v}{u}\times \left(-\frac{v}{w}\right)\\ m_{1.}{m}_2=\frac{v}{u}\times -\frac{u}{v}\\ m_{1.}{m}_2=-1\end{array}$

Therefore, $m_{1.}{m}_2=-1$ is proved.