Chapter 13.3, Problem 23WE

a.

To determine

To Prove, $\left(-\frac{c}{a}\right)\cdot \left(-\frac{c}{b}\right)=-1$ .By using Pythagorean theorem.

Explanation of Solution

Given information: If $\overleftrightarrow{TU}\perp \overleftrightarrow{US}$ the product of the slopes of $\overleftrightarrow{TU}$ and $\overleftrightarrow{US}$ equals $-1$ .

  

Proof:The required proved is obtained as,

  $TU\perp US$

That means-

  $\begin{array}{l}T{U}^2+U{S}^2=T{S}^2\\ T{U}^2={\left(a-0\right)}^2+{\left(0-c\right)}^2=a^2+c^2\\ U{S}^2={\left(0-b\right)}^2+{\left(c-0\right)}^2=b^2+c^2\\ T{S}^2={\left(a-b\right)}^2+0^2={\left(a-b\right)}^2\\ a^2+c^2+b^2+c^2={\left(a-b\right)}^2\\ a^2+b^2+2c^2=a^2+b^2-ab\\ a^2+2c^2=a^2-ab\\ 2c^2=-ab\\ c^2=-ab\\ \left(-c\right)\cdot \left(-c\right)=-ab\\ \left(-\frac{c}{a}\right)\cdot \left(-\frac{c}{b}\right)=-1\end{array}$

Hence, proved $\left(-\frac{c}{a}\right)\cdot \left(-\frac{c}{b}\right)=-1$ .

b.

To determine

To prove $\overleftrightarrow{TU}\perp \overleftrightarrow{US}$ , if $\left(-\frac{c}{a}\right)\cdot \left(-\frac{c}{b}\right)=-1$ . By using the converse of the Pythagorean theorem.

Explanation of Solution

Given information: The equation, $\left(-\frac{c}{a}\right)\cdot \left(-\frac{c}{b}\right)=-1$ .

  

Proof:The required proved is obtained as,

  $\begin{array}{l}\left(-\frac{c}{a}\right)\cdot \left(-\frac{c}{b}\right)=-1\\ c^2=-ab\end{array}$

Observe

  $\begin{array}{l}T{U}^2={\left(a-0\right)}^2+{\left(0-c\right)}^2=a^2+c^2\\ U{S}^2={\left(0-b\right)}^2+{\left(c-0\right)}^2=b^2+c^2\\ T{S}^2={\left(a-b\right)}^2+0^2={\left(a-b\right)}^2\\ T{S}^2=a^2+b^2-2ab\\ T{S}^2=a^2+b^2-2\left(-c^2\right)\left[\because ab=-c^2\right]\\ T{S}^2=a^2+b^2+2c^2\\ T{S}^2=a^2+c^2+b^2+c^2\\ T{S}^2=T{U}^2+U{S}^2\end{array}$

Which is, $\overleftrightarrow{TU}\perp \overleftrightarrow{US}$

Hence, proved $\overleftrightarrow{TU}\perp \overleftrightarrow{US}$ .