## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 2 - 2.1 - Linear Equations in Two Variables - 2.1 Exercises - Page 172: 106

#### Answer

$m_1=\frac{-1}{m_2}$

#### Work Step by Step

Use the Pythagorean theorem to find $d_1$ $d_1 = \sqrt {1^2+{m_1}^2}$ Use the Pythagorean theorem to find $d_2$ $d_2 = \sqrt {1^2+{m_2}^2}$ Use the Pythagorean theorem on $d_1$, $d_2$, and $m_1$-$m_2$ with $m_1$-$m_2$ as the hypotenuse. $(m_1-m_2)^2=(\sqrt {1+{m_1}^2})^2+(\sqrt {1+{m_2}^2})^2$ ${m_1}^2-2m_1*m_2+{m_2}^2=1+{m_1}^2+1+{m_2}^2$ $-2m_1*m_2=2$ $m_1=\frac{-1}{m_2}$

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