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.
$(m_1-m_2)^2=(\sqrt {1+{m_1}^2})^2+(\sqrt {1+{m_2}^2})^2$
${m_1}^2-2m_1*m_2+{m_2}^2=1+{m_1}^2+1+{m_2}^2$
$-2m_1*m_2=2$
$m_1=\frac{-1}{m_2}$
