Answer
(a) see explanations below.
(b) see explanations below.
(c) $8.13^\circ$
(d) see explanations below.
Work Step by Step
(a) Assume the line equation is given by $y=mx+b$, the x-intercept can be found as $x=-\frac{b}{m}$ which means that point $(-\frac{b}{m}, 0)$ is on the line. Let $P(x, y)$ on the line is another point to the right of the x-intercept, By definition, $tan\theta=\frac{\Delta y}{\Delta x}=\frac{y-0}{x+b/m}=\frac{my}{mx+b}$ Since $P(x,y)$ is one the line, $mx+b=y$, we have $tan\theta=\frac{my}{y}=m$
(b) Based on (a) above, we have $tan\theta_1=m_1$ and $tan\theta_2=m_2$. $tan\Phi=tan(\theta_2-\theta_1)=\frac{tan\theta_2-tan\theta_1}{1+tan\theta_1tan\theta_2}=\frac{m_2-m_1}{1+m_1m_2}$
(c) Given $m_1=\frac{1}{3}, m_2=\frac{1}{2}$, we have $tan\Phi=\frac{1/2-1/3}{1+1/6}=\frac{1}{7}$ Thus $\Phi=tan^{-1}\frac{1}{7}=8.13^\circ$
(d) When two lines are perpendicular to each other, we have $\Phi=\frac{\pi}{2}$ and $cot\Phi=0$ Since $cot\Phi=\frac{1}{tan\Phi}=\frac{1+m_1m_2}{m_2-m_1}$, we have $1+m_1m_2=0$ which gives $m_2=-\frac{1}{m_1}$
