Given the line $d : 2x + 3y + 4 = 0$, find the equation of a line $d1$ passing through the point $M_0(2, 1)$, in the following situations: (a) $d1$ is parallel with $d$; (b) $d1$ is orthogonal on $d$; (c) the angle determined by $d$ and $d1$ is $ϕ = π /4$.
Can somebody give me some tips please on how to solve these?
I've only managed to calculated tha $m_d= -\frac{2}{3}$.
In your case, you could use: \begin{equation*} \frac{y-y_0}{x-x_0}=m_{d1} \end{equation*}
Where:
\begin{align*} x_0&=2\\y_0&=1 \end{align*}
Then:
\begin{equation*} \frac{y-1}{x-2}=m_{d1} \end{equation*}
Case (a): $m_{d1}=-\dfrac{2}{3}$, Since $m_{d1}=m_{d}$
Case (b): $m_{d1}=\dfrac{3}{2}$, Since $m_{d1}m_{d}=-1$
Case (c): To find $m_{d1}$ you could use:
$\tan(\alpha+\pi/4)=\dfrac{\tan(\alpha)+\tan(\pi/4)}{1-\tan(\alpha)\tan(\pi/4)}$
being: $\tan(\alpha)=m_d$ and $\tan(\alpha+\pi/4)=m_{d1}$ we have:
$m_{d1}=\dfrac{m_d+1}{1-m_d}$
then:
$m_{d1}=\dfrac{2}{15}$