analytic geometryslope
What is the inclination of the line joining $(3, 0)$ and $(2, \sqrt{3}) $
Answer: $\frac{2\pi}{3}$
$m = \tan(\alpha)$
So,
$$m = \frac{y_1 – y_2}{x_1 – x_2}$$
$$\Rightarrow m = \frac{0 – \sqrt{3}}{ 3 – 2} = -\sqrt{3}$$
$$\Rightarrow -\sqrt{3} = \tan(\alpha)$$
How to get angle from negative slope?
HINT
The location of the points at which $AB$ substends a right angle is a $\color{red}{\text{circle}}$ of radius $2$ centered at $\color{red}{(1,0)}$ as shown below:
In the figure the black line through $P$ has two intersection point (green and blue) with the $\color{red}{\text{circle}}$ at the $\color{blue}{\text{point}}$ and at the $\color{green}{\text{point}}$.
The range of the straight lines form $P$ which intersect the $\color{red}{\text{circle}}$ are between the two white tangent lines. The tangent lines go through $P$ and the the intersection points of the white circle centered at $(3.5,0.5)$ and going through the center of the $\color{red}{\text{circle}}$. The center of the white circle is the midpoint of the interval joining $P$ and the center of the $\color{red}{\text{circle}}$.
Any point on $y^2=4x$ P$(t^2,2t)$
$$\dfrac{dy}{dx}_{(\text{ at }P)}=\dfrac4{2y}_{(\text{ at }P)}=\dfrac1t$$
$$\implies\dfrac1t=\dfrac{2t+1}{t^2+2}\iff t^2+t-2=0$$
If $t_1.t_2$ are the roots of the above equation, WLOG $t_1=1,t_2=-2$
So, $\tan\alpha=\left|\dfrac{\dfrac1{t_1}-\dfrac1{t_2}}{1+\dfrac1{t_1}\cdot\dfrac1{t_2}}\right|=\left|\dfrac{t_2-t_1}{t_1t_2+1}\right|$
Best Answer
There's no real problems here, we take \begin{align*} \theta &= \arctan(-\sqrt{3}) \\ &= -\arctan(\sqrt{3}) & \text{since } \arctan(x)=-\arctan(-x) \text{ for all } x \in \mathbb{R} \\ &= -\tfrac{\pi}{3}. \end{align*}
And, as we can see from the picture below, the angle should be negative:
The answer of $2\pi/3=\pi-|\theta|$ is instead the angle marked $\varphi$; this is the angle of inclination.