A line $L_0: 2x+5y=11$ rotates about a point $P(\alpha, \beta)$ on the line $L_0$ such that $\alpha$ and $\beta$ are integers and $|\alpha+\beta|$ is least, through an angle $(-1)^n.n^{\circ}$ in $n^{th}$ second. If $L_0$ becomes $L_n$ after $n$ seconds, find perpendicular distance of $L_{180}$ from $(0,0)$
Clearly, $(\alpha,\beta)=(-2,3)$
The line will rotate by $(-1)^{180} 180^{\circ}=180^{\circ}$
I know how to rotate axes, but I don’t know the formula for rotating a line
There is an answer for rotation of line in https://math.stackexchange.com/a/1064909/690228 But I didn’t understand the last part ie. what should be done with $A_x$ and $A_y$, which hasn’t been clearly defined by the user. That user is no longer present on the this site, otherwise I could have approached him/her directly.
Hint: the line won't rotate to 180. What is happening is that each second, it rotates by $n(-1)^n$ degrees. Hence, in 180 seconds, the final angle would be
$$\theta = -1 + 2 -3 +4 -5 +6.... +180 = 90$$
But in how to deal with rotation -
It helps when rotating a line, to express it in terms of the point of rotation. In this case, since it's $(-2,3)$, we re-write the line as
$$L_0: 2(x+2) +5(y-3) = 0$$
Now, let us consider that the origin is located at the pivot, and consider the line in the translated axes
$$L_1: 2X + 5Y = 0$$
Now, it is convenient for rotations to consider our line aligned to the $x$ axis, as it is standard to measure angles w.r.t $x$ axis. Now, let us take a sample point $(X', Y')$ to understand this better
Just like every point has a perpendicular distance to each axis, for any point, it has a perpendicular distance to each of two perpendicular lines. In our case, the sample point is $(X', Y')$ and the two lines are $2X + 5Y = 0$ and $5X - 2Y = 0$
Hence, for the given point, the perpendicular distances are
$$D_x = \frac{2X' + 5Y'}{\sqrt{2^2 + 5^2}}$$ $$D_y = \frac{5X' - 2Y'}{\sqrt{2^2 + 5^2}}$$
Hence, another way we can express the point $(X',Y')$ is through it's distances from these two lines, as the point
$$(X^*, Y^*) = \left(\frac{2X' + 5Y'}{\sqrt{2^2 + 5^2}}, \frac{5X' - 2Y'}{\sqrt{2^2 + 5^2}}\right)$$
Now in our new coordinate system, our old line would be $X^* = 0$. Now if we were to rotate this by $\theta$, then the line would become
$$Y^* = X^* \tan \theta$$
For angle 90, this would simply be the $Y^*$ axis, i.e $Y^* = 0$
Hence, for angle 90, the line is
$$Y* = 0 \\ \implies \frac{5X - 2Y}{\sqrt{2^2 + 5^2}} = 0 \\ \implies \frac{5(x+2) - 2(y-3)}{\sqrt{2^2 + 5^2}} = 0$$