I am working on this problem in my math book and need some help from you:
Let $l: -2x + 3y - 2 = 0$ be a line in $\mathbb{R_2}$ and let $P_1 = (4,-1)$ be a point. Find the coordinates of the point $P_2$ which is the mirror image of $P_1$ on the line $l$.
Like this:
My approach to this problem:
The first way is to look at the line equation for $l$. From the equation you can quickly see what the normal vector to the line would be: $\vec{u} = (-2, 3)$. So then we know that there is a line $l_1 = P_1 + t \vec{u}$ that is orthogonal to the line $l$.
That means that for some number $t_0$ the point $Q = (x_0, y_0) = P_1 + t_0\vec{u} = (4-2t_0, -1 + 3t_0)$ is on the line $l$.
Using the line equation this means that $-2(4-2t_0) +3(-1 + 3t_0) -2 = 0$ which gives $t_0 = 1$.
We can then find the point $P_2$ by using the following formula $P_2 = P_1 + 2t_0\vec{u}$.
The second way (this approach gives me problems)
Is to write the original line equation on parameter form (by setting $y$ to $t$):
$$x = \frac{3t}{2} - 1$$ $$y = t $$
The point $Q$ will then have the coordinates $Q = (\frac{3t_0}{2}-1, t_0)$
The vector $\vec{P_1Q}$ will have to coordinates $\vec{P_1Q} = (5-\frac{3t_0}{2}, -1 -t_0)$
To find the value of $t_0$ where $\vec{P_1Q}$ is orthogonal to $l$ we can calulate the dot product of the direction vector of the line $l$ and the vector $\vec{P_1Q}$. This gives us: $\vec{P_1Q} \cdot (\frac{3t_0}{2}-1, t_0) = 0$, this gives us:
$$(5-\frac{3t_0}{2}, -1 -t_0)\cdot(\frac{3t_0}{2}-1, t_0)$$
Now this $t_0$ should give us the same coordinates that the first approach gave us for the point $Q = (4-2,-1+3) = (2, 2)$. I can't seem to get this result, can someone please help me what I did wrong here?
Thank you!
