Suppose the line is non-vertical and has the equation $y=mx+b$, and let the point be $(x_0,y_0)$.
Then the distance from the point to the line is
$$d=\sqrt{(x-x_0)^2+(mx+b-y_0)^2}.$$
Since the square root is awkward to work with, we minimize $d^2$ instead.
$$f(x)=d^2=(x-x_0)^2+(mx+b,y_0)^2.$$
Now use the usual procedure with critical points and so on to minimize $f$.
Once you find the value of $x$ which minimizes $f$, you can obtain the point on the line which is closest to $(x_0,y_0)$. Of course, from there you can easily find the slope of the line through $(x_0,y_0)$ and the closest point.
I think you're over-thinking it. When you have the intersect point and another point, you just duplicate the difference to get the point on the other side.
$(\frac{6}{5}, -\frac{3}{5}) - (4, -2)$ is $(-\frac{14}{5}, \frac{7}{5})$
That is what you need to add to the point to get to your intersect point, so either add that do your intersect point, or double it and add it to the initial point to get the point on the other side:
$(-\frac{8}{5}, \frac{4}{5})$
Another option would just be to calculate the x difference and double it to get the opposite x coordinate and pop that into your perpendicular line equation.
Full:
Initial equation: $y = 2x - 3$
Point: $(4, -2)$
So any equation perpendicular to the initial one will have slope $-\frac{1}{2}$ as you suggest. You might have already found the full equation for the line to find your point, but plug in the $y$ value for the given point of -2 and solve for b:
$y = -\frac{1}{2}x$ + b
$b=y +\frac{1}{2}x$
$b=-2 + \frac{1}{2}4$
$b= 0$
So the formula for the perpendicular line passing through point $(4, -2)$ is $y = -\frac{1}{2}x$
Solving to find the intersection point to get the $x$ you set both equations to be equal and solve for x:
$2x - 3 = -\frac{1}{2}x$
$\frac{5x}{2} = 3$
$5x = 6$
$x = \frac{6}{5}$
Plugging that into either equation gives your intersect point:
$(\frac{6}{5},-\frac{3}{5})$
Ok, so now how do you find a point the same distance on the perpendicular line passing through $(4, -2)$? To get from $(4, -2)$ to $(\frac{6}{5}, -\frac{3}{5})$ you have to move $\frac{6}{5} - 4$ in the x direction and $-\frac{3}{5} - -2$ in the y direction, or $(-\frac{14}{5}, \frac{7}{5})$
Add that to the intersect point and you get the point on the opposite side, $(-\frac{8}{5}, \frac{4}{5})$. Plug the x value into your perpendicular equation:
$y = -\frac{1}{2}x$
$y = (-\frac{1}{2}) (-\frac{8}{5}$)
$y = \frac{8}{10}$
$y = \frac{4}{5}$
Best Answer
Let $D(x,y)$ be the distance from the point $(x,y)$ to our line. Call the quarter-disk $Q$. We calculate $$\iint_Q D(x,y) \,dA$$ and divide the result by the area of $Q$. Because of the symmetry, it seems best to use polar coordinates. But it would be worthwhile to compute using rectangular coordinates, and compare.
Suppose that point $P$ in our quarter-disk has polar coordinates $(r,\theta)$. Join the origin to $P$, and drop a perpendicular from $P$ to our line. The picture shows that the distance from $P$ to our line is $r\sin(\theta+\pi/4)$. So we want $$\iint_Q r^2 \sin(\theta+\pi/4)\,dr\,d\theta.$$
Remark: Add to your picture the point $P$, join it to the origin, and drop a perpendicular as suggested above. You will see from basic trigonometry that the distance from $P$ to the line $x+y=0$ is $r\sin(\theta+\pi/4)$. Then by the Addition Law for the sine function, $$r\sin(\theta+\pi/4)=r\sin(\pi/4)\cos\theta+r\cos(\pi/4)\sin \theta= \frac{1}{\sqrt{2}} \left(r\cos\theta+r\sin\theta\right)=\frac{1}{\sqrt{2}}(x+y).$$ This gives an explanation different from the Linear Algebra formula for the fact that the distance from $(x,y)$ to the line is $\frac{x+y}{\sqrt{2}}$.