The problem says: Find the point (coordinates $(x,y)=~?$) which is symmetrical to the point $(4,-2) $ considering the given equation $y=2x-3$
I have found the perpendicular line-slope $y=-~\frac{1}{2}x$ and the
intersection point which is shown in the graph $\left(\frac{6}{5},\frac{-3}{5}\right)$
I'm somehow unable to find the $x$ and $y$ I have found the distance based on the point and point distance equation:$$d(p_1,p_2)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{\left(\frac{6}{5}-4\right)^2+\left(-\frac{3}{5}+2\right)^2}= \frac{7\sqrt{5}}{5}$$
$$d_1=d_2=d(p_1,p_2)$$
So now I know the distance, the min. distance to the unknown point is the same.
What is the easiest way to find the symmetrical $x$ and $y$ coordinates?
(This is a high school problem)
Best Answer
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:
$y = -\frac{1}{2}x$
$y = (-\frac{1}{2}) (-\frac{8}{5}$)
$y = \frac{8}{10}$
$y = \frac{4}{5}$