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)

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  • 2
    $\begingroup$ The distance calculation is not fully necessary. I have edited my answer to propose a simple process $\endgroup$ – Laurent Duval Nov 12 '16 at 19:35

The point $C$ with coordinates $\left(\frac{6}{5},\frac{-3}{5}\right)$ is the midpoint of the segment defined by $(x,y)$ (called $U$ for unknown) and $(4,-2)$ (called $A$). So $x_C = (x+4)/2$, and the same for $y_C $.

To rewrite the process: $C$ verifies $y_c = 2x_c-3$. Also, the vector $\vec{AC}$ is orthogonal to the slope of this line, hence the scalar product gives $(x_c-4).1+(y_c+2).2=0$, or $x_c+2y_c=0$. You now just have to plug that into $x=2x_c-4$ and $y=2y_c+2$, which you get from $\vec{UC}=\vec{CA}$.


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.


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:


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:

$y = -\frac{1}{2}x$

$y = (-\frac{1}{2}) (-\frac{8}{5}$)

$y = \frac{8}{10}$

$y = \frac{4}{5}$


Alternative rapid way: calling $x_0$ and $y_0$ the coordinates of the symmetric point, the simplest method is to note that the differences in the $x $- and $y $- coordinates between the searched point and $( \frac{6}{5}, -\frac{3}{5})$ must equal those between this last point and $(4,-2) $. Thus:

$$ 4-\frac{6}{5} = \frac{6}{5} -x_0$$

$$ y_0 - \left(-\frac{3}{5}\right) = -\frac{3}{5} - \left(-2\right) $$

Now solve these equations to find the coordinates.


It is not as complex as you think after you found the mid point.

The midpoint is the average of the two end points.

So $(x+4, y-2) = 2*(6/5, -3/5)$

  • $\begingroup$ That's one equation with two unknowns. Besides $(\frac 65, - \frac{-3}5)$ is a point, and it doesn't make sense to multiply a point by $2$. $\endgroup$ – Namaste Nov 12 '16 at 19:14
  • $\begingroup$ Those are two equations, one for x and one for y. You can also divide (x+4,y-2) by two, but that gives the same result. $\endgroup$ – Pieter21 Nov 12 '16 at 19:16
  • $\begingroup$ It does not make sense to multiply a real number with a coordinate. $\endgroup$ – Namaste Nov 12 '16 at 19:17
  • $\begingroup$ Technically, I take the average of two vectors to find the midpoint.. $\endgroup$ – Pieter21 Nov 12 '16 at 19:22
  • $\begingroup$ And actually, the distance approach is incorrect, because that would result in a circle of point that satisfy (x,y) $\endgroup$ – Pieter21 Nov 12 '16 at 19:31

Let $A = (4,-2)$ and $B = (\frac65, -\frac35)$, and let $C$ denote the mirror image of $A$ with respect to $B$ that we wish to find.

The key observation is that not only are the distances $d(A,B)$ and $d(B,C)$ equal, but the vectors $\vec{AB}$ and $\vec{BC}$ are parallel and of equal length, and thus also equal:


Thus, we can simply calculate: $$ \vec{BC} = \vec{AB} = B-A = \left(\frac65, -\frac35\right)-(4,-2) = \left(-\frac{14}{5}, \frac75\right) $$ and then: $$ C = B + \vec{BC} = \left(\frac65, -\frac35\right) + \left(-\frac{14}{5}, \frac75\right) = \left(-\frac85, \frac45\right).$$

  • $\begingroup$ This is awesome! Tnx. $\endgroup$ – eugene_sunic Nov 13 '16 at 15:23

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