What is the third condition for finding the symmetric point of a line?

[Physicsissuef]

Homework Statement

Find the coordinates of the symmetric point of the point M(2,1,3) of the line

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

Homework Equations

The Attempt at a Solution

Out from here:

$$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$

$$M_1(x_1,y_1,z_1)$$

$$M_1(-2,-1,1) ; M_2(-1,1,0)$$

I got two conditions

lets say that the point we need to find is N.

M_1N=MM_1

and

M_2N=MM_2

How will I find the 3-rd condition? I tried also with normal distance from M to the line to be equal with the normal distance of N to the line... Please help... Thank you.

 

[Physicsissuef]

Ok I solved this, using the 3-rd condition MN=2*distance from the point to line... But I have another task:

Find point at equal distance from the points A(3,11,4) and B(-5,-13,-2) at the line
$$\left\{\begin{matrix} x+2y-z-1=0 & \\ 3x-y+4z-29=0 & \end{matrix}\right.$$

I find the line using x=0.

The equation of the line is:

$$\frac{x}{7}=\frac{y-\frac{33}{7}}{-7}=\frac{z-\frac{59}{7}}{-7}$$

Also I got:

$$\sqrt{(x-3)^2+(y-11)^2+(z-4)^2}=\sqrt{(x+5)^2+(y+13)^2+(z+2)^2}$$

And I put the conditions in one system:
$$\left\{\begin{matrix} \sqrt{(x-3)^2+(y-11)^2+(z-4)^2}=\sqrt{(x+5)^2+(y+13)^2+(z+2)^2} & \\ \frac{x}{7}=\frac{y-\frac{33}{7}}{-7}=\frac{z-\frac{59}{7}}{-7} & \end{matrix}\right.$$

I get that point $$(\frac{664}{77} ; \frac{-43}{11} ; \frac{-15}{77}$$

And in my textbook they got: $$(2,-3,5$$

Is my way correct?

 

[Physicsissuef]

tiny-tim, HallsofIvy, can you confirm me?