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Symmetric point of line

  • #1

Homework Statement



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

[tex]\frac{x+2}{1}=\frac{y+1}{2}=\frac{z-1}{-1}[/tex]


Homework Equations




The Attempt at a Solution




Out from here:

[tex]\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}[/tex]

[tex]M_1(x_1,y_1,z_1)[/tex]

[tex]M_1(-2,-1,1) ; M_2(-1,1,0)[/tex]

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.
 

Answers and Replies

  • #2
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
[tex]\left\{\begin{matrix}
x+2y-z-1=0 & \\
3x-y+4z-29=0 &
\end{matrix}\right.[/tex]

I find the line using x=0.

The equation of the line is:

[tex]\frac{x}{7}=\frac{y-\frac{33}{7}}{-7}=\frac{z-\frac{59}{7}}{-7}[/tex]

Also I got:

[tex]\sqrt{(x-3)^2+(y-11)^2+(z-4)^2}=\sqrt{(x+5)^2+(y+13)^2+(z+2)^2}[/tex]

And I put the conditions in one system:
[tex]\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.
[/tex]

I get that point [tex](\frac{664}{77} ; \frac{-43}{11} ; \frac{-15}{77}[/tex]

And in my text book they got: [tex](2,-3,5[/tex]

Is my way correct?
 
  • #3
tiny-tim, HallsofIvy, can you confirm me?
 

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