1. The problem statement, all variables and given/known data Find the coordinates of the symmetric point of the point M(2,1,3) of the line \(\frac{x+2}{1}=\frac{y+1}{2}=\frac{z-1}{-1}\) 2. Relevant equations 3. 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.
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 text book they got: \((2,-3,5)\) Is my way correct?
