The symmetric equation of the line passing through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is
A$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$
B$\dfrac{x+x_1}{x_2+x_1}=\dfrac{y+y_1}{y_2+y_1}=\dfrac{z+z_1}{z_2+z_1}$
C$\dfrac{x-x_2}{x_1}=\dfrac{y-y_2}{y_1}=\dfrac{z-z_2}{z_1}$
D$(x-x_1)+(y-y_1)+(z-z_1)=0$
Answer & Solution
Correct answer: A. $\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$
For a line through two points, the direction ratios are $(x_2-x_1,\; y_2-y_1,\; z_2-z_1)$. Substituting these into the point-direction form gives $\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$.
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