Equation of line parallel to the line  x−42=y+1−3=z+108and passing through the point −1,2,3is

# Equation of line parallel to the line  $\frac{x-4}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$and passing through the point $\left(-1,2,3\right)$is

1. A

$\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z-3}{8}$

2. B

$\frac{x+1}{2}=\frac{y-2}{-3}=\frac{z-3}{8}$

3. C

$\frac{x-4}{-1}=\frac{y+1}{2}=\frac{z+10}{3}$

4. D

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

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### Solution:

The equation of the line passing through the point  $\left({x}_{1},{y}_{1},{z}_{1}\right)$and parallel to the line $\frac{x-{x}_{2}}{a}=\frac{y-{y}_{2}}{b}=\frac{z-{z}_{2}}{c}$is$\frac{x-{x}_{1}}{a}=\frac{y-{y}_{1}}{b}=\frac{z-{z}_{1}}{c}$

Here the given point is  $\left(-1,2,3\right)$and the equation of the line is$\frac{x-4}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$

Therefore, the equation of the required line is  $\overline{)\frac{x+1}{2}=\frac{y-2}{-3}=\frac{z-3}{8}}$

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