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Question

Find the Cartesian equation of the line which passes through the point (-2,4,-5) and is parallel to the line $$\dfrac{x+3}{3} = \dfrac{4-y}{5} = \dfrac{z+8}{6}$$.


Solution

The ratio of the given line are (3, 5, 6). The required line passes through the point A(-2, 4, -5) and parallel to $$\frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}$$.
So, the cartesian equation of the required line are
$$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$
The equation is $$\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}$$

Mathematics
RS Agarwal
Standard XII

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