Question

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

Answer 1

It is given that the line passes through the point $(-2,4,-5)$ and is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ ....(1)
The direction ratios of the line (1) are $3,5$ and $6$.

The required line is parallel to equation (1).

Therefore, its direction ratios are $3k,5k$ and $6k$, where $k\ne 0$.

It is known that the equation of the line through the point $(x_1,y_1,z_1)$ and with direction ratios, $a,b,c$ is given by $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Therefore, the equation of the required line is,

$\frac{x+2}{3k}=\frac{y-4}{5k}=\frac{z+5}{6k}$

$\Rightarrow$ $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$