Question

Find the cartesian equation of the line that 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 $\frac{x+3}{3}$=$\frac{y-4}{5}$=$\frac{z+8}{6}$
The direction ratios of the line, $\frac{x+3}{3}$=$\frac{y-4}{5}$=$\frac{z+8}{6}$, are 3, 5, and 6.

The required line is parallel to $\frac{x+3}{3}$=$\frac{y-4}{5}$=$\frac{z+8}{6}$

Therefore, its direction ratios are 3k, 5k, and 6k, where k≠0
It is known that the equation of the line through the point (x1,y1,z1) 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
$\Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}=k$