Question

Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}.$

Answer 1

The equation of line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}$ can be re-written as

$\frac{x-{\displaystyle \frac{1}{2}}}{2}=\frac{y+{\displaystyle \frac{5}{3}}}{{\displaystyle \frac{2}{3}}}=\frac{z-2}{-3}$

The direction ratios of the line parallel to line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}$ are proportional to 2, $\frac{2}{3}$, $-$3.

Equation of the required line passing through the point ($-$1, 2, 1) having direction ratios proportional to 2, $\frac{2}{3}$, $-$3 is

$$\begin{gathered} \frac{x-\left(-1\right)}{2}=\frac{y-2}{{\displaystyle \frac{2}{3}}}=\frac{z-1}{-3} \\ =\frac{x+1}{2}=\frac{y-2}{{\displaystyle \frac{2}{3}}}=\frac{z-1}{-3} \end{gathered}$$