Answer to Question #222321 in Analytic Geometry for mike

Question

A line passes through the points (7,-4,8) and (9,0,5)

i) Determine the parametric and symmetric equations of the line

ii)Find any other point on the line

iii)Show that the line does not pass through the origin

Answer 1

i) The collinear vector of a line is $\overline{v}=(9-7,0-(-4),5-8)=(2,4,-3).$ Then the symmetric equation of the line is

$\frac{x-7}{2}=\frac{y+4}{4}=\frac{z-8}{-3}.$

It follows that the parametric equation of the line is

$\begin{cases}x=7+2t\\y=-4+4t\\z=8-3t\end{cases}.$

ii) For $t=-1$ we have that

$x=7+2(-1)=5,\ y=-4+4(-1)=-8,\ z=8-3(-1)=11.$

Then the point $(5,-8,11)$ is on the line.

iii) If $y=0,$ then $-4+4t=0,$ and hence $t=1.$ On the other hand for $t=1,$ we have that $x=9\ne 0,$ and hence the line does not pass through the origin.