On which of the following lines lies the point of intersection of the line x-42=y-52=z-31 …

Question

On which of the following lines lies the point of intersection of the line $\dfrac{x-4}{2}=\dfrac{y-5}{2}=\dfrac{z-3}{1}$ and the plane $x+y+z=2$?

Accepted Answer

Point on the line: $(4+2t,5+2t,3+t)$; substituting into the plane:
$$(4+2t)+(5+2t)+(3+t)=2\implies 12+5t=2\implies t=-2 \implies (0,1,1).$$
Testing option C: $\dfrac{0-1}{1}=-1,\ \dfrac{1-3}{2}=-1,\ \dfrac{1+4}{-5}=-1$ --- all equal, so the point lies on line C.

Solutions · 1