Question

show that the lines L1 and L2 are the same.
L1 is x=3-t and y=1+2t
L2 is x=-1+3t and y= 9-6t

Answer 1

......convert from parametric to regular than show they are the same...

Answer 2

so L1= P1(3,1)
P2 (2,3)
and L2= P1(-1,9)
P2( 2,3)

Answer 3

why are the P1's not the same

Answer 4

\[L1={{3}\choose{1}}+t{{-1}\choose{2}}\]
\[L2={{-1}\choose{9}}+t{{3}\choose{-6}}\]
\[L2=-\frac{1}{3}{{3}\choose{27}}-3t{{-1}\choose{2}}\]
you sure you aint got a typo in there?

Answer 5

if they are the same line we can scale L2 or L1 to get the other one, but as is there seems to be an issue

Answer 6

for L1 we have:\[\begin{align}
x&=3-t\implies t=3-x\\
y&=1+2t=1+2(3-x)=1+6-2x=7-2x
\end{align}\]for L2 we have:\[\begin{align}
x&=-1+3t\implies t=\frac{x+1}{3}\\
y&=9-6t=9-6(\frac{x+1}{3})=9-2(x+1)=9-2x-2=7-2x
\end{align}\]therefore L1 and L2 are the same line.