1. Intersection between two lines

Hi there. I got this exercise, where I must find the intersection between these two lines:

$L_1=2x-y=2$

$L_2=\begin{Bmatrix} x=-1+2\lambda & \mbox{ }& \\y=-1 & \mbox{}& \end{matrix}$

I've parametrized $L_1$

$P_0(1,0)\in{L_1}$ y $P_1(0,-2)\in{L_1}$

From these points of $L_1$ I found a vector parallel to L_1: $u=(-1,-2)$

And then: $L_1=\begin{Bmatrix} x=1-\lambda & \mbox{ }& \\y=-2\lambda & \mbox{}& \end{matrix}$

$L_1\cap{L_2}=\begin{Bmatrix} -1+2\lambda=1-\lambda & \mbox{ }& \\-1=-2\lambda & \mbox{}& \end{matrix}\Longrightarrow{\begin{Bmatrix} \lambda=\displaystyle\frac{3}{2} & \mbox{ }& \\\displaystyle\frac{1}{2}=\lambda & \mbox{}& \end{matrix}}$

Lambda should be in both equations $\displaystyle\frac{1}{2}$, but for x it gives me $\displaystyle\frac{3}{2}$, and I don't know where the error is.

Any help would be appreciated.

Well, I don't know where is the error on my latex code neither

2. Originally Posted by Ulysses
Hi there. I got this exercise, where I must find the intersection between these two lines:

$L_1=2x-y=2$

$L_2=\begin{Bmatrix} x=-1+2\lambda & \mbox{ }& \\y=-1 & \mbox{}& \end{matrix}$

I've parametrized $L_1$

$P_0(1,0)\in{L_1}$ y $P_1(0,-2)\in{L_1}$

From these points of $L_1$ I found a vector parallel to L_1: $u=(-1,-2)$

And then: $L_1=\begin{Bmatrix} x=1-\lambda & \mbox{ }& \\y=-2\lambda & \mbox{}& \end{matrix}$

$L_1\cap{L_2}=\begin{Bmatrix} -1+2\lambda=1-\lambda & \mbox{ }& \\-1=-2\lambda & \mbox{}& \end{matrix}\Longrightarrow{\begin{Bmatrix} \lambda=\displaystyle\frac{3}{2} & \mbox{ }& \\\displaystyle\frac{1}{2}=\lambda & \mbox{}& \end{matrix}}$

Lambda should be in both equations $\displaystyle\frac{1}{2}$, but for x it gives me $\displaystyle\frac{3}{2}$, and I don't know where the error is.

Any help would be appreciated.

Well, I don't know where is the error on my latex code neither

L1: 2x1-y1=2 or y1=2x1-2

L2: x=-1+2t
y=-1

To find the intersection point you need to do this: y1=y or 2x1-2=-1 and x1=1/2, so the intersection point is (1/2,-1).

Regards.