Question

Draw any 3 lines to have a point of concurrency.

Answer 1

the green point is point of concurrency where 3 lines meet

Answer 2

Answer:

We can draw the lines

$r: \lambda(0,1), ~ \lambda \in \mathbb{R}$

$s: \lambda(1,0), ~ \lambda \in \mathbb{R}$

$t: \lambda(1,1), ~ \lambda \in \mathbb{R}$

Step-by-step explanation:

Remember that given two points in the plane, we can draw a line that pass through them. So we only need to set one fixed point (let's say, the origin) and them give three other points (let's say, $(0,1), ~(1,0)$ and $(1,1)$) to draw the three lines.

The line $r$ that pass through (0,0) and (1,0) is given by $\lambda((1,0)-(0,0) = \lambda(1,0), ~ \lambda \in \mathbb{R}$.

The line $s$ that pass through (0,0) and (0,1) is given by $\lambda((0,1)-(0,0) = \lambda(0,1), ~ \lambda \in \mathbb{R}$.

The line $t$ that pass through (0,0) and (1,1) is given by $\lambda((1,1)-(0,0) = \lambda(1,1), ~ \lambda \in \mathbb{R}$.

These three lines have a point of concurrency at $(0,0)$ (to check this, one just need to set $\lambda = 0$.