If we move two straight lines $ f(x, y) = C $ and $ g(x, y) = D $ so that they intersect at (0, 0), they become $ f(x, y) = 0 $ and $g(x, y) = 0 $

Question

Let us start with these two equations of two lines:

$$ x + y = 4 $$ $$ x - y = 2 $$

They intersect at $ (x, y) = (3, 1) $.

Let us now translate (move) both lines so that they intersect at $ (0, 0) $. We need to move both lines by $ -3 $ along $ x $-axis and by $-1$ along $y$-axis. So the equations of the lines become.

$$ (x + 3) + (y + 1) = 4 $$ $$ (x + 3) - (y + 1) = 2 $$

This is equivalent to

$$ x + y = 0 $$ $$ x - y = 0 $$

Why do the RHS become 0 for both equations? This happens no matter which two intersecting lines we begin with. What is the geometrical interpretation of this?

Answer 1

The RHS is always zero for any line that goes through the origin.

Think about it this way, if $$ax + by = c$$ for $c\neq 0$ then $(0,0)$ isn't a solution to this equation since $$a(0)+b(0) = 0 \neq c.$$ Therefore the only way for the point (0,0) to be on your line is for the RHS to be 0.