Y-intercept

The two two parallel lines each have a slope of $1$ , which makes an angle of $\pi/4$ radians with the $x-$ axis. If the distance between their $y-$ intercepts is $4$ , then the normal distance between the lines is $4 \cdot \cos(\pi/4) = 2\sqrt{2}$ . The triangle's altitude from the origin to it base coincides with the line $y=x$ with length $\Rightarrow \frac{1}{2}(2\sqrt{2})h = 6 \Rightarrow h = 3\sqrt{2}$ . The foot of this altitude lies on the point $(x_{0},x_{0})$ , and $\sqrt{x_{0}^2 + x_{0}^2} = \sqrt{2}x_{0} = 3\sqrt{2} \Rightarrow x_{0} = 3.$ The equation of the line through $(3,3)$ with slope = $-1$ computes to:

$y-3 = -(x-3) \Rightarrow \boxed{y = -x+6}$

with a $y$ -intercept of $6.$

distance between two paraller lines is 2root2 and from (0,0) to line AB distance is 6/root2 and now assum line x+y=c from AB and distance 6/root2 ans is c/root2=6/root2