Coordinate geometry #1

Using the Intersecting Chords Theorem (the $x$ -axis and the $y$ -axis are chords of the circle that meet at the origin), we deduce that $ac = bd$ . Since $a^2 + c^2 = b^2 + d^2$ , we deduce that $(a+c)^2 = (b+d)^2$ and $(a-c)^2 = (b-d)^2$ , so that $a + c \; = \; b + d \hspace{2cm} aa - c \; = \; \pm(b - d)$ If $a - c = b - d$ we deduce that $a=b$ and $c=d$ . This means that both lines have gradient $-1$ . If the two lines are distinct, they never meet each other. If they are identical, then they only meet the $x$ - and $y$ -axes at two points, not four . Thus this case must be excluded (even though it is the case that has the lines intersecting at points like $(1,-1)$ and $(2,-2)$ .

Thus we deduce that $a-c = d-b$ , so that $a=d$ and $b=c$ . It is now easy to see that these two lines meet in the first quadrant somewhere along the line $x=y$ , and so (by suitable scaling) the points $(1,1)$ and $(3,3)$ are possible points of intersection. Thus the only possible solutions are $\boxed{\mathrm{A\;and\;D}}$ .