The line $l$ has gradient $-2$ and passes through $A=(3,5)$. $B$ is a point on $l$ such that $\overline{AB}=6\sqrt5$. Find coordinates of each possible points of $B$.
There are two $x$-coordinates and two $y$-coordinates and so far I have got to $(3-x)^2+(5-y)^2=180$. What do I do from here?
If you didn't want to go through the formal processes of solving a system of non-linear equations, you can recognize that moving two points down and one point to the right from $A$ gives a point on the line $\sqrt5$ units from $A$. Therefore, the points $6\sqrt5$ points from $A$ would be $(3\pm1\cdot6, 5\pm(-2)\cdot6)=\{(9,-7),(-3,17)\}$.
Hint:
If $B(h,k)$
$$\dfrac{k-5}{h-3}=-2$$
$$180=(k-5)^2+(h-3)^2=\{-2(h-3)\}^2+(h-3)^2=(h-3)^2(4+1)$$
