Points $A$ and $B$ are on the number plane. The vector $\overrightarrow{AB}$ is $4\vec{i} + \vec{j}$.
Point C is chosen so that the area of triangle ABC is $\dfrac{17}{2}$ square units and $\left|AC\right| = \sqrt{34}$.
Find all possible vectors of $\overrightarrow{AC}$.
Sorry the question and working is in keyboard script, I haven't been able to get MathJax working.
I've tried using the cosine rule and area rule to answer this question, $c^2 = a^2 + b^2 - 2ab\cos{\theta}$ but I can't figure out how to get simultaneous equations going with this way of thinking.
If you can get me on the right path, or if I'm just doing it completely wrong any help is greatly appreciated. Also sorry again for the crude use of keyboard script.
Take $A$ at the origin $(0,0)$ and $B$ at $(4,1)$.
Think of $AB$ as the base of the triangle. It has length $\sqrt{17}$, so you need a triangle with height $\sqrt{17}$. That means that $C$ must lie on one of the two lines parallel to $AB$ and a distance $\sqrt{17}$ from it. The requirement that $|AC|=\sqrt{34}$ then gives us two points on each line.
One passes through $(-1,4)$ and the other through $(1,-4)$. [Since the height is $|AB|$, we just need to rotate $AB$ through $\pm90^o$.]
The line through $(-1,4)$ is $(-1+4t,4+t)$. We need the distance $AC$ to be $\sqrt{34}$, so we need $t=\pm1$ giving the two points $(3,5)$ and $(-5,3)$.
Similarly, the line through $(1,-4)$ is $(1+4t,-4+t)$, giving the two points $(5,-3),(-3,-5)$.