THE QUESTION:
$A$ is the point $(-1,5)$. Let $(x,y)$ be any point on the line $y = 3x$.
a) Write an equation in terms of $x$ for the distance between $(x,y)$ and $A(-1,5)$.
b) Find the coordinates of the two points, $B$ and $C$ on the line $y = 3x$ which are a distance of root $74$ from $(-1,5)$.
I drew a diagram but that didn't really help me.
For part a) I tried to form an equation using pythagoras's theorem but I got a really obscure quadratic equation.
For part b) I am thinking that the root $74$ is the hypothenuse, then I tried to break down root $74$ but couldn't find any square numbers that would fit into it (only root $37$ and root $2$).
I might be thinking on the wrong lines so any help would be appreciated!
(Just for extra info, I started A-Level Maths in September).
Hint: use the Distance formula
Write distance between $A$ and point $P(x,y)$ in terms of $x$ and $y$, then, replace $y$ with $x$ using the relation for $x$ and $y$ ($y=3x$).
Now, once you've got this relation, equate it to $\sqrt{74}$ and solve for $x$. Again relate $x=y/3$ to get $y$.
This way, you'll get two $x$ values, and two corresponding values of $y$, giving you two points $B$ and $C$.
The distance is: $$\sqrt{(-1-x)^2+(5-y)^2}=\sqrt{(-1-x)^2+(5-3x)^2}$$
Now, equating and solving:
$$\sqrt{(-1-x)^2+(5-3x)^2}=\sqrt{74}$$ implies $$10x^2 -28x -48=0$$ which has solutions:
$$x=4,-6/5$$ Corresponding $y$ are:
$$y=12,-18/5$$