Coordinate of a Point Inside a Triangle

Question

Coordinates to my triangle are $A(-1,10)$, $B(-7,1)$, $C(5,2)$ using trigonometry I can find the distance between $A-B$, $B-C$, and $C-A$. There is a point inside this triangle which is $7$ units from point $A$, $5.5$ units from $B$ and $9$ units from $C$. How do I find this point?

Answer 1

Let the point you want to locate be $P(x,y)$, then you have been given the distances $PA, PB$ and $PC$ as $7, 5.5$ and $9$ respectively. So \begin{align*} PA^2&=(x+1)^2+(y-10)^2=7^2\\ PB^2&=(x+7)^2+(y-1)^2=(5.5)^2\\ PC^2&=(x-5)^2+(y-2)^2=9^2 \end{align*} From these equations you can get (by subtraction etc.) two linear equations in $x$ and $y$. If they have a solution, then you get your point $P$, otherwise the point doesn't exist.

Answer 2

There is a point inside this triangle which is $7$ units from point $A$, $5.5$ units from $B$ and $9$ units from $C$. How do I find this point?

Hint: All points which are $7$ units away from point $A$ form a $\dotsc$ with $\dotsc$ $=7$.