Find the relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3, 6)$ and $(-3, 4)$

Question

Find the relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3, 6)$ and $(-3, 4)$.

Accepted Answer

Let the given point be $P(x, y)$, and the two fixed points be $A(3, 6)$ and $B(-3, 4)$. According to the problem, point $P$ is equidistant from $A$ and $B$, which means: $$PA = PB$$ 
Squaring both sides to eliminate the square roots, we get: $$PA^2 = PB^2$$ 
Using the distance formula, we write: $$(x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2$$ $$(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2$$ 
Expand each binomial using algebraic identities: $$(x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16)$$ 
Cancel out the common terms ($x^2, y^2, 9$) from both sides: $$-6x - 12y + 36 = 6x - 8y + 16$$ 
Rearrange all terms to one side: $$6x + 6x - 8y + 12y + 16 - 36 = 0$$ $$12x + 4y - 20 = 0$$ 
Divide the entire equation by $4$ to simplify: $$3x + y - 5 = 0$$ 
Thus, the required relation between $x$ and $y$ is $3x + y - 5 = 0$.

Coordinate Geometry

❓ Sibling Questions in this Chapter