Given points $A(7,2)$, $B(-4,2)$, $P(x,2x)$ such that the distance from $A$ to $P$ is equal to the distance from $B$ to $P$, find $x$

Question

I have $A(7,2)$ and $B(-4,2)$, these 2 points in my graph.

I have another point $P(x,2x)$.

The distance between points $A$ and $P$ is equal to the distance between points $B$ and $P$.

Find $x$.

Answer 1

Let's define the distance between two points in a plane. If we have the points $A(x_1, y_1)$ and $B(x_2,y_2)$, then the distance between them if defined by: $$d(A,B) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ In this problem we are said that the distance between $A$ and $P$ is the same as the distance between $B$ and $P$, so applying the definition and equaling we get $$\sqrt{(7- x)^2 + (2 - 2x)^2} = \sqrt{(-4 - x)^2 + (2 - 2x)^2}$$ Square both sides, calcel the second sumand of each side and appling the square identity for subtraction: $$49 - 14x + x^2 = 16 + 8x + x^2$$ And solving this first degree equation (notice quatratic terms cancel each other out) we get: $$x = \frac32$$