Question

Find the coordinates of the point equidistant from points $A(1,2),B(3,-4)$, and $C(5,-6)$.

Answer 1

Step 1: Assume the co-ordinates of point $P$ according to the given condition

$A(1,2),$$B(3,-4)$$C(5,-6).$

Let $P(x,y)$ be the point equidistant from these three points.

$\Rightarrow$ $PA=PB=PC$

Step 2 : Apply distance formula

According to the distance formula , the distance between two points $M\left(x_1,y_1\right)$ and $N\left(x_2,y_2\right)$ is given as

$MN=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Applying distance formula to given co-ordinates we get,

$\sqrt{{(x-1)}^2+{(y-2)}^2}=\sqrt{{(x-3)}^2+{(y+4)}^2}=\sqrt{{(x-5)}^2+{(y+6)}^2}$

Squaring throughout the equation we get,

${(x-1)}^2+{(y-2)}^2$$={(x-3)}^2+{(y+4)}^2={(x-5)}^2+{(y+6)}^2$

$\Rightarrow x^2+1–2x+y^2+4–4y=x^2+9–6x+y^2+16+8y=x^2+25–10x+y^2+36+12y$

$\Rightarrow –2x–4y+5=-6x+8y+25=–10x+12y+61$

Step 3 :Comparing two of the three equations at a time

Let us take $PA=PB$

$$\begin{gathered} \Rightarrow –2x–4y+5=-6x+8y+25 \\ \Rightarrow –2x–4y+5+6x–8y–25=0 \\ \Rightarrow 4x–12y–20=0 \\ \Rightarrow x–3y–5=0-----------\left(1\right) \end{gathered}$$

Let us take PA = PC

$$\begin{gathered} \Rightarrow –2x–4y+5=–10x+12y+61 \\ \Rightarrow -2x–4y+5+10x–12y–61=0 \\ \Rightarrow 8x–16y–56=0 \\ \Rightarrow x–2y–7=0----------\left(2\right) \end{gathered}$$

Step 4: Solve the obtained equations simultaneously to find co-ordinates of $P$

Solving $\left(1\right)$ and $\left(2\right)$we get

$$\begin{gathered} x=11 \\ y=2 \end{gathered}$$

Hence, the coordinates of the point equidistant from points $A(1,2),B(3,-4)$, and $C(5,-6)$ are $P(11,2)$