Question

Find the coordinate of the points equidistant from three given points $A(5,1),B(-3,-7),andC(7,-1)$.

Answer 1

Step 1: Given data:

Let, the point $P(x,y)$ is equidistant from the points $A(5,1),B(-3,-7),\&C(7,-1)$

We know that the distance between two points $(x_1,y_1)\&(x_2,y_2)$ is given the formula, $S=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Step 2: Finding $x$ & $y$:

Now, $PA=PB$

$\sqrt{{(x-5)}^2+{(y-1)}^2}=\sqrt{{\left[x-(-3)\right]}^2+{\left[y-(-7)\right]}^2}$

Squaring both sides

$x^2-10x+25+y^2-2y+1=x^2+6x+9+y^2+14y+49$

$\Rightarrow -16x-16y=32$

$\Rightarrow x+y=-2$ $—\left(\mathrm{i}\right)$

Also, $PB=PC$

$\sqrt{{\left[x-(-3)\right]}^2+{\left[y-(-7)\right]}^2}=\sqrt{{(x-7)}^2+{\left[y-(-1)\right]}^2}$

Squaring on both sides

$x^2+6x+9+y^2+14y+49=x^2-14x+49+y^2+2y+1$

$\Rightarrow 20x+12y=-8$

$\Rightarrow 5x+3y=-2$ $—\left(\mathrm{ii}\right)$

Step 3: Solving the linear equation with two variables:

$x+y=-2$ $—\left(\mathrm{i}\right)\times 3$

$5x+3y=-2$ $—\left(\mathrm{ii}\right)$

Solving the linear equation (i) & (ii) and we get x and y.

Equation $\left(ii\right)-3\times \left(i\right)$

$$\begin{gathered} 5x+3y-3x-3y=-2-(-6) \\ 2x=-2+6 \\ 2x=4 \\ \therefore x=2 \end{gathered}$$

Putting the value of x in equation $\left(i\right)$:

$$\begin{gathered} x+y=-2 \\ 2+y=-2 \\ y=-4 \end{gathered}$$

Hence, the point $\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{)}$ is equidistant from the $\mathbf{A}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{B}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{7}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{C}\mathbf{(}\mathbf{7}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}$.