Question

Find the coordinates of the point equidistant from point $A\left( {1,2} \right)$, $B\left( {3, - 4} \right)$ and $C\left( {5, - 6} \right)$.

Answer 1

Here in this question, we have to find the coordinate $P\left( {x,y} \right)$ which is equidistant from the given 3 point $A$, $B$ and C i.e., $PA = PB = PC$. First, we will find the distance between the given points and unknown point $P$ by using the distance formula and further simplify by the elimination method of the system of linear equations to get the required coordinate $P\left( {x,y} \right)$.

Complete step by step answer:
The distance between two points is the length of the interval joining the two points. If the two points lie on the same horizontal or same vertical line. In general, the distance can be found by subtracting the coordinates that are not the same. The distance between two points of the $xy$ -plane can be found using the distance formula. An ordered pair $\left( {x,{\text{ }}y} \right)$ represents co-ordinate of the point, where x-coordinate (or abscissa) is the distance of the point from the centre and y-coordinate (or ordinate) is the distance of the point from the centre.

Formula to find Distance Between Two Points in 2d plane. Consider two points, $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ on the given coordinate axis.
The distance between these points is given as: $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$ ----(1)
Now consider, the given points $A\left( {1,2} \right)$, $B\left( {3, - 4} \right)$ and $C\left( {5, - 6} \right)$ are equidistant from the $P\left( {x,y} \right)$, the we have to find the value of coordinate $P\left( {x,y} \right)$.

Given, there points $A$,$B$and $C$ are equidistant from$P$, so
$PA = PB = PC$ ----(2)
Consider,
$\Rightarrow \,\,\,PA = PB$
On applying the distance formula, we have
$\Rightarrow \,\,\,\sqrt {{{\left( {1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2}} = \sqrt {{{\left( {3 - x} \right)}^2} + {{\left( { - 4 - y} \right)}^2}}$
Taking square on both side, then
$\Rightarrow \,\,\,{\left( {1 - x} \right)^2} + {\left( {2 - y} \right)^2} = {\left( {3 - x} \right)^2} + {\left( { - 4 - y} \right)^2}$

We know the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, then
$\Rightarrow \,\,\,{1^2} + {x^2} - 2\left( 1 \right)\left( x \right) + {2^2} + {y^2} - 2\left( 2 \right)\left( y \right) = {3^2} + {x^2} - 2\left( 3 \right)\left( x \right) + {\left( { - 4} \right)^2} + {y^2} - 2\left( { - 4} \right)\left( y \right)$
$\Rightarrow \,\,\,1 + {x^2} - 2x + 4 + {y^2} - 4y = 9 + {x^2} - 6x + 16 + {y^2} + 8y$
$\Rightarrow \,\,\,{x^2} + {y^2} - 2x - 4y + 5 = {x^2} + {y^2} - 6x + 8y + 25$
Take all the RHS term to LHS, then
$\Rightarrow \,\,\,{x^2} + {y^2} - 2x - 4y + 5 - {x^2} - {y^2} + 6x - 8y - 25 = 0$
On simplification we get
$\Rightarrow \,\,\,4x - 12y - 20 = 0$
Divide whole equation by 4, then we get
$\Rightarrow \,\,\,x - 3y - 5 = 0$ -------(3)
Now, consider
$\Rightarrow \,\,\,PA = PC$

On applying the distance formula, we have
$\Rightarrow \,\,\,\sqrt {{{\left( {1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2}} = \sqrt {{{\left( {5 - x} \right)}^2} + {{\left( { - 6 - y} \right)}^2}}$
Taking square on both side, then
$\Rightarrow \,\,\,{\left( {1 - x} \right)^2} + {\left( {2 - y} \right)^2} = {\left( {5 - x} \right)^2} + {\left( { - 6 - y} \right)^2}$
We know the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, then
$\Rightarrow \,\,\,{1^2} + {x^2} - 2\left( 1 \right)\left( x \right) + {2^2} + {y^2} - 2\left( 2 \right)\left( y \right) = {5^2} + {x^2} - 2\left( 5 \right)\left( x \right) + {\left( { - 6} \right)^2} + {y^2} - 2\left( { - 6} \right)\left( y \right)$
$\Rightarrow \,\,\,1 + {x^2} - 2x + 4 + {y^2} - 4y = 25 + {x^2} - 10x + 36 + {y^2} + 12y$
$\Rightarrow \,\,\,{x^2} + {y^2} - 2x - 4y + 5 = {x^2} + {y^2} - 10x + 12y + 61$
Take all the RHS term to LHS, then
$\Rightarrow \,\,\,{x^2} + {y^2} - 2x - 4y + 5 - {x^2} - {y^2} + 10x - 12y - 61 = 0$

On simplification we get
$\Rightarrow \,\,\,8x - 16y - 56 = 0$
Divide whole equation by 8, then we get
$\Rightarrow \,\,\,x - 2y - 7 = 0$ -------(4)
Solve equation (3) and (4) to get the value of $x$ and $y$ of coordinate $P\left( {x,y} \right)$
$x - 3y - 5 = 0$
$\Rightarrow x - 2y - 7 = 0$
Since the coefficients of $x$ and $y$ are the same, we change the sign by the alternate sign.

\+ x - \,\,\,3y - 5 = 0 \\\ \mathop + \limits_{( - )} x\mathop - \limits_{( + )} 2y\mathop - \limits_{( + )} 7 = 0 }\\\

Now we cancel the $x$ term and simplify other terms, so we have

$$\underline { \+ x - \,\,\,3y - 5 = 0 \\\ \mathop + \limits_{( - )} x\mathop - \limits_{( + )} 2y\mathop - \limits_{( + )} 7 = 0 \\\ } \\\ \- y + 2 = 0 \\\$$

$\Rightarrow \,\, - y + 2 = 0$
Subtract 2 on both side, then
$\Rightarrow \,\, - y = - 2$
Cancel $'\, - '$ sign on both side, then we get
$\therefore \,\,y = 2$
We have found the value of $y$, now we can find the value of $x$ by substituting the value $y$to any one of the equations (3) or (4) . we will substitute the value of $y$to equation (3).
Therefore, we have $x - 3y - 5 = 0$
$\Rightarrow x - 3\left( 2 \right) - 5 = 0$
$\Rightarrow x - 6 - 5 = 0$
$\Rightarrow x - 11 = 0$
Add 6 on both side, then we get
$\therefore \,\,x = 11$

Hence, the required coordinate or point $P\left( {x,y} \right) = \left( {11,2} \right)$.

Note: The distance is a length between the two points. In geometry we have a formula to determine the distance between the points. While determining the distance between the points we consider the both values of $x$ and the value of $y$. Where $x$ and $y$ are the coordinates and in the elimination method to eliminate the term we must be aware of the sign where we change the sign by the alternate sign and we have made the one variable term have the same coefficient such that it will be easy to solve the equation.