Question

Find relation between \[x\] and \[y\] such that the point \[(x,y)\] is equidistant from the point \[(7,1)\] and \[(3,5)\].

Answer 1

Hint: Here, we have to find the relation between \[x\] and \[y\]. For that, we will first find the distance of the point \[(x,y)\] from the point \[(7,1)\]by applying distance formula and then we will find the distance of the point \[(x,y)\] from the point \[(3,5)\]. We will equate the two distances we obtained using distance formulas to get the relation between \[x\] and \[y\].

Formula used:
We will use the distance formula, \[\sqrt {\left[ {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \right]} \] , where \[({x_1},{y_1})\] and \[({x_2},{y_2})\] are two distinct points.

Complete step-by-step answer:
It is given that the distance of point \[(x,y)\] from the point \[(7,1)\] and the distance of the point \[(x,y)\] from the point\[(3,5)\]are equal.
We will first find the distance between the points \[(x,y)\] and \[(7,1)\] by applying the distance formula.
We know the distance formula is \[\sqrt {\left[ {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \right]} \].
To find the distance between \[(x,y)\] and \[(7,1)\] , we will replace \[{x_1}\] by \[x\], \[{y_1}\] by \[y\], \[{x_2}\] by 7 and \[{y_2}\] by 1.
Distance of the point \[(x,y)\] from \[(7,1)\] \[ = \sqrt {\left[ {{{\left( {7 - x} \right)}^2} + {{\left( {1 - y} \right)}^2}} \right]} \]
Applying exponents on the bases, we get
\[ \Rightarrow \] Distance of the point \[(x,y)\] from \[(7,1)\] \[ = \sqrt {\left[ {49 + {x^2} - 2 \times 7 \times x + 1 + {y^2} - 2 \times 1 \times y} \right]} \]
On simplifying the equation further, we get
\[ \Rightarrow \] Distance of the point \[(x,y)\] from \[(7,1)\] \[ = \sqrt {\left[ {{x^2} + {y^2} - 2y - 14x + 50} \right]} \]
We will apply the same method for finding the distance between \[(x,y)\] and \[(3,5)\].
We have to find the distance between the points \[(x,y)\] and \[(3,5)\]. For that, we will replace \[{x_1}\] by \[x\], \[{y_1}\] by \[y\], \[{x_2}\] by 3 and \[{y_2}\] by 5.
Distance of the point \[(x,y)\] from \[(3,5)\] \[ = \sqrt {\left[ {{{\left( {3 - x} \right)}^2} + {{\left( {5 - y} \right)}^2}} \right]} \]
Applying exponents on the bases, we get
\[ \Rightarrow \] Distance of the point \[(x,y)\] from \[(3,5)\] \[ = \sqrt {\left[ {9 + {x^2} - 2 \times 3 \times x + 25 + {y^2} - 2 \times 5 \times y} \right]} \]
On simplifying the equation further, we get
\[ \Rightarrow \] Distance of the point \[(x,y)\] from \[(3,5)\] \[ = \sqrt {\left[ {{x^2} + {y^2} - 10y - 6x + 34} \right]} \]
Now, we will equate the distance of point \[(x,y)\] from the point \[(7,1)\] with the distance of the point \[(x,y)\] from the point\[(3,5)\]. Thus,
\[\sqrt {\left[ {{x^2} + {y^2} - 2y - 14x + 50} \right]} = \sqrt {\left[ {{x^2} + {y^2} - 10y - 6x + 34} \right]} \]
Squaring both sides, we get
\[ \Rightarrow {\left( {\sqrt {\left[ {{x^2} + {y^2} - 2y - 14x + 50} \right]} } \right)^2} = {\left( {\sqrt {\left[ {{x^2} + {y^2} - 10y - 6x + 34} \right]} } \right)^2}\]
\[ \Rightarrow {x^2} + {y^2} - 2y - 14x + 50 = {x^2} + {y^2} - 10y - 6x + 34\]
Taking all the terms to the left hand side, we get
\[ \Rightarrow {x^2} + {y^2} - 2y - 14x + 50 - \left( {{x^2} + {y^2} - 10y - 6x + 34} \right) = 0\]
Now, we will open the parenthesis and multiply the negative sign to each term. Therefore, we get
\[ \Rightarrow {x^2} + {y^2} - 2y - 14x + 50 - {x^2} - {y^2} + 10y + 6x - 34 = 0\]
On further simplifying the equation, we get
\[ \Rightarrow 8y - 8x + 16 = 0\]
Taking 8 common from the equation, we get
\[ \Rightarrow y - x + 2 = 0\]
This is the required relation between \[x\] and \[y\].

Note: The formula, which we have used here, for finding the distance between the two points is in two dimensional.
The distance formula for three dimensional is different.
Say we have to find the distance between the points \[({x_1},{y_1},{z_1})\] and \[({x_2},{y_2},{z_2})\]
So the distance between these two points is
\[\sqrt {\left[ {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{({z_2} - {z_1})}^2}} \right]} \]