Question

Find a relation between $ x $ and $ y $ such that the point $ \left( {x,y} \right) $ is equidistant from the point $ \left( {3,6} \right) $ and $ \left( { - 3,4} \right) $ .

Answer 1

Hint: In the given question, we are required to find the distance of the point $ \left( {x,y} \right) $ from the points $ \left( {3,6} \right) $ and $ \left( { - 3,4} \right) $ . We will use the distance formula, that is,
 $ d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Where, $ {x_2} = $ x-coordinate of second point
 $ {x_1} = $ x-coordinate of first point
 $ {y_2} = $ y-coordinate of second point
 $ {y_1} = $ y-coordinate of first point
By using this formula, we can find the distance between the point $ \left( {x,y} \right) $ and the given two points. Then we can equate the two equations that will be obtained as they are equidistant to get the
required relation between $ x $ and $ y $ .

Complete step-by-step answer:
Let us name the given points as, $ P\left( {3,6} \right) $ and $ Q\left( { - 3,4} \right) $ .
Now, we are to find their distance from the point $ R\left( {x,y} \right) $ .
Now, by using the distance formula between two points, i.e., $ d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ , we get the distance between $ P $ and $ R $ as,
 $ PR = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 6} \right)}^2}} $
Now, using the formula, $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ , we get,
 $ \Rightarrow PR = \sqrt {\left( {{x^2} - 6x + 9} \right) + \left( {{y^2} - 12y + 36} \right)} $
 $ \Rightarrow PR = \sqrt {{x^2} + {y^2} - 6x - 12y + 45} $
And, now, we get the distance between the points $ Q $ and $ R $ as,
 $ QR = \sqrt {{{\left( {x - \left( { - 3} \right)} \right)}^2} + {{\left( {y - 4} \right)}^2}} $
 $ \Rightarrow QR = \sqrt {{{\left( {x + 3} \right)}^2} + {{\left( {y - 4} \right)}^2}} $
Now, using the formula, $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ , we get,
 $ \Rightarrow QR = \sqrt {\left( {{x^2} + 6x + 9} \right) + \left( {{y^2} - 8y + 16} \right)} $
 $ \Rightarrow QR = \sqrt {{x^2} + {y^2} + 6x - 8y + 25} $
Now, given the points are equidistant, i.eThe distance between $ P $ and $ R $ is equal to $ Q $ and $ R $ .
Therefore, $ PR = QR $
Now, substituting the values, we get,
 $ \Rightarrow \sqrt {{x^2} + {y^2} - 6x - 12y + 45} = \sqrt {{x^2} + {y^2} + 6x - 8y + 25} $
Now, squaring both sides, we get,
 $ \Rightarrow {x^2} + {y^2} - 6x - 12y + 45 = {x^2} + {y^2} + 6x - 8y + 25 $
Now, cancelling the same terms from both sides, we get,
 $ \Rightarrow - 6x - 12y + 45 = 6x - 8y + 25 $
Now, making right hand side $ 0 $ , by subtracting $ 6x $ , adding $ 8y $ and subtracting $ 25 $ on both sides, we get,
 $ \Rightarrow - 6x - 6x - 12y + 8y + 45 - 25 = 0 $
Simplifying the equation, we get,
 $ \Rightarrow - 12x - 4y + 20 = 0 $
Simplifying more by dividing both sides by $ 4 $ , we get,
 $ \Rightarrow - 3x - y + 5 = 0 $
Now, multiplying both sides by $ - 1 $ , we get,
 $ \Rightarrow 3x + y - 5 = 0 $
 $ \Rightarrow 3x + y = 5 $
Therefore, the relation between $ x $ and $ y $ is $ 3x + y = 5 $ .
So, the correct answer is “$ 3x + y = 5 $”.

Note: -The equation that is obtained by solving the problem is the equation of a line that will contain all the points those will be equidistant from the given points $ \left( {3,6} \right) $ and $ \left( { - 3,4} \right) $ . This is the locus of the point that is equidistant from these points. Using this equation we can find every point that will be equidistant from the given points.