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.