Question

The equation to the locus of S point which is equidistant from the points $\left( {2,3} \right),\left( { - 2,5} \right)$is
A.$2x - y + 4 = 0$
B.$2x - y - 1 = 0$
C.$2x + y - 4 = 0$
D.$2x + y + 1 = 0$

Answer 1

In this question we have given point S at equidistant from the points $\left( {2,3} \right),\left( { - 2,5} \right)$i.e. point S is the mid-point of these two points. Therefore, the distance from point S to these two points will be the same. So, using this fact we can equate distance from point S to both the points given in the question and by simplifying that we will get the desired answer.

Complete step-by-step answer:
We have given point S is equidistant from the points$\left( {2,3} \right),\left( { - 2,5} \right)$.
And we have to find the equation to the locus of point S.
Let A$= \left( {2,3} \right)$, B $= \left( { - 2,5} \right)$and assume $S = \left( {x,y} \right)$
As S is equidistant from the points$\left( {2,3} \right),\left( { - 2,5} \right)$.
Which implies that s is the midpoint of the points $\left( {2,3} \right)$and$\left( { - 2,5} \right)$.
Which clearly means that distance between point S and A is the same as distance between point S and B.
i.e. ${\left( {SA} \right)^2} = {\left( {SB} \right)^2}$ $\ldots (1)$
As we know that formula for a distance between two points is $\sqrt {{{\left( {{a_2} - {a_1}} \right)}^2} + {{\left( {{b_2} - {b_1}} \right)}^2}}$.
Now, using the above formula in equation (1) we get,
${\left( {\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2}} } \right)^2} = {\left( {\sqrt {{{\left( {x - \left( { - 2} \right)} \right)}^2} + {{\left( {y - 5} \right)}^2}} } \right)^2}$
$\Rightarrow$ ${\left( {\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2}} } \right)^2} = {\left( {\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {y - 5} \right)}^2}} } \right)^2}$
On further simplifying the above expression we get,
$\Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = {\left( {x + 2} \right)^2} + {\left( {y - 5} \right)^2}$
Now, simplify the above expression using the formulae ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
$\Rightarrow$ $\left( {{x^2} - 4x + 4} \right) + \left( {{y^2} - 6y + 9} \right) = \left( {{x^2} + 4x + 4} \right) + \left( {{y^2} - 10y + 25} \right)$
After opening the brackets we get,
${x^2} - 4x + 4 + {y^2} - 6y + 9 = {x^2} + 4x + 4 + {y^2} - 10y + 25$
Now, write all the terms to the left side of the equal sign and equate it with zero.
$\Rightarrow {x^2} - 4x + 4 + {y^2} - 6y + 9 - {x^2} - 4x - 4 - {y^2} + 10y - 25 = 0$
On simplifying further we get,
$- 8x + 4y - 16 = 0$
Now, take 4 common throughout the equation
$\Rightarrow 4\left( { - 2x + y - 4} \right) = 0$
$\Rightarrow - 2x + y - 4 = 0$
Now, multiply the above expression with $\left( { - 1} \right)$
$\Rightarrow 2x - y + 4 = 0$
Thus, the equation to the locus of s point which is equidistant from the points $\left( {2,3} \right),\left( { - 2,5} \right)$is $2x - y + 4 = 0$.
Hence, option A. $2x - y + 4 = 0$ is the correct answer.

Note: Locus of a point: A locus is a set of all points (commonly, a line, a line segment, a curve or a surface), whose location satisfies or is determined by one or more satisfied conditions.
In two dimensions, the locus of points equidistant from two given (different) points is their perpendicular bisector.