Question

If the point (x, y) is equidistant from the points (a+b, b-a) and (a-b, a+b), then prove that bx = ay.

Answer 1

Hint: The distance between any two points \[\left( {{x}_{1}},{{y}_{1}} \right)\]and \[\left( {{x}_{2}},{{y}_{2}} \right)\]is given as \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]. Apply this formula for finding the distances between (x, y) and (a+b, b-a), c accordingly and equate them as they are equidistant.

Complete step-by-step answer:

Given that the point (x, y) is equidistant from the points (a+b, b-a) and (a-b, a+b)..

Now, let us find the distance between the point (x, y) and (a+b, b-a).

The distance is given as:

\[\sqrt{{{\left( a+b-x \right)}^{2}}+{{\left( b-a-y \right)}^{2}}}\]

(Since the distance between any two points \[\left( {{x}_{1}},{{y}_{1}} \right)\]and \[\left( {{x}_{2}},{{y}_{2}} \right)\]is computed using the formula\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\])

Opening the bracket, we get

\[=\sqrt{{{a}^{2}}+{{b}^{2}}+{{x}^{2}}-2ax-2ab+2bx+{{a}^{2}}+{{b}^{2}}+{{y}^{2}}-2ay+2ab-2by}\]

Cancelling the like terms, we get

\[=\sqrt{2{{a}^{2}}+2{{b}^{2}}+{{x}^{2}}+{{y}^{2}}-2ax+2bx-2ay-2by}.............(1)\]

Now let us find the distance between the points (x, y) and (a+b, b-a).

Let us employ the same distance formula that is used in the earlier case,

The distance is given as:

\[\sqrt{{{\left( a-b-x \right)}^{2}}+{{\left( a+b-y \right)}^{2}}}\]

Opening the bracket, we get

\[=\sqrt{{{a}^{2}}+{{b}^{2}}+{{x}^{2}}-2ax+2ab-2bx+{{a}^{2}}+{{b}^{2}}+{{y}^{2}}+2ay-2ab-2by}\]

Cancelling the like terms, we get

\[=\sqrt{2{{a}^{2}}+2{{b}^{2}}+{{x}^{2}}+{{y}^{2}}-2ax-2bx+2ay-2by}.............(2)\]

Since it was mentioned in the given question that they are equidistant, let us equate the distances.

So, equating the equations (1) and (2), we get

\[\sqrt{2{{a}^{2}}+2{{b}^{2}}+{{x}^{2}}+{{y}^{2}}-2ax+2bx-2ay-2by}=\sqrt{2{{a}^{2}}+2{{b}^{2}}+{{x}^{2}}+{{y}^{2}}-2ax-2bx+2ay-2by}\]

Now, let us do squaring on both sides, we get

\[2{{a}^{2}}+2{{b}^{2}}+{{x}^{2}}+{{y}^{2}}-2ax+2bx-2ay-2by=2{{a}^{2}}+2{{b}^{2}}+{{x}^{2}}+{{y}^{2}}-2ax-2bx+2ay-2by\]

Up on cancelling the like terms, we have:

\[4bx=4ay\]

Dividing throughout by ‘4’, we get

\[\Rightarrow bx=ay\].

Therefore, we have proved that \[bx=ay\]when the point \[\left( x,y \right)\]is equidistant from the points \[\left( a+b,b-a \right)\] and \[\left( a-b,a+b \right)\].


Note: Finding out the locus means nothing but figuring out the relation between ‘x’ and ’y’ coordinates. You must consider a variable point and impose the given condition to find any locus. You must be attentive while expanding the square in this question, as it has some negative coefficients.