# Find the equation to the locus of a point which is always equidistant from the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$.

Last updated date: 15th Mar 2023

•

Total views: 305.7k

•

Views today: 7.85k

Answer

Verified

305.7k+ views

Hint: Assume a point whose coordinate is $\left( x,y \right)$. Since this assumed point is equidistant to the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$, we can apply distance formula equate the distance between $\left( x,y \right)$ and $\left( a+b,a-b \right)$ to the distance between $\left( x,y \right)$ and $\left( a-b,a+b \right)$.

Before proceeding with the question, we must know the distance formula. The distance formula is used to calculate the distance between any two points on the cartesian plane. Let us consider two points $\left( x,y \right)$ and $\left( x',y' \right)$ on the cartesian plane. From distance formula, the distance $d$ between the points $\left( x,y \right)$ and $\left( x',y' \right)$ is given by,

$d=\sqrt{{{\left( x-x' \right)}^{2}}+{{\left( y-y' \right)}^{2}}}...........................\left( 1 \right)$

In this question, we have to find the locus of a point which is always equidistant from the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$. Let us assume a point $\left( x,y \right)$ which is always equidistant from the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$.

We will first find the distance ${{d}_{1}}$ i.e. the distance between the points $\left( x,y \right)$ and $\left( a+b,a-b \right)$. Using distance formula in equation $\left( 1 \right)$, we get,

$\begin{align}

& {{d}_{1}}=\sqrt{{{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( a-b \right) \right)}^{2}}} \\

& \Rightarrow {{d}_{1}}^{2}={{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( a-b \right) \right)}^{2}}...................\left( 2 \right) \\

\end{align}$

Now, we will find the distance ${{d}_{2}}$ i.e. the distance between the points $\left( x,y \right)$ and $\left( a-b,a+b \right)$. Using distance formula in equation $\left( 1 \right)$, we get,

$\begin{align}

& {{d}_{2}}=\sqrt{{{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}}} \\

& \Rightarrow {{d}_{1}}^{2}={{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}}...................\left( 3 \right) \\

\end{align}$

Since, the assumed point i.e. $\left( x,y \right)$ is always equidistant to the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$, we can equate ${{d}_{1}}$ to ${{d}_{2}}$.

$\begin{align}

& \Rightarrow ~{{d}_{1}}={{d}_{2}} \\

& \Rightarrow ~{{d}_{1}}^{2}={{d}_{2}}^{2}.....................\left( 4 \right) \\

\end{align}$

Substituting ${{d}_{1}}^{2}$ from equation $\left( 2 \right)$ and ${{d}_{2}}^{2}$ from equation $\left( 3 \right)$ in the equation $\left( 4 \right)$, we get,

$\begin{align}

& {{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( a-b \right) \right)}^{2}}={{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}} \\

& \Rightarrow {{x}^{2}}+{{\left( a+b \right)}^{2}}-2x\left( a+b \right)+{{y}^{2}}+{{\left( a-b \right)}^{2}}-2y\left( a-b \right)={{x}^{2}}+{{\left( a-b \right)}^{2}}-2x\left( a-b \right)+{{y}^{2}}+{{\left( a+b \right)}^{2}}-2y\left( a+b \right) \\

\end{align}$

Cancelling similar terms on both the sides of equality in the above equation, we get,

$\begin{align}

& -2x\left( a+b \right)-2y\left( a-b \right)=-2x\left( a-b \right)-2y\left( a+b \right) \\

& \Rightarrow x\left( a+b \right)+y\left( a-b \right)=x\left( a-b \right)+y\left( a+b \right) \\

& \Rightarrow x\left( a+b \right)-x\left( a-b \right)+y\left( a-b \right)-y\left( a+b \right)=0 \\

& \Rightarrow x\left( a+b-\left( a-b \right) \right)+y\left( a-b-\left( a+b \right) \right)=0 \\

& \Rightarrow x\left( a+b-a+b \right)+y\left( a-b-a-b \right)=0 \\

& \Rightarrow 2bx-2by=0 \\

& \Rightarrow 2b\left( x-y \right)=0 \\

& \Rightarrow x-y=0 \\

\end{align}$

Hence, the locus of a point which is always equidistant from the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$ is a straight line having the equation $x-y=0$.

Note: If we are asked to find the locus of a point satisfying a particular condition, then we have to answer the nature of the curve on which the above point is lying on and not the equation of the curve. For example in this question, If we are asked to find the locus of the point, we will report our answer as the locus of the point is a straight line and not as the locus of the point is $x-y=0$.

Before proceeding with the question, we must know the distance formula. The distance formula is used to calculate the distance between any two points on the cartesian plane. Let us consider two points $\left( x,y \right)$ and $\left( x',y' \right)$ on the cartesian plane. From distance formula, the distance $d$ between the points $\left( x,y \right)$ and $\left( x',y' \right)$ is given by,

$d=\sqrt{{{\left( x-x' \right)}^{2}}+{{\left( y-y' \right)}^{2}}}...........................\left( 1 \right)$

In this question, we have to find the locus of a point which is always equidistant from the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$. Let us assume a point $\left( x,y \right)$ which is always equidistant from the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$.

We will first find the distance ${{d}_{1}}$ i.e. the distance between the points $\left( x,y \right)$ and $\left( a+b,a-b \right)$. Using distance formula in equation $\left( 1 \right)$, we get,

$\begin{align}

& {{d}_{1}}=\sqrt{{{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( a-b \right) \right)}^{2}}} \\

& \Rightarrow {{d}_{1}}^{2}={{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( a-b \right) \right)}^{2}}...................\left( 2 \right) \\

\end{align}$

Now, we will find the distance ${{d}_{2}}$ i.e. the distance between the points $\left( x,y \right)$ and $\left( a-b,a+b \right)$. Using distance formula in equation $\left( 1 \right)$, we get,

$\begin{align}

& {{d}_{2}}=\sqrt{{{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}}} \\

& \Rightarrow {{d}_{1}}^{2}={{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}}...................\left( 3 \right) \\

\end{align}$

Since, the assumed point i.e. $\left( x,y \right)$ is always equidistant to the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$, we can equate ${{d}_{1}}$ to ${{d}_{2}}$.

$\begin{align}

& \Rightarrow ~{{d}_{1}}={{d}_{2}} \\

& \Rightarrow ~{{d}_{1}}^{2}={{d}_{2}}^{2}.....................\left( 4 \right) \\

\end{align}$

Substituting ${{d}_{1}}^{2}$ from equation $\left( 2 \right)$ and ${{d}_{2}}^{2}$ from equation $\left( 3 \right)$ in the equation $\left( 4 \right)$, we get,

$\begin{align}

& {{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( a-b \right) \right)}^{2}}={{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}} \\

& \Rightarrow {{x}^{2}}+{{\left( a+b \right)}^{2}}-2x\left( a+b \right)+{{y}^{2}}+{{\left( a-b \right)}^{2}}-2y\left( a-b \right)={{x}^{2}}+{{\left( a-b \right)}^{2}}-2x\left( a-b \right)+{{y}^{2}}+{{\left( a+b \right)}^{2}}-2y\left( a+b \right) \\

\end{align}$

Cancelling similar terms on both the sides of equality in the above equation, we get,

$\begin{align}

& -2x\left( a+b \right)-2y\left( a-b \right)=-2x\left( a-b \right)-2y\left( a+b \right) \\

& \Rightarrow x\left( a+b \right)+y\left( a-b \right)=x\left( a-b \right)+y\left( a+b \right) \\

& \Rightarrow x\left( a+b \right)-x\left( a-b \right)+y\left( a-b \right)-y\left( a+b \right)=0 \\

& \Rightarrow x\left( a+b-\left( a-b \right) \right)+y\left( a-b-\left( a+b \right) \right)=0 \\

& \Rightarrow x\left( a+b-a+b \right)+y\left( a-b-a-b \right)=0 \\

& \Rightarrow 2bx-2by=0 \\

& \Rightarrow 2b\left( x-y \right)=0 \\

& \Rightarrow x-y=0 \\

\end{align}$

Hence, the locus of a point which is always equidistant from the points whose coordinates are $\left( a+b,a-b \right)$ and $\left( a-b,a+b \right)$ is a straight line having the equation $x-y=0$.

Note: If we are asked to find the locus of a point satisfying a particular condition, then we have to answer the nature of the curve on which the above point is lying on and not the equation of the curve. For example in this question, If we are asked to find the locus of the point, we will report our answer as the locus of the point is a straight line and not as the locus of the point is $x-y=0$.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE