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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 $x-y=0$.

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