Only Approach Matters Answer Don't

So by Dropping Perpendicular from P(x,y) to give line's L1 and L2 we should get

$X\quad =\quad \left| \cfrac { x+y+\lambda }{ \sqrt { 2 } } \right| \quad \\ Y\quad =\quad \left| \cfrac { x-y+\mu }{ \sqrt { 2 } } \right|$ .

Since Area is Independent Of Reference Frame So , And These lines are Perpendicular lines So we Switch To New Co-ordinate system Such that :

$X\quad =\quad \cfrac { x+y+\lambda }{ \sqrt { 2 } } \\ Y\quad =\quad \cfrac { x-y+\mu }{ \sqrt { 2 } }$ .

This Question is analogous To : (I will Generalise it , So let 'a=2' and 'b=4' )

$\cfrac { a }{ \sqrt { 2 } } \quad \le \quad \left| X \right| +\left| Y \right| \quad \le \quad \cfrac { b }{ \sqrt { 2 } }$ .

So required Area = Shaded Region

In New Co-ordinate $A(\cfrac { a }{ \sqrt { 2 } } ,0)\quad ,\quad B(\cfrac { b }{ \sqrt { 2 } } ,0)\\ C(0,\cfrac { a }{ \sqrt { 2 } } )\quad ,\quad D(0,\cfrac { b }{ \sqrt { 2 } } )$ .

$\boxed { { A }_{ required }\quad =\quad { b }^{ 2 }\quad -\quad { a }^{ 2 } }$ .

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Now Beautiful Thing in this question is that Answer is Independent of $\lambda$ and $\mu$ Can You Guess why ?