Question

Solve :
$(x+y-4{)}^2+(x-y-2{)}^2=0$

• A. $x=1,y=1$
• B. $x=4,y=1$
• C. $x=3,y=1$
• D. $x=5,y=1$

Answer 1

The correct option is C $x=3,y=1$

Given, $(x+y-4{)}^2+(x-y-2{)}^2=0$

$\Rightarrow (x^2+y^2+4^2+2xy-8y-8x)+(x^2+y^2+2^2-2xy+4y-4x)=0$

$\Rightarrow (x^2+y^2+16+2xy-8y-8x)+(x^2+y^2+4-2xy+4y-4x)=0$

$\Rightarrow 2x^2+2y^2-4y-12x+20=0$

$\Rightarrow x^2+y^2-2y-6x+10=0$

$\Rightarrow x^2-6x+9+y^2-2y+1=0$

$\Rightarrow (x-3{)}^2+(y-1{)}^2=0$

$\Rightarrow (x-3{)}^2=0,(y-1{)}^2=0$

$\Rightarrow x-3=0,y-1=0$

$\therefore x=3,y=1$