Question

If the four vertices of a square taken in order are (0,0) (0,5), (x,y) and (5,0), then (x,y) is

• A. (-5,-5)
• B. (5,-5)
• C. (5,5)
• D. (-5,5)

Answer 1

The correct option is C (5,5)


$AB=BC=CD=AD$
$BC=CD$
$\Rightarrow (\sqrt{(o-x{)}^2+(5-y{)}^2}{)}^2=(\sqrt{(x-5{)}^2+(y-0{)}^2}{)}^2$
$\Rightarrow x^2+25+y^2-10y=x^2+25-10x+y^2$
$\Rightarrow -10y=-10x$
$\therefore y=x$.....(1)
$AD=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$AD=\sqrt{(5-0{)}^2+(0-0{)}^2}$
$AD=\sqrt{5^2}$
$AD=5$
Thus, $AB=BC=CD=AD=5$
In $\Delta ABC,\angle B={90}^0$
By pythagoras theorem,
$A{B}^2+B{C}^2=A{C}^2$
$5^2+5^2=A{C}^2$
$A{C}^2=25+25=50$
Now, $AC=\sqrt{(x-0{)}^2+(x-0{)}^2}$
$A{C}^2=(x-0{)}^2+(x-0{)}^2$
$50=2(x{)}^2$
$x^2=25$
$x=5$
Therefore, $y=x=5$.
Thus, $(x,y)=(5,5)$
So, option c is correct.