Question

Show that $O(0,0),C(a,0),B(a,b),C(0,b)$ are the vertices of a rectangle. If $P(x,y)$ is a point in the coordinates plane, then prove using formula that $P{Q}^2+P{B}^2=P{A}^2+P{C}^2$.

Answer 1

$O(o,o),A(o,o),B(a,b),c(a,b)$

$OA=BC,OC=AB$

$OA=2,BC=a$

$OC=b,AB=b$

$\therefore OA=BC$ & $OC=AB$

eqn of $OC,x=0$

$OA\Rightarrow y=0$

$AB\Rightarrow x=a$

$BC\Rightarrow y=b$

Let $P(x,y)$ be in co-ordinates plane

To prove

$P{O}^2+P{B}^2=P{A}^2+P{C}^2$

$PO=\sqrt{(x-0{)}^2+(y-0{)}^2}$

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

$PB=\sqrt{(x-0{)}^2+(y-0{)}^2}$ ......... $\left(2\right)$

$PA=\sqrt{(x-a{)}^2+(y-0{)}^2}$

$=\sqrt{(x-a{)}^2+y^2}$ ........... $\left(3\right)$

$PC=\sqrt{(x-0{)}^2+(y-0{)}^2}$

$=\sqrt{x^2+(y-b{)}^2}$ .......... $\left(4\right)$

$P{O}^2+P{B}^2=x^2+y^2+(x-a{)}^2+(y-b{)}^2$

$P{A}^2+P{C}^2=(x-a{)}^2+y^2+x^2+(y-b{)}^2$

$\therefore P{O}^2+P{B}^2=P{A}^2+P{C}^2$

Hence proved