Question

At what point the origin be shifted so that the equation $x^2+xy-3x-y+2=0$ does not contain any first degree term and constant term?

Answer 1

Let (h, k) be the required point

Then $x=X+h$, $y=Y+k$

Substituting x & y in the equation $x^2+xy-3x-y+2=0$,

We get

$(X+h{)}^2+(X+h\left)\right(Y+k)-3(X+h)-(Y+k)+2=0$

$X^2+h^2+2Xh+XY+kX+hY+hk-3X-3h-Y-k+2=0$

$X^2+XY+2Xh+kx-3X+hY-Y+h^2+hk-3h-k+2=0$

$X^2+XY+X(2h+k-3)+Y(h-1)+h^2+hk-3h-k+2=0$

Given after origin is shifted equation should not contain any first degree term and constant term

$\Rightarrow 2h+k-3=0$ …….$\left(1\right)$

$h-1=0$

$\Rightarrow h=1$

$h^2+hk-3h-k+2=0$

Put $h=1$ in equation $\left(1\right)$

$2+k-3=0$

$k=1$

$\therefore$ The required point is $(1,1)$.