Question

A line is drawn through the origin intersecting the lines $2x+y=2$ and $x-2y=2$ in $A$ and $B$. Then the locus of the mid points of $AB$ is

• A. $2x^2-3xy-2y^2=0$
• B. $2x^2-3xy-2y^2+x+3y=0$
• C. $2x^2-3xy+2y^2-3x+y=0$
• D. none of these

Answer 1

The correct option is C $2x^2-3xy+2y^2-3x+y=0$

Let the equation of line is $L:y=mx$

At point $A(x_1,y_1)$ on $L_1$:

$y_1=m{x}_1$ $(1)$

$2x_1+y_1=2$ $(2)$

Eliminating $y_1$ from Eq. (1) and (2):

$x_1=\frac{2}{2+m}$

At point $B(x_2,y_2)$ on $L_2$:

$y_2=m{x}_2$ $(3)$

$x_2-2y_2=2$ $(4)$

Eliminating $y_2$ from Eq. (3) and (4):

$x_2=\frac{2}{1-2m}$

Mid- point $P(X,Y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$Y=mX$ $(5)$

Now substitute $x_1$ and $x_2$ in $X=\frac{x_1+x_2}{2}$

$X=\frac{\frac{2}{2+m}+\frac{2}{1-2m}}{2}$

$X=\frac{3-m}{(2+m)(1-2m)}$ $(6)$

Put $m=\frac{Y}{X}$ in Eq. (6):

$X=\frac{3-\frac{Y}{X}}{(2+\frac{Y}{X})(1-2\frac{Y}{X})}$

$X(2+\frac{Y}{X})(1-2\frac{Y}{X})=3-\frac{Y}{X}$

$(2X+Y)(X-2Y)=3X-Y$

$2X^2-3XY-2Y^2-3X+Y=0$

The locus of midpoint $P(X,Y)$ is:

$2x^2-3xy-2y^2-3x+y=0$

Option $D$