The pairs of straight lines $x^2-3 x y+2 y^2=0$ and $x^2-3 x y+2 y^2+x-2=0$ form a

The pairs of straight lines $x^2-3 x y+2 y^2=0$ and $x^2-3 x y+2 y^2+x-2=0$ form a
  1. square but not rhombus
  2. rhombus
  3. parallelogram
  4. rectangle but not a square

Solution

Given pair of lines are $x^2-3 x y+2 y^2=0$ and $\begin{array}{ll} & x^2-3 x y+2 y^2+x-2=0 . \\ \therefore & (x-2 y)(x-y)=0 \end{array}$ $\begin{aligned} & \text { and } \quad(x-2 y+2)(x-y-1)=0 \\ & \Rightarrow x-2 y=0, \quad x-y=0 \text { and } x-2 y+2=0, \\ & \quad x-y-1=0 \end{aligned}$ Since, the lines $x-2 y=0, x-2 y+2=0$ and $x-y=0, x-y-1=0$ are parallel. Also, angle between $x-2 y=0$ and $x-y=0$ is not $90^{\circ}$. $\therefore$ It is a parallelogram.

Asked in: AP EAMCET 2009

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