Q87 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

Let $P$ be the pair of lines represented by $2 x^2-5 x y+2 y^2+6 x-3 y=0$ and consider the following independent statements (i) $\alpha$ is the $x$ coordinate of the point of intersection of the pair of lines $P$. (ii) $\beta$ is the slope of one of the lines of $P$ passing through origin. (iii) $\gamma$ is the constant term in the equation of the pair of angular bisectors of $P$. Then,

  • A
    $\beta < \gamma < \alpha$
  • B
    $\alpha < \beta=\gamma$
  • C
    $\alpha=\beta < \gamma$
  • Correct answer
    D
    $\gamma < \alpha < \beta$

Solution

Given pairs of lines $2 x^2-5 x y+2 y^2+6 x-3 y=0$ $\Rightarrow \quad(2 y-x-3)(y-2 x)=0$ $\therefore$ Pairs of line are $2 y-x-3=0 \text { and } y-2 x=0$ Solving these equation we get, $x=1, y=2$ $\therefore \quad \alpha=1$ Equation of line passing through origin is $y-2 x=0$ $\because$ Slope of line $m=2$ $\therefore \quad \beta=2$ Equation of angular bisector of line $2 y-x-3=0$ and $y-2 x=0$ is $\frac{2 y-x-3}{\sqrt{5}}= \pm \frac{y-2 x}{\sqrt{5}}$ $2 y-x-3=y-2 x$ or $2 y-x-3=-y+2 x$ $\because \quad x+y-3=0$ or $3 x-3 y+3=0$ $x+y-3=0$ or $x-y+1=0$ $\gamma=-3$ $\therefore \gamma < \alpha < \beta$ i.e., $-3 < 1 < 2$

Hence, the correct answer is option: 'D'.

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