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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,
Options:
(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,
Solution:
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Verified Answer
The correct answer is:
$\gamma < \alpha < \beta$
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., $-3 < 1 < 2$
$2 x^2-5 x y+2 y^2+6 x-3 y=0$
$\Rightarrow \quad(2 y-x-3)(y-2 x)=0$
$\therefore$ 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., $-3 < 1 < 2$
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