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If one of the lines given by the equation $2 x^2+a x y+3 y^2=0$ coincide with one of those given by the equation $2 x^2+b x y-3 y^2=0$, while the other two lines are perpendicular to each other, then the values of $a$ and $b$ are
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The correct answer is:
$a=-5$ and $b=-1$
Given, equation of pair of straight line
$2 x^2+a x y+3 y^2=0$
and $2 x^2+b x y-3 y^2=0$
Let the slope of line $m$ and $m_1$ of the equation
$2 x^2+a x y+3 y^2=0$
$\therefore \quad m+m_1=\frac{-a}{3}$ and $m m_1=\frac{2}{3}$
Slope of the other line is $m$ and $m_2$.
Then, $m+m_2=\frac{b}{3}$ and $m m_2=-2 / 3$
Given, $\quad m_1 m_2=-1$ ...(i)
$\therefore \quad \frac{m m_1}{m m_2}=-1$
$\Rightarrow \quad m_1=-m_2$
$\Rightarrow \quad m_1+m_2=0$ ...(ii)
From Eqs. (i) and (ii), we get
$m_1=1$ and $m_2=-1$
$\therefore \quad m=2 / 3$
$m+m_1=-a / 3$
$\Rightarrow \quad \frac{2}{3}+1=-a / 3 \Rightarrow a=-5$
and $m+m_2=b / 3$
$\Rightarrow \quad \frac{2}{3}-1=b / 3 \Rightarrow b=-1$
$\therefore \quad a=-5$ and $b=-1$
$2 x^2+a x y+3 y^2=0$
and $2 x^2+b x y-3 y^2=0$
Let the slope of line $m$ and $m_1$ of the equation
$2 x^2+a x y+3 y^2=0$
$\therefore \quad m+m_1=\frac{-a}{3}$ and $m m_1=\frac{2}{3}$
Slope of the other line is $m$ and $m_2$.
Then, $m+m_2=\frac{b}{3}$ and $m m_2=-2 / 3$
Given, $\quad m_1 m_2=-1$ ...(i)
$\therefore \quad \frac{m m_1}{m m_2}=-1$
$\Rightarrow \quad m_1=-m_2$
$\Rightarrow \quad m_1+m_2=0$ ...(ii)
From Eqs. (i) and (ii), we get
$m_1=1$ and $m_2=-1$
$\therefore \quad m=2 / 3$
$m+m_1=-a / 3$
$\Rightarrow \quad \frac{2}{3}+1=-a / 3 \Rightarrow a=-5$
and $m+m_2=b / 3$
$\Rightarrow \quad \frac{2}{3}-1=b / 3 \Rightarrow b=-1$
$\therefore \quad a=-5$ and $b=-1$
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