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(D) Let the slopes of the lines represented by $2x^2 + axy + 3y^2 = 0$ be $m$ and $m_1$. Then $m + m_1 = -a/3$ and $mm_1 = 2/3$. Let the slopes of the

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$a = -5$ and $b = -1$

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(D) Let the slopes of the lines represented by $2x^2 + axy + 3y^2 = 0$ be $m$ and $m_1$. Then $m + m_1 = -a/3$ and $mm_1 = 2/3$. Let the slopes of the lines represented by $2x^2 + bxy - 3y^2 = 0$ be $m$ and $m_2$. Then $m + m_2 = -b/(-3) = b/3$ and $mm_2 = 2/(-3) = -2/3$. Since the other two lines are perpendicular,$m_1m_2 = -1$. Dividing the product of slopes: $(mm_1) / (mm_2) = (2/3) / (-2/3) = -1$. Thus,$m_1 / m_2 = -1$,which implies $m_1 = -m_2$ or $m_1 + m_2 = 0$. From $mm_1 = 2/3$ and $mm_2 = -2/3$,we have $m_1 = 2/(3m)$ and $m_2 = -2/(3m)$. Substituting into $m_1m_2 = -1$: $(2/(3m)) 	imes (-2/(3m)) = -1$ $\Rightarrow -4/(9m^2) = -1$ $\Rightarrow m^2 = 4/9$ $\Rightarrow m = 2/3$ (taking $m = 2/3$). Then $m_1 = (2/3) / (2/3) = 1$ and $m_2 = (-2/3) / (2/3) = -1$. Using $m + m_1 = -a/3$: $2/3 + 1 = -a/3$ $\Rightarrow 5/3 = -a/3$ $\Rightarrow a = -5$. Using $m + m_2 = b/3$: $2/3 - 1 = b/3$ $\Rightarrow -1/3 = b/3$ $\Rightarrow b = -1$.

If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincides with one of those given by the equation $2x^2 + bxy - 3y^2 = 0$,while the other two lines are perpendicular to each other,then the values of $a$ and $b$ are

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