Suppose the pairs of straight lines 2x^2 + ax | vedclass

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(C) Let the common line have slope $m$. Let the slopes of the other two lines be $m_1$ and $m_2$. For the pair $2x^2 + axy + 3y^2 = 0$,the slopes sati

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$5, 1$

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(C) Let the common line have slope $m$. Let the slopes of the other two lines be $m_1$ and $m_2$. For the pair $2x^2 + axy + 3y^2 = 0$,the slopes satisfy $m + m_1 = -a/3$ and $m \cdot m_1 = 2/3$. For the pair $2x^2 + bxy - 3y^2 = 0$,the slopes satisfy $m + m_2 = -b/(-3) = b/3$ and $m \cdot m_2 = 2/(-3) = -2/3$. Since the other two lines are perpendicular,$m_1 \cdot m_2 = -1$. From the product equations,$m_1 = 2/(3m)$ and $m_2 = -2/(3m)$. Substituting into the perpendicularity condition: $(2/(3m)) \cdot (-2/(3m)) = -1$ $\Rightarrow -4/(9m^2) = -1$ $\Rightarrow m^2 = 4/9$ $\Rightarrow m = \pm 2/3$. Case $1$: $m = 2/3$. Then $m_1 = (2/3) / (2/3) = 1$ and $m_2 = (-2/3) / (2/3) = -1$. $a = -3(m + m_1) = -3(2/3 + 1) = -3(5/3) = -5$. $b = 3(m + m_2) = 3(2/3 - 1) = 3(-1/3) = -1$. Case $2$: $m = -2/3$. Then $m_1 = (2/3) / (-2/3) = -1$ and $m_2 = (-2/3) / (-2/3) = 1$. $a = -3(m + m_1) = -3(-2/3 - 1) = -3(-5/3) = 5$. $b = 3(m + m_2) = 3(-2/3 + 1) = 3(1/3) = 1$. Thus,$(a, b)$ can be $(-5, -1)$ or $(5, 1)$. Given the options,$(5, 1)$ is the correct choice.

Suppose the pairs of straight lines $2x^2 + axy + 3y^2 = 0$ and $2x^2 + bxy - 3y^2 = 0$ are such that they have one common line,and the other two lines are perpendicular. Then the values of $a$ and $b$ are respectively:

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