Question

{"given":"A homogeneous equation $6x^2 - xy + 4cy^2 = 0$ represents a pair of straight lines, and one of the lines is $3x + 4y = 0$. We need to find the value of parameter $c$.","key_observation":"For a homogeneous equation $ax^2 + hxy + by^2 = 0$ representing two lines with slopes $m_1$ and $m_2$, the relationships are: $m_1 + m_2 = -\\frac{h}{b}$ and $m_1 m_2 = \\frac{a}{b}$. Since we know one line, we can find its slope and use these relationships to determine $c$.","option_analysis":[{"label":"(A)","text":"$1$","verdict":"incorrect","explanation":"If $c = 1$, then $m_1 + m_2 = \\frac{1}{4}$ and $m_1 m_2 = \\frac{3}{2}$. With $m_1 = -\\frac{3}{4}$, we get $m_2 = \\frac{1}{4} + \\frac{3}{4} = 1$. But then $m_1 m_2 = -\\frac{3}{4} \\cdot 1 = -\\frac{3}{4} \\neq \\frac{3}{2}$, so this is inconsistent."},{"label":"(B)","text":"$-1$","verdict":"incorrect","explanation":"If $c = -1$, then $b = -4$, giving $m_1 + m_2 = -\\frac{1}{4}$ and $m_1 m_2 = -\\frac{3}{2}$. With $m_1 = -\\frac{3}{4}$, we get $m_2 = -\\frac{1}{4} + \\frac{3}{4} = \\frac{1}{2}$. Then $m_1 m_2 = -\\frac{3}{4} \\cdot \\frac{1}{2} = -\\frac{3}{8} \\neq -\\frac{3}{2}$, so this is also inconsistent."},{"label":"(C)","text":"$3$","verdict":"incorrect","explanation":"If $c = 3$, then $m_1 + m_2 = \\frac{1}{12}$ and $m_1 m_2 = \\frac{1}{2}$. With $m_1 = -\\frac{3}{4}$, we get $m_2 = \\frac{1}{12} + \\frac{3}{4} = \\frac{10}{12} = \\frac{5}{6}$. Then $m_1 m_2 = -\\frac{3}{4} \\cdot \\frac{5}{6} = -\\frac{5}{8} \\neq \\frac{1}{2}$, creating another inconsistency."},{"label":"(D)","text":"$-3$","verdict":"correct","explanation":"Step 1: From the given line $3x + 4y = 0$, the slope is $m_1 = -\\frac{3}{4}$.\nStep 2: For the equation $6x^2 - xy + 4cy^2 = 0$, we have $a = 6$, $h = -1$, $b = 4c$.\nStep 3: Using $m_1 + m_2 = -\\frac{h}{b} = \\frac{1}{4c}$ and $m_1 m_2 = \\frac{a}{b} = \\frac{6}{4c} = \\frac{3}{2c}$.\nStep 4: From $m_1 m_2 = \\frac{3}{2c}$: $(-\\frac{3}{4}) \\cdot m_2 = \\frac{3}{2c}$, so $m_2 = -\\frac{2}{c}$.\nStep 5: Substituting into $m_1 + m_2 = \\frac{1}{4c}$: $-\\frac{3}{4} + (-\\frac{2}{c}) = \\frac{1}{4c}$.\nStep 6: Solving: $-\\frac{3c + 8}{4c} = \\frac{1}{4c}$, which gives $-3c - 8 = 1$, so $c = -3$."}],"answer":"(D)","formula_steps":[]}