If the sum of the slopes of the lines given by $x^2-2cxy-7y^2=0$ is four times their product, find the value of $c$.

Question

A line passing through the point $ A(9, 0) $ makes an angle of $ 30^\circ $ with the positive direction of the x-axis. If this line is rotated about $ A $ through an angle of $ 15^\circ $ in the clockwise direction, then its equation in the new position is:

Answer 1

Step 1: Recall the slope relations for a pair of lines.
For $ax^2+2hxy+by^2=0$, if $m_1,m_2$ are the slopes then $m_1+m_2=-\dfrac{2h}{b}$ and $m_1m_2=\dfrac{a}{b}$.
Step 2: Read the coefficients.
From $x^2-2cxy-7y^2=0$ we have $a=1$, $2h=-2c$ so $h=-c$, and $b=-7$.
Step 3: Compute the sum of slopes.
\[ m_1+m_2=-\frac{2(-c)}{-7}=-\frac{2c}{7}. \]
Step 4: Compute the product of slopes.
\[ m_1m_2=\frac{1}{-7}=-\frac17. \]
Step 5: Apply the given condition.
Sum equals four times the product, so $-\dfrac{2c}{7}=4\left(-\dfrac17\right)=-\dfrac47$.
Step 6: Solve for $c$.
Cancelling the $-\tfrac17$ gives $2c=4$, so $c=2$, which is option 1.
\[ \boxed{c=2} \]

If the sum of the slopes of the lines given by \(x^2-2cxy-7y^2=0\) is four times their product, find the value of \(c\).

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The Correct Option is A

Solution and Explanation

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