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

Let the given pair of straight lines passing through the origin be represented by the homogeneous quadratic equation:

$$ x^2 - 2cxy - 7y^2 = 0 $$

We can rewrite this equation in the standard form of a pair of straight lines:

$$ ax^2 + 2hxy + by^2 = 0 $$

By comparing the coefficients of the given equation with the standard form, we can identify the following values:

$$ a = 1 $$

$$ 2h = -2c \implies h = -c $$

$$ b = -7 $$

Let the individual slopes of the two lines represented by the joint equation be denoted as m1 and m2. According to the theory of pairs of straight lines, the formulas for the sum and product of the slopes are:

$$ m_1 + m_2 = -\frac{2h}{b} $$

$$ m_1 \cdot m_2 = \frac{a}{b} $$

Substitute the specific values of a, h, and b into these two relationships:

$$ m_1 + m_2 = -\frac{-2c}{-7} = -\frac{2c}{7} $$

$$ m_1 \cdot m_2 = \frac{1}{-7} = -\frac{1}{7} $$

The problem states a specific condition relating the sum and the product of the slopes, which is that the sum of the slopes is four times their product:

$$ m_1 + m_2 = 4(m_1 \cdot m_2) $$

Substitute the fractional expressions for the sum and product into this given condition:

$$ -\frac{2c}{7} = 4\left(-\frac{1}{7}\right) $$

$$ -\frac{2c}{7} = -\frac{4}{7} $$

Since both sides have a common denominator of 7, we can multiply the entire equation by -7 to isolate the variable term:

$$ 2c = 4 $$

Divide both sides by 2 to find the final value of c:

$$ c = 2 $$

Final Answer:

The value of c is 2.