Question

If slopes of lines represented by $kx^2+5xy+y^2=0$ differ by 1 then $k=...$

• A. 2
• B. 3
• C. 6
• D. 8

Answer 1

Answer: (C)

6

Answer 2

The given equation of the pair of straight lines is $kx^2+5xy+y^2=0$. This is a homogeneous equation of degree 2, which represents two straight lines passing through the origin. The general form of such an equation is $ax^2 + 2hxy + by^2 = 0$.

Comparing the given equation with the general form, we have: $a = k$ $2h = 5 \implies h = \frac{5}{2}$ $b = 1$

Let the slopes of the two lines be $m_1$ and $m_2$. For a pair of straight lines represented by $ax^2 + 2hxy + by^2 = 0$, the sum and product of the slopes are given by: Sum of slopes: $m_1 + m_2 = -\frac{2h}{b}$ Product of slopes: $m_1 m_2 = \frac{a}{b}$

Substituting the values of $a, h, b$ from our equation: $m_1 + m_2 = -\frac{5}{1} = -5$ $m_1 m_2 = \frac{k}{1} = k$

The problem states that the slopes of the lines differ by 1. This means: $|m_1 - m_2| = 1$

Squaring both sides of this equation: $(m_1 - m_2)^2 = 1^2$ $(m_1 - m_2)^2 = 1$

We know the algebraic identity: $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$. Substitute the values from Equation 1 and Equation 2 into this identity: $1 = (-5)^2 - 4(k)$ $1 = 25 - 4k$

Now, solve for $k$: $4k = 25 - 1$ $4k = 24$ $k = \frac{24}{4}$ $k = 6$

Thus, the value of $k$ is 6.