If 4x^2-5xy+y^2=0 represents a pair of lines | vedclass

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(C) The given equation of the pair of lines is $4x^2-5xy+y^2=0$. Dividing by $x^2$,we get the quadratic equation in terms of $m = \frac{y}{x}$: $m^2-5

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$3$

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(C) The given equation of the pair of lines is $4x^2-5xy+y^2=0$. Dividing by $x^2$,we get the quadratic equation in terms of $m = \frac{y}{x}$: $m^2-5m+4=0$. Here,$m_1$ and $m_2$ are the roots of this equation. Thus,the sum of slopes is $m_1+m_2 = 5$ and the product of slopes is $m_1m_2 = 4$. We know that $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2}$. Substituting the values,we get: $|m_1-m_2| = \sqrt{5^2-4(4)} = \sqrt{25-16} = \sqrt{9} = 3$.

If $4x^2-5xy+y^2=0$ represents a pair of lines with slopes $m_1$ and $m_2$,then the value of $|m_1-m_2|$ equals

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