If the slope of one of the pair of lines repr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation is $2x^2 + 3xy + Ky^2 = 0$. Dividing by $Ky^2$ (assuming $K 
eq 0$),we get the quadratic in terms of slope $m = \frac{y}{x}$ a

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) The given equation is $2x^2 + 3xy + Ky^2 = 0$. Dividing by $Ky^2$ (assuming $K 
eq 0$),we get the quadratic in terms of slope $m = \frac{y}{x}$ as $Km^2 + 3m + 2 = 0$. Since one slope $m_1 = 2$,it must satisfy the equation: $K(2)^2 + 3(2) + 2 = 0$ $\Rightarrow 4K + 8 = 0$ $\Rightarrow K = -2$. Substituting $K = -2$ into the equation,we get $2x^2 + 3xy - 2y^2 = 0$. Factoring this,we get $(2x - y)(x + 2y) = 0$. The slopes of the two lines are $m_1 = 2$ and $m_2 = -\frac{1}{2}$. Since $m_1 	imes m_2 = 2 	imes (-\frac{1}{2}) = -1$,the lines are perpendicular to each other. Therefore,the angle between the pair of lines is $	heta = \frac{\pi}{2}$.

If the slope of one of the pair of lines represented by $2x^2 + 3xy + Ky^2 = 0$ is $2$,then the angle between the pair of lines is

Similar Questions

In polymeric $(BeCl_2)_n$,there are

Which of the following cylindrical rods will conduct the most heat when their ends are maintained at the same steady temperatures?

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

The solution of $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$ is

$\int \frac{dx}{\sqrt{x-x^2}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module