The product of the perpendiculars drawn from | vedclass

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

Question

(D) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Let the two lines be $L_1: l_1x + m_1y + n_1 = 0$ and

Vedclass · Accepted Answer

$\frac{c}{\sqrt{(a - b)^2 + 4h^2}}$

Vedclass · Answer

(D) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Let the two lines be $L_1: l_1x + m_1y + n_1 = 0$ and $L_2: l_2x + m_2y + n_2 = 0$.The product of the perpendiculars from the origin $(0, 0)$ to these lines is given by the formula $\left| \frac{n_1}{\sqrt{l_1^2 + m_1^2}} ight| 	imes \left| \frac{n_2}{\sqrt{l_2^2 + m_2^2}} ight|$.For the given equation,the product of the perpendiculars from the origin is $\frac{|c|}{\sqrt{(a-b)^2 + 4h^2}}$.

The product of the perpendiculars drawn from the origin to the lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is:

Similar Questions

The auxiliary equation of the lines passing through the origin and having slopes $\sqrt{3}+1$ and $\sqrt{3}-1$ is

The joint equation of two lines passing through the origin,each of which makes an angle of $30^{\circ}$ with the line $x+y=0$,is

If one of the lines given by the equation $x^{2}+kxy+2y^{2}=0$ is $x+2y=0$,then $k=$

If one of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is $y = mx$,then

If $ax^2 - y^2 + 4x - y = 0$ represents a pair of lines,then $a = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module