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 joint equation of the pair of lines passing through $(3, -2)$ and parallel to the lines represented by $x^{2} - 4xy + 3y^{2} = 0$ is:

If the slope of one line is twice the slope of the other in the pair of straight lines $ax^2+2hxy+by^2=0$,then $8h^2$ is equal to (in $ab$)

For $\ell \in R$,the equation $(2 \ell-3) x^2+2 \ell xy-y^2=0$ represents a pair of distinct lines

The product of the perpendicular distances from $(1, -1)$ to the pair of lines $x^2 - 4xy + y^2 = 0$ is:

If $m$ is the slope of one of the lines represented by $ax^{2}+2hxy+by^{2}=0$,then $(h+bm)^{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module