The product of the perpendicular distances dr | vedclass

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

Question

(C) The given equation is $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$.First,we factorize the homogeneous part $6x^2 - 5xy - 6y^2$:$6x^2 - 9xy + 4xy - 6y^2 =

Vedclass · Accepted Answer

$\frac{1}{13}$

Vedclass · Answer

(C) The given equation is $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$.First,we factorize the homogeneous part $6x^2 - 5xy - 6y^2$:$6x^2 - 9xy + 4xy - 6y^2 = 3x(2x - 3y) + 2y(2x - 3y) = (3x + 2y)(2x - 3y)$.Let the lines be $(3x + 2y + c_1)(2x - 3y + c_2) = 0$.Comparing the coefficients with $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$:$c_2(3x + 2y) + c_1(2x - 3y) = x + 5y \implies (3c_2 + 2c_1)x + (2c_2 - 3c_1)y = x + 5y$.Solving $3c_2 + 2c_1 = 1$ and $2c_2 - 3c_1 = 5$,we get $c_1 = -1$ and $c_2 = 1$.Thus,the lines are $3x + 2y - 1 = 0$ and $2x - 3y + 1 = 0$.The perpendicular distance from $(0, 0)$ to $3x + 2y - 1 = 0$ is $d_1 = \frac{|-1|}{\sqrt{3^2 + 2^2}} = \frac{1}{\sqrt{13}}$.The perpendicular distance from $(0, 0)$ to $2x - 3y + 1 = 0$ is $d_2 = \frac{|1|}{\sqrt{2^2 + (-3)^2}} = \frac{1}{\sqrt{13}}$.The product of the distances is $d_1 	imes d_2 = \frac{1}{\sqrt{13}} 	imes \frac{1}{\sqrt{13}} = \frac{1}{13}$.

The product of the perpendicular distances drawn from the origin to the pair of straight lines $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$ is

Similar Questions

The lines represented by the equation $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$ and the line $4x - 3y - 2 = 0$:

The equation of the pair of lines passing through the point $(2,1)$ and perpendicular to the pair of lines $4xy + 2x + 6y + 3 = 0$ is:

The combined equation of a possible pair of adjacent sides of a square with area $16 \text{ square units}$ whose centre is the point of intersection of the lines $x+2y-3=0$ and $2x-y-1=0$ is

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

If the pair of lines $3x^2 - 5xy + py^2 = 0$ and $6x^2 - xy - 5y^2 = 0$ have one line in common,then $p =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module