The distance between the pair of lines repres | vedclass

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

Question

(C) The given equation is $x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we ge

Vedclass · Accepted Answer

$\sqrt{\frac{5}{2}}$

Vedclass · Answer

(C) The given equation is $x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 1, h = -3, b = 9, g = 3/2, f = -9/2, c = -4$.The distance between two parallel lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.Substituting the values,we get $d = 2\sqrt{\frac{(3/2)^2 - (1)(-4)}{1(1 + 9)}} = 2\sqrt{\frac{9/4 + 4}{10}} = 2\sqrt{\frac{25/4}{10}} = 2\sqrt{\frac{25}{40}} = 2\sqrt{\frac{5}{8}} = 2 	imes \frac{\sqrt{5}}{2\sqrt{2}} = \sqrt{\frac{5}{2}}$.

The distance between the pair of lines represented by the equation $x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0$ is

Similar Questions

The distance between the pair of parallel lines represented by $x^2+2xy+y^2-8ax-8ay-9a^2=0$ is $...$ units.

The distance between the lines represented by the equation $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$ is

$\triangle OAB$ is formed by the lines $x^2-4xy+y^2=0$ and the line $AB$. The equation of line $AB$ is $2x+3y-1=0$. Then the equation of the median of the triangle drawn from the origin is

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

The line $x + y = 3$ intersects the pair of straight lines $x^2 - 3xy + y^2 = 0$ at points $A$ and $B$. Then the coordinates of the mid-point of $AB$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module