The distance between the lines represented by | vedclass

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

Question

(A) The given equation is $x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 + 2gx + 2fy +

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) The given equation is $x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a=1, h=\sqrt{3}, b=3, g=-3/2, f=-3\sqrt{3}/2, c=-4$.First,check if the lines are parallel by verifying $h^2 - ab = (\sqrt{3})^2 - (1)(3) = 3 - 3 = 0$. Since $h^2 - ab = 0$,the lines are parallel.The formula for the distance between two parallel lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.Substituting the values: $d = 2\sqrt{\frac{(-3/2)^2 - (1)(-4)}{1(1 + 3)}} = 2\sqrt{\frac{9/4 + 4}{4}} = 2\sqrt{\frac{25/4}{4}} = 2\sqrt{\frac{25}{16}} = 2 	imes \frac{5}{4} = \frac{5}{2} = 2.5$.

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

Similar Questions

Find the value$(s)$ of $k$ such that the distance between the two parallel lines represented by $(x-2y)^2 + k(x-2y) = 0$ is $3$ units.

The lines joining the origin to the points of intersection of the line $3x - 2y = 1$ and the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ are:

Let the line $2x - 3y - 1 = 0$ intersect the curve $x^2 + 2xy + 5y^2 + 2x + 3y - 1 = 0$ at distinct points $A$ and $B$. If $O$ is the origin,then $\cos \angle AOB =$

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

The line $3x + 4y - 5 = 0$ cuts the curve $2x^2 + 3y^2 = 5$ at $A$ and $B$. If $O$ is the origin,then $\angle AOB =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module