The distance between the two parallel lines r | vedclass

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

Question

(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$. We can rewrite the quadratic part as $2(4x^2 - 12xy + 9y^2) = 2(2x - 3y)^2$. Let th

Vedclass · Accepted Answer

$\frac{7}{2\sqrt{13}}$

Vedclass · Answer

(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$. We can rewrite the quadratic part as $2(4x^2 - 12xy + 9y^2) = 2(2x - 3y)^2$. Let the lines be of the form $(2x - 3y + c_1)(2x - 3y + c_2) = 0$. Expanding this,we get $4x^2 - 12xy + 9y^2 + 2(c_1 + c_2)x - 3(c_1 + c_2)y + c_1c_2 = 0$. Comparing this with the given equation $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$,we divide the given equation by $2$: $4x^2 - 12xy + 9y^2 - 3x + 4.5y - 2.5 = 0$. Here,$c_1 + c_2 = -3/2$ and $c_1c_2 = -2.5$. The distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Using $(c_1 - c_2)^2 = (c_1 + c_2)^2 - 4c_1c_2 = (-1.5)^2 - 4(-2.5) = 2.25 + 10 = 12.25$. So,$|c_1 - c_2| = \sqrt{12.25} = 3.5 = 7/2$. The distance is $\frac{7/2}{\sqrt{2^2 + (-3)^2}} = \frac{7/2}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.

The distance between the two parallel lines represented by the equation $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$ is

Similar Questions

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

The equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+xy+y^2+x+3y+1=0$ and the line $x+y+2=0$ is

$A$ pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4$ with $x+y=a$. The set containing the value of $a$ is

The straight lines joining the origin to the points of intersection of the line $2x + y = 1$ and the curve $3x^2 + 4xy - 4x + 1 = 0$ include an angle of:

If the lines joining the origin to the points of intersection of the curve $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ and the line $x + 2y = k$ are at right angles,then $k^2$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module