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 $4x^2 + 20xy + 25y^2 + 2x + 5y - 12 = 0$. This can be written as $(2x + 5y)^2 + (2x + 5y) - 12 = 0$. Let $t = 2x + 5y$. Then

Vedclass · Accepted Answer

$\frac{7}{\sqrt{29}}$

Vedclass · Answer

(A) The given equation is $4x^2 + 20xy + 25y^2 + 2x + 5y - 12 = 0$. This can be written as $(2x + 5y)^2 + (2x + 5y) - 12 = 0$. Let $t = 2x + 5y$. Then the equation becomes $t^2 + t - 12 = 0$. Factoring the quadratic: $(t + 4)(t - 3) = 0$. So,$t = -4$ or $t = 3$. This gives us two parallel lines: $2x + 5y + 4 = 0$ and $2x + 5y - 3 = 0$. The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Here,$A = 2, B = 5, C_1 = 4, C_2 = -3$. $d = \frac{|4 - (-3)|}{\sqrt{2^2 + 5^2}} = \frac{|7|}{\sqrt{4 + 25}} = \frac{7}{\sqrt{29}}$.

The distance between the lines represented by $4x^2 + 20xy + 25y^2 + 2x + 5y - 12 = 0$ is equal to

Similar Questions

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

The distance between the pair of lines $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$ is

The angle between the lines joining the origin to the points of intersection of the line $y = 3x + 2$ and the curve $x^{2} + 2xy + 3y^{2} + 4x + 8y - 11 = 0$ is:

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

If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

Vedclass Test Series

Exam Paper Generator

Online Exam Module