If the distance between the lines represented | vedclass

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

Question

(A) The given equation is $(x - 2y)^2 + k(x - 2y) = 0$.This can be factored as $(x - 2y)(x - 2y + k) = 0$.Thus, the two lines are $L_1: x - 2y = 0$ an

Vedclass · Accepted Answer

$x - 2y = 0$ and $x - 2y + 3\sqrt{5} = 0$

Vedclass · Answer

(A) The given equation is $(x - 2y)^2 + k(x - 2y) = 0$.This can be factored as $(x - 2y)(x - 2y + k) = 0$.Thus, the two lines are $L_1: x - 2y = 0$ and $L_2: x - 2y + k = 0$.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, $a = 1$, $b = -2$, $c_1 = 0$, and $c_2 = k$.Given $d = 3$, we have $3 = \frac{|0 - k|}{\sqrt{1^2 + (-2)^2}}$.$3 = \frac{|-k|}{\sqrt{1 + 4}} = \frac{k}{\sqrt{5}}$ (since $k > 0$).$k = 3\sqrt{5}$.Substituting $k$ back into the equation for $L_2$, we get $x - 2y + 3\sqrt{5} = 0$.Therefore, the lines are $x - 2y = 0$ and $x - 2y + 3\sqrt{5} = 0$.

If the distance between the lines represented by $(x - 2y)^2 + k(x - 2y) = 0$ is $3$ units and $k > 0$, then the equations of the lines are:

Similar Questions

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

Let $3x^2+8xy-3y^2=0$ represent the lines $L_1, L_2$ and $3x^2+8xy-3y^2+2x-4y-1=0$ represent the lines $L_3, L_4$. Let $L$ be the line joining the points of intersection of $L_1, L_3$ and $L_2, L_4$. Then,the area (in sq units) of the triangle formed by $L$ with the coordinate axes is

The lines joining the points of intersection of the curve $5x^2 + 12xy - 8y^2 + 8x - 4y + 12 = 0$ and the line $x - y = 2$ to the origin make angles with the axes that are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module