The perpendicular distance between the lines | vedclass

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

Question

(A) The given equation is $(x-2y+1)^2 + k(x-2y+1) = 0$. Factoring this,we get $(x-2y+1)(x-2y+1+k) = 0$. This represents two parallel lines: $L_1: x-2y

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The given equation is $(x-2y+1)^2 + k(x-2y+1) = 0$. Factoring this,we get $(x-2y+1)(x-2y+1+k) = 0$. This represents two parallel lines: $L_1: x-2y+1 = 0$ $L_2: x-2y+(1+k) = 0$. 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,$A=1, B=-2, C_1=1, C_2=1+k$. Given $d = \sqrt{5}$,so $\sqrt{5} = \frac{|1-(1+k)|}{\sqrt{1^2+(-2)^2}}$. $\sqrt{5} = \frac{|-k|}{\sqrt{5}}$. $|-k| = \sqrt{5} 	imes \sqrt{5} = 5$. $|k| = 5$,so $k = \pm 5$. Since the options only provide $5$,the correct option is $A$.

The perpendicular distance between the lines given by $(x-2y+1)^2 + k(x-2y+1) = 0$ is $\sqrt{5}$,then $k=$

Similar Questions

If the lines joining the origin to the points of intersection of the line $fx - gy = \lambda$ and the curve $x^2 + hxy - y^2 + gx + fy = 0$ are mutually perpendicular,then:

Let $L$ be the line joining the origin to the point of intersection of the lines represented by $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. If $L$ is perpendicular to the line $kx + y + 3 = 0$,then $k$ is equal to

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

If the pair of lines joining the origin to the points of intersection of the line $x+y=1$ with the curve $x^2+y^2+2hxy+gx+fy+1=0$ are at right angles,then the point $(g, f)$ lies on the line

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module