Let d be the distance between the parallel li | vedclass

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

Question

(A) The distance $d$ between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For $3x

Vedclass · Accepted Answer

$(3x - 2y)^2 + 24(3x - 2y) + 143 = 0$

Vedclass · Answer

(A) The distance $d$ between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For $3x - 2y + 5 = 0$ and $3x - 2y + 5 + 2\sqrt{13} = 0$,$d = \frac{|5 - (5 + 2\sqrt{13})|}{\sqrt{3^2 + (-2)^2}} = \frac{2\sqrt{13}}{\sqrt{13}} = 2$.Given $L_1: 3x - 2y + k_1 = 0$ is at distance $\frac{4d}{\sqrt{13}} = \frac{8}{\sqrt{13}}$ from $3x - 2y + 5 = 0$,we have $\frac{|k_1 - 5|}{\sqrt{13}} = \frac{8}{\sqrt{13}} \Rightarrow |k_1 - 5| = 8$. Since $k_1 > 0$,$k_1 - 5 = 8 \Rightarrow k_1 = 13$.Given $L_2: 3x - 2y + k_2 = 0$ is at distance $\frac{3d}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ from $3x - 2y + 5 = 0$,we have $\frac{|k_2 - 5|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \Rightarrow |k_2 - 5| = 6$. Since $k_2 > 0$,$k_2 - 5 = 6 \Rightarrow k_2 = 11$.The combined equation is $(3x - 2y + 13)(3x - 2y + 11) = 0$.Let $u = 3x - 2y$. Then $(u + 13)(u + 11) = u^2 + 24u + 143 = 0$.Substituting back,we get $(3x - 2y)^2 + 24(3x - 2y) + 143 = 0$.

Similar Questions

The coordinates of a point on the line $x+y+1=0,$ which is at a distance of $\frac{1}{5}$ unit from the line $3x+4y+2=0,$ are

The number of lines which pass through the point $(2, -3)$ and are at a distance of $8$ from the point $(-1, 2)$ is:

The coordinates of the points on the line $2x - y = 5$ which are at a distance of $1$ unit from the line $3x + 4y = 5$ are:

If the line $x-y+1=0$ cuts the lines $2x+2y+3=0$ and $3x+3y+2=0$ at the points $A$ and $B$ respectively,then $AB=$

If the perpendicular distances from the points $(2, 3)$,$(4, a)$ and $(\alpha, \beta)$ to the line $3x + 4y - 3 = 0$ are equal and $4\alpha - 3\beta + 1 = 0$,then the sum of all possible values of $a$,$\alpha$,and $\beta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module