A straight line is parallel to the lines 3x - | vedclass

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

Question

(A) Let the given lines be $L_1: 3x - y - 3 = 0$ and $L_2: 3x - y + 5 = 0$. Since the required line $L$ is parallel to these lines,its equation is of

Vedclass · Accepted Answer

$3x - y = 0$

Vedclass · Answer

(A) Let the given lines be $L_1: 3x - y - 3 = 0$ and $L_2: 3x - y + 5 = 0$. Since the required line $L$ is parallel to these lines,its equation is of the form $3x - y + K = 0$. The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. The distance between $L_1$ and $L$ is $d_1 = \frac{|K - (-3)|}{\sqrt{3^2 + (-1)^2}} = \frac{|K + 3|}{\sqrt{10}}$. The distance between $L$ and $L_2$ is $d_2 = \frac{|5 - K|}{\sqrt{3^2 + (-1)^2}} = \frac{|5 - K|}{\sqrt{10}}$. Given the ratio $d_1 : d_2 = 3 : 5$,we have $\frac{K + 3}{5 - K} = \frac{3}{5}$. $5(K + 3) = 3(5 - K) \implies 5K + 15 = 15 - 3K$. $8K = 0 \implies K = 0$. Thus,the equation of the line is $3x - y = 0$.

$A$ straight line is parallel to the lines $3x - y - 3 = 0$ and $3x - y + 5 = 0$ and lies between them. If its distance from these lines is in the ratio $3 : 5$,then its equation is:

Similar Questions

Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$,for which the points $(a^2, a+1)$ lie in $R$,is :

The vertex of an equilateral triangle is $(2, -1)$ and the equation of its base is $x + 2y - 1 = 0$. The length of its side is

If the point $(1, a)$ lies between the lines $x + y = 1$ and $2(x + y) = 3$,then in which interval does $a$ lie?

Let the lines be $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$. For the point $A(3, -5)$,which of the following is true?

If $12a + 5b = 9$,where $a, b \in \mathbb{R}$,then the minimum value of $a^2 + b^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module