The distance between the lines 3x + 4y = 9 an | vedclass

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

Question

(B) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$.First,rewrite the equations in the same form by multiplying the first equation by $2$:$6x + 8

Vedclass · Accepted Answer

$3/10$

Vedclass · Answer

(B) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$.First,rewrite the equations in the same form by multiplying the first equation by $2$:$6x + 8y = 18$ and $6x + 8y = 15$.These are parallel lines of the form $ax + by = c_1$ and $ax + by = c_2$.The distance $d$ between two parallel lines $ax + by = c_1$ and $ax + by = c_2$ is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Here,$a = 6$,$b = 8$,$c_1 = 18$,and $c_2 = 15$.Substituting these values into the formula:$d = \frac{|18 - 15|}{\sqrt{6^2 + 8^2}} = \frac{3}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10}$.

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is

Similar Questions

Let the expression $E = 8^a + 8^b - 3 \cdot 2^{a+b}$ take its minimum value $p$ at $a = \alpha$ and $b = \beta$. Then,the perpendicular distance of the point $P(\alpha, \beta)$ from the line $x + y + 2p = 0$ is

Find the points on the $x$-axis,whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are $4$ units.

$A$ 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:

Find the length of the perpendicular drawn from the origin to the line $\sqrt{3}x - y + 2 = 0$.

$A$ straight line makes an intercept on the $Y$-axis twice as long as that on the $X$-axis and is at a unit distance from the origin. Then the line is represented by the equations:

Vedclass Test Series

Exam Paper Generator

Online Exam Module