The distance between the lines 4x + 3y = 11 a | vedclass

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

Question

(C) The given lines are $4x + 3y = 11$ and $8x + 6y = 15$. First,we rewrite the second equation to have the same coefficients for $x$ and $y$ as the f

Vedclass · Accepted Answer

$\frac{7}{10}$

Vedclass · Answer

(C) The given lines are $4x + 3y = 11$ and $8x + 6y = 15$. First,we rewrite the second equation to have the same coefficients for $x$ and $y$ as the first equation by dividing by $2$: $4x + 3y = \frac{15}{2}$. 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,$a = 4$,$b = 3$,$c_1 = 11$,and $c_2 = \frac{15}{2}$. $D = \frac{|11 - \frac{15}{2}|}{\sqrt{4^2 + 3^2}} = \frac{|\frac{22 - 15}{2}|}{\sqrt{16 + 9}} = \frac{\frac{7}{2}}{\sqrt{25}} = \frac{\frac{7}{2}}{5} = \frac{7}{10}$.

The distance between the lines $4x + 3y = 11$ and $8x + 6y = 15$ is:

Similar Questions

Two points $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$. If $P$ is the point of intersection of the lines $4x + 3y + k = 0$ and $3x + 4y + k = 0$,then the range of $b$ is

The system of equations $x+2y=3$ and $3x+6y=a-2$ has no solution if:

If two sides of a square are $4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$,then the area of the square 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

Find the distance between the parallel lines $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module