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

$A$ straight line makes equal negative intercepts on the coordinate axes. If the perpendicular distance from the origin to the line is $4 \text{ units}$, then the equation of the line is:

$P$ is a point on $x+y+5=0$,whose perpendicular distance from $2x+3y+3=0$ is $\sqrt{13}$. Then the coordinates of $P$ are:

$A$ 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:

If the length of the perpendicular drawn from the origin to the line whose intercepts on the axes are $a$ and $b$ is $p$,then

The equation of the base of an equilateral triangle is $12x+5y-65=0$. If one of its vertices is $(2,3)$,then the length of the side is

Vedclass Test Series

Exam Paper Generator

Online Exam Module