The line L is given by frac{x}{5} + frac{y}{b | vedclass

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

Question

(C) Since the line $L$ passes through $(13, 32)$,we have:$\frac{13}{5} + \frac{32}{b} = 1$ $\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}

Vedclass · Accepted Answer

$\frac{23}{\sqrt{17}}$

Vedclass · Answer

(C) Since the line $L$ passes through $(13, 32)$,we have:$\frac{13}{5} + \frac{32}{b} = 1$ $\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$ $\Rightarrow b = -20$.Thus,the equation of line $L$ is $\frac{x}{5} - \frac{y}{20} = 1$,which simplifies to $4x - y = 20$.Since line $K$ is parallel to $L$,its equation is of the form $4x - y = k$.Given the equation of $K$ is $\frac{x}{c} + \frac{y}{3} = 1$,we can rewrite it as $y = -\frac{3}{c}x + 3$,or $3x + cy = 3c$. Comparing slopes,the slope of $L$ is $4$,so the slope of $K$ is $-\frac{3}{c} = 4$,which gives $c = -\frac{3}{4}$.Substituting $c$ into the equation $\frac{x}{c} + \frac{y}{3} = 1$,we get $\frac{x}{-3/4} + \frac{y}{3} = 1$ $\Rightarrow -\frac{4x}{3} + \frac{y}{3} = 1$ $\Rightarrow -4x + y = 3$,or $4x - y = -3$.The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Here,$A = 4, B = -1, C_1 = -20, C_2 = 3$.$d = \frac{|-20 - 3|}{\sqrt{4^2 + (-1)^2}} = \frac{|-23|}{\sqrt{16 + 1}} = \frac{23}{\sqrt{17}}$.

The line $L$ is given by $\frac{x}{5} + \frac{y}{b} = 1$ and passes through the point $(13, 32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$. Then the distance between $L$ and $K$ is

Similar Questions

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

The vertices of $\Delta OBC$ are $(0, 0)$,$(-3, 1)$,and $(-1, -3)$ respectively. $A$ line parallel to $BC$ intersects $OB$ and $OC$ at a distance of $1/2$ from $O$. Find the equation of this line.

The vertices of a triangle $OBC$ are $(0,0)$,$(-3,-1)$,and $(-1,-3)$ respectively. The equation of the line parallel to $BC$ which is at a distance of $\frac{1}{2}$ unit from the origin and cuts $OB$ and $OC$ is:

Find the equation of the line equidistant from the lines $x = 3$ and $x = 8$.

If $p_{1}$ and $p_{2}$ are the lengths of perpendiculars from the origin to the lines $x \sin \theta + y \cos \theta = 5 \cos 2 \theta$ and $x \operatorname{cosec} \theta + y \sec \theta - 5 = 0$ respectively,then $p_{1}^{2} + 4 p_{2}^{2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module