The line L given by frac{x}{5}+frac{y}{b}=1 p | vedclass

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

Question

(D) Line $L$ passes through $(13,32)$.$\frac{13}{5}+\frac{32}{b}=1$$\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$\Rightarrow b = -20$So

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Line $L$ passes through $(13,32)$.$\frac{13}{5}+\frac{32}{b}=1$$\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$\Rightarrow b = -20$So,the equation of $L$ is $\frac{x}{5}-\frac{y}{20}=1$,which simplifies to $4x-y=20$.The slope of $L$ is $m_1=4$.The slope of line $K$ given by $\frac{x}{c}+\frac{y}{3}=1$ is $m_2=-\frac{3}{c}$.Since $L$ and $K$ are parallel,$m_1=m_2$.$4 = -\frac{3}{c} \Rightarrow c = -\frac{3}{4}$.The equation of line $K$ is $-\frac{4x}{3}+\frac{y}{3}=1$,which simplifies to $4x-y=-3$.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 $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ 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

If $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line $x+y+3=0$ such that each of them is at a distance of $\sqrt{5}$ units from the line $x+2y+2=0$,then the value of $|x_1-x_2|$ is:

If a point $(\alpha, \beta)$ on the line $3x + y = 0$ and the point $(3, 4)$ lie on opposite sides of the line $3x - 4y - 8 = 0$,then which of the following is correct?

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is (in $\text{ units}$)

Reduce the equation $x-y=4$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module