Consider the lines x(3 lambda+1)+y(7 lambda+2 | vedclass

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

Question

(A) The given equation of the family of lines is $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$. Rearranging the terms,we get $(x+2y-5) + \lambda(3x+7y-

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The given equation of the family of lines is $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$. Rearranging the terms,we get $(x+2y-5) + \lambda(3x+7y-17) = 0$. This family of lines passes through the intersection of $x+2y-5=0$ and $3x+7y-17=0$. Solving these equations: $x = 5-2y$,substituting into the second: $3(5-2y)+7y-17=0 \implies 15-6y+7y-17=0 \implies y=2$. Thus,$x = 5-2(2) = 1$. The fixed point is $P(1, 2)$. The line $L$ that is farthest from the origin $(0, 0)$ is the line perpendicular to the vector $\vec{OP}$,where $O$ is the origin. The slope of $OP$ is $m_{OP} = \frac{2-0}{1-0} = 2$. The slope of line $L$ is $m_L = -\frac{1}{m_{OP}} = -\frac{1}{2}$. The equation of line $L$ passing through $P(1, 2)$ is $y-2 = -\frac{1}{2}(x-1) \implies 2y-4 = -x+1 \implies x+2y-5=0$. We need to find the distance $d$ of line $L$ from the point $Q(3, 6)$. $d = \frac{|1(3) + 2(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3+12-5|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$. Therefore,$d^2 = (2\sqrt{5})^2 = 4 	imes 5 = 20$.

Consider the lines $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$,where $\lambda$ is a parameter. All these lines pass through a fixed point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$,then the value of $d^2$ is

Similar Questions

The polar coordinates of $P$ are $\left(2, \frac{\pi}{6}\right)$. If $Q$ is the image of $P$ about the $X$-axis,then the polar coordinates of $Q$ are...

$A$ person standing at the junction of two straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. Find the equation of the path that he should follow.

$A$ ray of light along $x + \sqrt{3}y = \sqrt{3}$ gets reflected upon reaching the $x$-axis. The equation of the reflected ray is:

If $A(2, -1)$ and $B(6, 5)$ are two points,the ratio in which the foot of the perpendicular from $P(4, 1)$ to $AB$ divides $AB$ is:

The image of the line $x+y-2=0$ in the $y$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module