Question

Consider a family of st lines $(x+y)+\lambda (2x+y+1)=0$. Equation of the line of family that is farthest from (1,-3) is

• A. $3x+6y=7$
• B. $13x-6y=7$
• C. $15x+6y=7$
• D. $15y-6x=7$

Answer 1

Answer: (D)

$15y-6x=7$

Solve: Given

$(x+y)+\lambda (2x+y+1)=0$

First we have to find the point from where all the lines of family Passed

$\Rightarrow L_1:x+y=0-\left(i\right)$

$L_2:2x-y=-1-\left(ii\right)$

on adding equation (i) and (ii) we get,

$3x=-1$

$\Rightarrow x=-\frac{1}{3}$

and from equation (i)

$y=\frac{1}{3}$

For farthest distance from $Q(1,-3)$ Required

line must be perpendicular to

line $P\theta$

slope of $PQ=\frac{-3-\frac{1}{3}}{1-(-\frac{1}{3})}=\frac{-5}{2}$

$\Rightarrow$ Slope of required line, $L:=\frac{2}{5}$

because the product of slop of

two perpendicular line is $-1$

so equation of line =

$\frac{2}{5}=\frac{y-\frac{1}{3}}{x+1/3}$

$\Rightarrow 5y-\frac{5}{3}=2x+\frac{2}{3}$

$\Rightarrow 5y-\frac{7}{3}=2x\Rightarrow 15y-6x=7$