Question

A family of lines is given by $\left( 1+2\lambda \right)x+\left( 1-\lambda \right)y+\lambda =0$ , $\lambda $ being the parameter. The line belonging to this family at the maximum distance from the point $\left( 1,4 \right)$ is
(a)$4x-y+1=0$
(b)$33x+12y+7=0$
(c)$12x+33y=7$
(d)None of these

Answer 1

Hint: Rewrite the family of lines equation as $\left( x+y \right)+\lambda \left( 2x-y+1 \right)=0$ then find intersection of lines $\left( x+y \right)=0$ and $\left( 2x-y+1 \right)=0$ .Find slope with the intersection point $\left( 1,4 \right)$ and the point of intersection. Then find the slope of the perpendicular line with a point of intersection. Thus, find the equation of the straight line.

Complete step-by-step answer:
In the question a family of lines is represented by $\left( 1+2\lambda \right)x+\left( 1-\lambda \right)y+\lambda =0$ , where $\lambda $ is a parameter. Now, we have to find a line belonging to the family of lines such that it should have a maximum distance from the point $\left( 1,4 \right)$ .
We are given that the family of lines is represented by $\left( 1+2\lambda \right)x+\left( 1-\lambda \right)y+\lambda =0$ which can be reframed as,
$x+2\lambda x+y-\lambda y+\lambda =0$
$\Rightarrow \left( x+y \right)+\lambda \left( 2x-y+1 \right)=0$
So, the family of straight lines can be reframed as $\left( x+y \right)+\lambda \left( 2x-y+1 \right)=0$ .
Now, we have to find the numbers which are straight. So, it should pass through the intersection of
$x+y=0$ ………………….(i) and $2x-y+1=0$ ……………..(ii)
So, $y=-x$ by the equation (i)
Now substituting $y=-x$ in equation (ii) so, we get
$2x-\left( -x \right)+1=0$
$\Rightarrow 3x+1=0$
$\Rightarrow x=\dfrac{-1}{3}$
Then $y=-\left( x \right)=+\dfrac{1}{3}$
So, the point is $\left( \dfrac{-1}{3},\dfrac{1}{3} \right)$
The required line passes through the point $\left( \dfrac{-1}{3},\dfrac{1}{3} \right)$ and it should also be perpendicular to the line joining the points $\left( \dfrac{-1}{3},\dfrac{1}{3} \right)$ and $\left( 1,4 \right)$ .
At first we will find the slope using formula $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ where points are $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ .
So, for points $\left( \dfrac{-1}{3},\dfrac{1}{3} \right)$ and $\left( 1,4 \right)$ slope will be,
$m=\dfrac{4-\dfrac{1}{3}}{1-\left( \dfrac{-1}{3} \right)}=\dfrac{\dfrac{11}{3}}{\dfrac{4}{3}}=\dfrac{11}{4}$
So, slope $\left( m \right)=\dfrac{11}{4}$
The line perpendicular t, the line joining points $\left( \dfrac{-1}{3},\dfrac{1}{3} \right)$ and $\left( 1,4 \right)$ its slope will be $\dfrac{-1}{\dfrac{11}{4}}$ which can find using formula,
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
The line passes through $\left( \dfrac{-1}{3},\dfrac{1}{3} \right)$ with slope equals to $\dfrac{-4}{11}$ which can find using formula,
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
Where $\left( {{x}_{1}},{{y}_{1}} \right)$ is the point through which it is passing and m is the slope.
So, the equation of line is
$y-\dfrac{1}{3}=\dfrac{-4}{11}\left( x+\dfrac{1}{3} \right)$
Hence, on cross multiplication, we get
$11y-\dfrac{11}{3}=-4x-\dfrac{4}{3}$
Now by multiplying with 3 throughout the equation we get,
$33y-11=-12x-4$
Now on rearranging we get,
$12x+33y=7$
So, the correct option is ‘C’.

Note: In the equation of the family of straight lines $\lambda $ is generally a parameter for some value of $\lambda $ there will be a different straight line. So, the one equation represents an infinite number of straight lines for an infinite number of values of $\lambda $ .