Family of Straight line : Consider a family of straight lines $(x+y) +\lambda (2x-y +1) =0$. Find the equation of the straight ....

Question

Problem : Consider a family of straight lines $(x+y) +\lambda ( 2x-y +1) =0$.

Find the equation of the straight line belonging to this family that is farthest from $(1,-3)$.

Solution:

Let the point of intersection of the family of lines be P. If solve :

$$\left\{\begin{matrix} x+y=0 & \\ 2x-y+1=0 & \end{matrix}\right.$$ We get the point of intersection which is $P \left(-\dfrac{1}{3}, \dfrac{1}{3} \right)$

Now let us denote the point $(1,-3)$ as $Q$. So, now how to find $\lambda$ so that this will be fartheset from $Q$.

If we see the slope of $PQ = m_{PQ} = -\dfrac{5}{2}$

Any line perpendicular to $PQ$ will have slope $\dfrac{2}{5}$ Please suggest further.. thanks.

Answer 1

HINT:

We can rewrite the equation as $$x(1+2\lambda)+y(1-\lambda)+\lambda=0$$

If $d$ is the perpendicular distance from $(1,-3)$

$$d^2=\frac{\{1(1+2\lambda)+(-3)(1-\lambda)+\lambda\}^2}{(1+2\lambda)^2+(1-\lambda)^2}$$

We need to maximize this which can be done using the pattern described here or here

Answer 2

The point of intersection of the family of lines is $(-1/3, 1/3)$.

Thus, the required straight line is the line that is perpendicular to the join of $(-1/3,1/3)$ and $(1,-3)$ which is $15y-6x-7=0$.