Question

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).

Answer 1

Consider a family of straight line

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

This family consists of two lines

$x+y=0$ and $2x-y+1=0$

Add the two equations

$\begin{array}{c}3x+1=0\\ 3x=-1\\ x=-\frac{1}{3}\end{array}$

Now put this value in $x+y=0$

$\begin{array}{c}-\frac{1}{3}+y=0\\ y=\frac{1}{3}\end{array}$

So, we have $\left(x,y\right)=A\left(-\frac{1}{3},\frac{1}{3}\right)$

Let the given point be $B(1,-3)$ and that the required line is perpendicular to line $AB$

The slope of the line $AB$ is

$\begin{array}{c}{m}_1=\frac{-3-\frac{1}{3}}{1+\frac{1}{3}}\\ =\frac{-9-1}{3+1}\\ =\frac{-10}{4}\\ =-\frac{5}{2}\end{array}$

So, the slope of the required line is

$\begin{array}{c}{m}_2=-\frac{1}{m_1}\\ =-\frac{1}{\frac{-5}{2}}\\ =\frac{2}{5}\end{array}$

Using point slope form to find the required equation of line.

$\begin{array}{c}y-y_1=m_2\left(x-x_1\right)\\ \left(y-\frac{1}{3}\right)=\frac{2}{5}\left(x+\frac{1}{3}\right)\\ \left(3y-1\right)=\frac{2}{5}\left(3x+1\right)\\ 5\left(3y-1\right)=2\left(3x+1\right)\\ 15y-5=6x+2\\ 15y-6x-7=0\end{array}$

Hence, the required equation of line is $15y-6x-7=0$