# Equation of all lines passing through the intersection of lines 2x + 3y -5=0 and x+y-2=0 is given by (2x + 3y-5) + lambda(x+y-2) = 0, of all these lines one line is farthest from point (7,4) the value of lambda is?

Feb 12, 2018

$\lambda = - 4$

#### Explanation:

Let us evaluate the coordinates of the point of intersection of the lines

$2 x + 3 y - 5 = 0$
and

$x + y - 2 = 0$
The equations in s and y are:

$2 x + 3 y = 5 - - - - - - - 1$
and

$x + y = 2 - - - - - - - - 2$

Multiplying--2 by 2

$2 x + 2 y = 4$

Subtracting from ----1
$= y = - 1$
or
$y = 1$
--2 becomes
$x + 1 = 2$
$x = 2 - 1$
$x = 1$

All the lines pass through the point
$\left(x , y\right) = \left(1 , 1\right)$

The point $\left(7 , 4\right)$ is joined with this intersection point to get the normal.

Slope of this normal is
$m 1 = \frac{4 - 1}{7 - 1}$
$m 1 = \frac{3}{6} = \frac{1}{2}$

The slope of the line passing through $\left(1 , 1\right)$ needs to be perpendicular to the normal for having the farthest distance from $\left(7 , 4\right)$

Slope of the line farthest from $\left(7 , 4\right)$ is
$m 2 = - \frac{1}{m 1} = - \frac{1}{\frac{1}{2}} = - 2$

The farthest line passes through the point $\left(1 , 1\right)$

Slope of the farthest line is $m = - 2$

Equation of the line is
$\left(2 x + 3 y - 5\right) + \lambda \left(x + y - 2\right) = 0$

Simplifying
$\left(2 + \lambda\right) x + \left(3 + \lambda\right) y + \left(- 5 - 2 \lambda\right) = 0$
$\left(\lambda + 2\right) x + \left(\lambda + 3\right) y - \left(2 \lambda + 5\right) = 0$

Slope of the line
$a x + b y + c = 0$ is given by
$m = - \frac{a}{b}$

Here,
$a = \lambda + 2$
$b = \lambda + 3$

Slope, $m = - \frac{\lambda + 2}{\lambda + 3}$

Substituting for the slope of the line is
$m = - 2$

We have
$- \frac{\lambda + 2}{\lambda + 3} = - 2$
ie
$\frac{\lambda + 2}{\lambda + 3} = 2$
$\lambda + 2 = 2 \left(\lambda + 3\right)$
$\lambda + 2 = 2 \lambda + 6$

$2 \lambda - \lambda = 2 - 6$
$\lambda = - 4$