Question

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?

Answer 1

`lambda=-4`

Explanation:

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

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

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

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

`x+y=2--------2`

Multiplying--2 by 2

`2x+2y=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
`(x,y)=(1,1)`

The point `(7,4)` is joined with this intersection point to get the normal.

Slope of this normal is
`m1=(4-1)/(7-1)`
`m1=3/6=1/2`

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

Slope of the line farthest from `(7,4)` is
`m2=-1/(m1)=-1/(1/2)=-2`

The farthest line passes through the point `(1,1)`

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

Equation of the line is
`(2x+3y-5)+lambda(x+y-2)=0`

Simplifying
`(2+lambda)x+(3+lambda)y+(-5-2lambda)=0`
`(lambda+2)x+(lambda+3)y-(2lambda+5)=0`

Slope of the line
`ax+by+c=0` is given by
`m=-a/b`

Here,
`a=lambda+2`
`b=lambda+3`

Slope, `m=-(lambda+2)/(lambda+3)`

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

We have
`-(lambda+2)/(lambda+3)=-2`
ie
`(lambda+2)/(lambda+3)=2`
`lambda+2=2(lambda+3)`
`lambda+2=2lambda+6`

`2lambda-lambda=2-6`
`lambda=-4`