Let the given point  be $A=(−5,4)$ and the given lines be $l _1 \rightarrow x+y+1=0$ and  $l _2\rightarrow x+y-1=0$

The  point $A$ lies on $l _1$

If segment $AM\perp l _2$ and $M$ lies on $l _2$, then, the distance $ AM$ is given by,

$\Rightarrow AM=\dfrac{|−5+4−1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt 2}=\sqrt 2$

$\Rightarrow $ This means that if $B$ is any point  on $l _2$ then  $AB>AM$. No line other than $AM$ cuts off an intercept of

length $\sqrt 2$ between $l _1$ and $l _2$.

$\Rightarrow $To determine the equation of $AM$, we need to find the co-ordinates of the Point $M$

Since, $AM\perp l _2$ and  the slope $l _2$ is $−1$, the slope of$AM$ must be $1$.  Also  $A(−5,4)$ lies on $AM$

By the point slope formula, the equation of the required line is

$\Rightarrow  y−4=1(x−(−5))$

$\Rightarrow y-4=x+5$

$\Rightarrow x−y+9=0$

Let the given point be $A=(-5,4)$ and the given lines be,

Let the given point be $A\left( {{x} _{1}},{{y} _{1}} \right)=\left( -5,4 \right)$ and the given lines be

$ x+y+1=0\,\,......\,\,\left( 1 \right) $

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

Observe that,

From equation (1) to,

Let AM is perpendicular distance

Then, $AM=\dfrac{\left| -5+4-1 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

Let B is any point on equation $(2)$to,

From equation (2)

$ x+y-1=0 $

$ y=-x+1 $

On comparing that,

$y=mx+c$

Then, $m=-1$

Slope of perpendicular line is

${{m} _{1}}=\dfrac{-1}{m}=1$

Then, equation of line is

$ y-{{y} _{1}}=m\left( x-{{x} _{1}} \right) $

$ \Rightarrow y-4=1\left( x+5 \right) $

$ \Rightarrow y-4=x+5 $

$ \Rightarrow x-y+5+4=0 $

$ \Rightarrow x-y+9=0 $

Hence, this is the answer.

As the line $x+y=1$ intersect the curve,$\displaystyle x^{2}+y^{2}+x-2y-m=0$

Joint equation of the lines joining the origin and the point of intersection of the line $y =mx + 2$ and
the curve $x^{2} + y^{2}=1$ is
$ \displaystyle x^{2} +y^{2} =\left( \frac{y-mx}{2}\right)^{2} $
$ x^{2}(4 -m^{2} )+2mxy +3y^{2} = 0$
Since these lines are at right angles
$4 -m^{2} + 3 =0 \Rightarrow m^{2} = 7$

Homogenizing the curve with the help of the straight line.

By application of the method of homogenization we get
$3x^2+mxy-4x(2x+y)+1(2x+y)^{2}$
$=3x^2+mxy-8x^2-4xy+4x^2+y^2+4xy$
$=-x^2+mxy+y^2$
$=0$
Hence the equations of the lines is given by $ x^2-mxy-y^2=0$
Hence the lines are perpendicular for all values of $m.$

Given equations of curve and the line are $2x^2 + 3xy -4x + 1 = 0$ and $3x + y = 1$.
Homogenising the curve with the line gives
$2x^2+3xy-4x(3x+y)+(3x+y)^2=0$
$\Rightarrow 2x^2+3xy-12x^2-4xy+9x^2+y^2+6xy=0$
$\Rightarrow y^2-x^2+5xy=0$
$\therefore$ The equation of the lines joining the origin to the points of intersection of the curve and hte line is $x^2-5xy-

y^2=0$
Hence, option A.