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.