Question

The equation of a straight line passing through a point $(-5,4)$ and which cuts off an intercept of $\sqrt{2}$ units between the lines $x+y+1=0$ and $x+y-1=0$ is

• A. $x-2y-13=0$
• B. $2x-y+14=0$
• C. $x-y+9=0$
• D. $x-y+10=0$

Answer 1

The correct option is C $x-y+9=0$

Let the given point be $A=(-5,4)$ and the given lines be $l_1\to x+y+1=0$ and $l_2\to x+y-1=0$

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,

$\Rightarrow AM=\frac{|-5+4-1|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\Rightarrow$ 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

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

$\Rightarrow y-4=x+5$

$\Rightarrow x-y+9=0$