Question

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

• A. $2x-y+14=0$
• B. $3x+y+11=0$
• C. $x-y+9=0$
• D. $4x+5y=0$

Answer 1

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

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

$L_1:x+y+1=0$ and

$L_2:x+y-1=0$

Observe that, $A\in L_1$.

If segment $AM\perp L_2,$ $M\in L_2,$ then, the distance $AM$ is given by,

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

This means that if $B$ is any point on $L_2,$ then, $AB>AM.$

In other words, no line other than $AM$ cuts off an intercept of length $\sqrt{2}$ between $L_1,$ and $L_2$ or $AM$ is the required line.

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 of $L_2$ is $-1,$ the slope of $AM$ must be $1.$

Further, $A(-5,4)\in AM$

By the slope-point form the equation of the required line is,

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

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

$\Rightarrow$ $x-y+9=0$