Question

The equation of a straight line passing through the point (-5, 4) and which cuts off in 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 $A(x_1,y_1)=(-5,4)$ 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, $AM=\frac{|-5+4-1|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Let B is any point on equation $\left(2\right)$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

$m_1=\frac{-1}{m}=1$

Then, equation of line is

$y-y_1=m(x-x_1)$

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

$\Rightarrow y-4=x+5$

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

$\Rightarrow x-y+9=0$

Hence, this is the answer.