Question

The equations of the lines which cuts off an intercept -$1$ from $y$-axis and equally inclined to the axes are

• A. $x-y+1=0,x+y+1=0$
• B. $x-y-1=0,x+y-1=0$
• C. $x-y-1=0,x+y+1=0$
• D. None of the above

Answer 1

The correct option is C

$x-y-1=0,x+y+1=0$

Explanation for the correct answer

There will be two lines passing through $\left(0,-1\right)$ which makes $45°$ and $135°$ with the $x$-axis.

In the given figure,

Slope of the line, ${\mathrm{m}}_1=\tan 45°=1$

Slope of the another line, ${\mathrm{m}}_2=\tan 135°=-1$

The standard equation of a line passing through $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ with slope m is $\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)$

When slope $=1$, required equation becomes

$\begin{array}{rcl}\mathrm{y}+1& =& 1\left(\mathrm{x}-0\right)\\ \mathrm{x}-\mathrm{y}-1& =& 0\end{array}$

When slope $=-1$, required equation becomes

$\begin{array}{rcl}\mathrm{y}+1& =& -1\left(\mathrm{x}-0\right)\\ \mathrm{x}+\mathrm{y}+1& =& 0\end{array}$

Hence option $\left(C\right)$ is the answer.