$\begin {array} {1 1} (A)\;x+y-1=0 & \quad (B)\;x-y+1=0 \\ (C)\;x-y-1=0 & \quad (D)\;x+y+1=0 \end {array}$

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- Equation of a line in its intercept form is $\large\frac{x}{a}$$ + \large\frac{y}{b}$$=1$ where $a$ and $b$ are the intercepts on the coordinate axes.

Step 1 :

Given $ a = b$

Hence the equation of the line is $ \large\frac{x}{a} $$+\large\frac{y}{a}$$=1$

$ \Rightarrow x+y=a$

It passes through the point (1, -2)

$ \therefore 1-2=a$

$ \Rightarrow a = -1 $

Hence equation of the line is $ \large\frac{x}{-1}$$+\large\frac{y}{-1}$$=1$

$ \Rightarrow x+y=-1 $ or $ x+y+1=0$

Hence the equation of the required line is $ x+y+1=0$

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