$\begin {array} {1 1} (A)\;y = x, y + x = 1 & \quad (B)\;y = x, x + y = 2 \\ (C)\;2y = x, y + x =\large\frac{1}{3} & \quad (D)\;y = 2x, y + 2x = 1 \end {array}$

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- Equation of line whose join of points is $(x_1,y_1)$ and $(x_2, y_2)$ is $ \large\frac{y-y_1}{y_2-y_1}$$=\large\frac{x-x_1}{x_2-x_1}$

Step 1 :

The equation of the square formed by the lines are $x=0, y=0, x=1 \: and \: y=1$ respectively.

Hence the vertices of the square are $0(0,0), A(1,0), B (1,1), C (0,1)$ respectively

Hence the equation of the diagonals 0B is

$ \large\frac{y-y_1}{y_2-y_1}$$= \large\frac{x-x_1}{x_2-x_1}$

(i.e) $ \large\frac{y-0}{1-0}$$=\large\frac{x-0}{1-0}$

$ \Rightarrow y=x$--------(1)

equation of the diagonal AC is

$ \large\frac{y-0}{1-0}$$ = \large\frac{x-1}{0-1}$

$ \Rightarrow -y=x-1$

$ \Rightarrow x+y=1$

Hence 'A' is the correct answer.

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