A very Happy New to all

Solution

The given equations are : a x + b y - (a - b) = 0 b x - a y - (a + b) = 0 The above equations are of the form, a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0 Now, by cross multiplication, we know that, \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}} By cross multiplication, we have \frac{x}{- b (a + b) - a (a - b)} = \frac{y}{- b (a - b) + a (a + b)} = \frac{1}{- a^{2} - b^{2}} \Rightarrow \frac{x}{- a b - b^{2} - a^{2} + a b} = \frac{y}{- a b + b^{2} + a^{2} + a b} = \frac{1}{- (a^{2} + b^{2})} \Rightarrow \frac{x}{- (a^{2} + b^{2})} = \frac{y}{(a^{2} + b^{2})} = \frac{1}{- (a^{2} + b^{2})} \Rightarrow \frac{x}{- (a^{2} + b^{2})} = \frac{1}{- (a^{2} + b^{2})} and \frac{y}{(a^{2} + b^{2})} = \frac{1}{- (a^{2} + b^{2})} \Rightarrow x = 1 and y = - 1 so, x = 1 and y = - 1 is the solution .

Option (d) is correct.