Finding the equation of a straight line that passes through the points $(\alpha, \beta)$ and whose x & y intercepts are equal Answers given by my book:
$x+y=\alpha+\beta...(1)$, $x-y=\alpha-\beta...(2)$
Answer given by me:
We know that the straight line passes through the point $(\alpha, \beta)$. Let, the x & y intercepts are $a$ & $b$ respectively. We know,
$$a=b...(i)$$
$$\frac{x}{a}+\frac{y}{b}=1...(ii)$$
Putting the values of $a$ from (i) in and putting the values of the point $(\alpha, \beta)$ in (ii),
$$\frac{\alpha}{a}+\frac{\beta}{a}=1$$
$$\implies a=\alpha + \beta...(iii)$$
Now, putting the value of $a$ in (ii),
$$\frac{x}{\alpha + \beta}+\frac{y}{\alpha + \beta}=1[\because a=b]$$
$$x+y=\alpha+\beta$$
This is the same as one of the answers(1) given in my book. However, I can't seem to derive the second answer, (2). How can I derive the second answer given by my book? Is my book wrong?

There are $3$ possible cases:
Case 1: both intercepts, and $\alpha$ and $\beta$ are $0$

*

*Then the straight line has equation of the form $$y=mx$$ for some
nonzero real $m.$
Case 2: both intercepts are $0,$ and $\alpha, \beta$ are nonzero

*

*Then the straight line has equation of the form $y=\frac\beta\alpha
   x,$ i.e., $$\alpha y=\beta x.$$
Case 3: neither intercept is $0$

*

*Then the straight line has gradient $-1$ and equation
$y-\beta=(-1)(x-\alpha),$ i.e., $$x+y=\alpha+\beta.$$ (This is just an alternative to your correct working & answer for this case.)

