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?

For equal intercepts the equation of straight line is
$$ x+y= c = \alpha + \beta $$
and if passing through point $\alpha,  \beta$ then we must have
$$ x+y=  \alpha + \beta $$
This takes care algebraically  of the sign of segments as well... i.e., the segments should have the same sign.
If we interpret that the segments have equal absolute value / length, but can have opposite signs, then the second case in the book is also admissible.
