Question regarding slope and intercepts I have been doing GRE practice and saw following reasoning on one of the question

When x and y intercepts are of same sign, the intercept would be negative

Now this is confusing me
slope can be expressed as
y = mx + b
assuming b = 0 here
y = mx
or 
m = y/x
now if both x and y intercepts are negative then
m = -y/-x => m = + y/x
similarly, if both x and y are positive then
m = y/x => m = + y/x
then how come the statement i quoted above is true ? 
 A: Two good proofs have been posted already, so I will offer just an attempt to develop an intuition about the fact, rather than a proof.
The "$x$ intercept" means where the line crosses the $x$ axis, that is, where $y = 0$.
The line can only cross the $x$ axis once. This could happen somewhere to the right of the origin, that is, where $x$ is positive (such as at a point labeled $(1, 0)$ or $(2, 0)$ or $(719,0)$, for example), or it could happen somewhere to the left of the origin (such as at a point labeled $(-1, 0)$ or $(-2, 0)$ or $(-683,0)$, for example).
The "$y$ intercept" means where the line crosses the $y$ axis, that is, where $x = 0$.
Now this could be on the upper part of the $y$ axis, above the origin (for example at $(0,1)$ or $(0,3)$ or $(0,59)$), or it might be on the lower part of the $y$ axis, below the origin (for example at $(0,-1)$ or $(0,-3)$ or $(0,-83)$).
Now suppose we have a line that crosses the upper (positive) part of the $y$ axis and the rightward (positive) part of the $x$ axis. So we know one point on that line--the $y$ intercept--is directly upward from $(0,0)$ and another point--the $x$ intercept--is directly to the right of $(0,0)$. So to get from the $y$ intercept to the $x$ intercept we have to go down (because we started above the $x$ axis) and to the right.
You can confirm an example of this with a drawing.
Going down as we go to the right $\implies$ negative slope when the
intercepts are both positive.
On the other hand, suppose we have a line that crosses the leftward (negative) part of the $x$ axis and the lower (negative) part of the $y$ axis. So we know one point on that line--the $x$ intercept--is directly to the left of $(0,0)$ and another point--the $y$ intercept--is directly downward from $(0,0)$. So to get from the $x$ intercept to the $y$ intercept we have to go right (because we started to the left of the $x$ axis) and downward.  This can be illustrated by a drawing.
Again, we're going down as we go to the right $\implies$ negative slope when the
intercepts are both negative.
Note that we aren't considering any lines parallel to either axis (because they would have no intersections with the parallel axis, or too many) or any line that goes directly through the origin (the point $(0,0)$).
Your review book is talking about other lines you might encounter while taking the GRE,
such that the "intercepts have same sign" makes sense and means "both positive" or "both negative".
A: Given the intercepts $x_i$ and $y_i$, you know that the change in $y$ as $x$ goes from $0$ to $x_i$ will be $-y_i$, since the points $(x_i,0)$ and $(0,y_i)$ are on the line.
Thus, the slope is equal to $-\frac{y_i}{x_i}$, and if the signs of $x_i$ and $y_i$ are equal,
this is negative.
Also, using point-slope form of a line, you can write it as $y - y_i = mx$ and $y = m(x - x_i)$.
Combining these equations gives $m(x - x_i) - y_i = mx$, or $-mx_i - y_i = 0$. Solving for $m$ then yields that $m = -\frac{y_i}{x_i}$.
(This assumes that the intercepts are nonzero. If both are zero, then nothing can be said about the slope, and if $x_i$ is zero and $y_i$ is nonzero, then you no longer have a function.)
A: Do you mean that if the intercepts are the same sign, then the slope is negative? Clearly you can have a line where both intercepts are positive; e.g. $x+y=1$. 
If so, then suppose the line has $x$-intercept $(a,0)$ and $y$-intercept $(0,b)$ where $a$ and $b$ are either both positive or both negative. Then the slope of this line is $m=\frac{b-0}{0-a}=-\frac{b}{a}<0$ since $\frac{b}{a}>0$.
