Show that the line with intercepts (a,0) and (0,b) has this equation?

First suppose the line is not vertical; then the line will have the form y=mx+B.

Now, the y intercept is (0,b), so y=mx+b, the slope is b-0/(0-a) = -b/a,

now we have

y=-bx/a + b. Now we know that b cannot be 0, since this would mean a=0 and a vertical line, against our assumption. Thus we can divide by b, to get y/b = -x/a + 1

Now adding x/a to both sides we get

x/a + y/b = 1

Suppose instead the line is vertical. Then the equation is actually not valid. Then the line would be x=0, thus a=0, and the equation does not account for the case where a is zero.

Now, here's the important thing, the equation x/a + y/b = 1 always has intercepts (a,0) and (0,b), but not every line with intercepts (a,0) and (0,b) has the equation x/a + y/b = 1.