Question 41518
For determination of the equation of a straight line passing through two points, whose coordinates are given, see my answer to this question.


http://www.algebra.com/cgi-bin/jump-to-question.mpl?question=41158


For the slope-intercept form of the straight line obtained as above procede as follows. If the equation be ax + by + c = 0 (in general form), then take the whole part of the equation excepting the y-part on the right. Thus you have by = -ax - c. Now divide both sides by the coefficient of y. Thus we get {{{y = -(a/b)*x - c/b}}}. This is the slope intercept form of the st. line ax + by + c = 0 and its slope is {{{-a/b}}} and point of intersection with y-axis is (0,{{{-c/a}}}). To see an example refer to my solution to this question.


http://www.algebra.com/cgi-bin/jump-to-question.mpl?question=40798


For expressing the equation of a straight line thus obtained in intercept form to find out the intercepts on the coordinate axes see my solution to this question.


http://www.algebra.com/cgi-bin/jump-to-question.mpl?question=41324


Now, to find the equation of a st. line passing through a given point and having a given slope, follow the procedure below.
Here, we are reqd. to find a st. line whose slope is {{{1/2}}} and which passes through the point ({{{1/2}}},{{{-1/2}}}).
Let the equation be y = mx + k where m = slope and k = intercept on y-axis.
So m = {{{1/2}}}.
Hence y = {{{1/2}}}x + k 
or 2y = x + 2k
Now, this st. line passes through the point ({{{1/2}}},{{{-1/2}}}) so its equation must be satisfied by the coordinates of this point.
Hence, {{{2*(-1/2) = 1/2 + 2k}}} 
or {{{2k = -1/2 - 1 = -3/2}}} 
or {{{k = (-3/2)/2 = -3/4}}}
Thus the reqd. equation of the given st. line is 
{{{2y = x + 2(-3/4)}}}
or {{{2y = x - 3/2}}}
or 2x - 4y = 3
Now, express this equation in intercept form and get the answer.


Draw the graphs yourself.