A challenging question, but you may find Lesson 9, question 7 in my Lecture Problems of some help.
Part (a) is just a matter of following their instructions to generate the x-intercept Q and y-intercept
R.
Part (b) is just telling you to make a graph with x-axis and y-axis, plot the points and draw the line through Q and R.
For part (d), note that the x, and y coefficients of the given line tell you the vector normal to the line. Which is to say if given a line ax + by = c, then n = (a,b).
Then, rather than do what I do in my question 7, you can find the distance they want
in part (e) by computing the projection of arrow QP onto n. The length of that projection vector is the distance you desire.
I strongly recommend that you use the (x,y) graph to help. You can then visualize the distance from P to the line by drawing a line from P perpendicular to the line. Perhaps that will help you understand why the length of the projection vector gives you the
distance in part (e). Literally, you are projecting the arrow QP onto n, thus making a vector that is collinear with n but precisely the length you need, since that is the perpendicular distance from the point to the line.