Could anyone help me with these questions please..?

Question

a) the shortest distance from point (4,-3) to a straight line is 2 units. The straight line has a negative slope and intersects the y-axis at the point (0, - 5/2). Find the equation of a straight line.

b) Find the perpendicular distance between the parallel lines

6x - 8y = 15 and 6x - 8y = 20.

Answer 1

a)  Let (A, B) be a point on the line you want.  Specifically, let it be the point closest to (4, -3).  Then you know from the Distance Formula

sqrt((A - 4)^2 + (B + 3)^2) = 2

Since that's also the shortest distance, you know that the slope of the line connecting (4, -3) to (A, B) is the negative reciprocal of the slope of the line you want (because they must be perpendicular).  Let the slope you want be M.  That means

(B + 3)/(A - 4) = -1/M

Lastly, you know that the line you want will have the form

y = Mx - 5/2

from the slope-intercept form.  Plug in (A, B) and you have

B = AM - 5/2

You have three unknowns: A, B and M.  You have three equations.  Solve them.  When you find M, you'll have your equation.

(b)  Pick a point on one of the lines:  say, (1, -9/8) for the first.  Pick a general point (A, B) on the second line.  Let that point be the point on the second line closest to (1, -9/8) on the first one.  So:  you have two unknowns, A and B.  You know the point (A, B) solves the second equation.  There's one equation.  You also know that the slope of the line joining (1, -9/8) and (A, B) is the negative reciprocal of the slope of your parallel lines (which I presume you can find).  There's your second equation.  You can solve for (A, B).  Once you know the coordinates, you can use the Distance Formula to find their distance apart.

Note:  The back of my mind is tickling me that there is an easier way to do (b), but it isn't being very specific.  The approach I detailed is the first one to occur to me.