Question

. Suppose the slope of a straight line L is -3/4 and P is a given point on L. If Q is a point on L lying 2 units to the right of P, then Q is situated 3/2 units below P.

Answer 1

is it true or false?

Answer 2

The slope of our line is -3/4. Lets set this guy up as an equation in slope-intercept form.

y=mx+b
y=-3/4x (b is irrelevant in our case)
So if we define our point, P, as being x=1 on our graph, y will equal -3/4

y=-3/4(1)

Now we need to define Q. Q is 2 units away from P. P = 1 so Q = 3

Lets plug in for Q:

y=-3/4(3)
y=-9/4

Now we need to find out if the distance (upwards or downwards) between our two points is 1.5 and the way we do that is subtraction. If this doesn't quite make sense, think about it. You have $5 and your friend has $10. You want to know how far away you are from having $10. So what do you do? Subtract the numbers! It doesn't really matter which number comes first, as distance is always positive. Remember that!

(-3/4) - (-9/4)
-3/4 + 9/4
6/4
3/2 or 1.5

Answer 3

The second question is true (wow had to think about that for a second). If you put the standard equation into slope-intercept form, you'll see:

y=(-Ax/B) - (C/B)

Your y-intercept is defined when you set x=0 (which will remove x and anything attached to him from the equation). This is a very handy thing, as it allows you to view your graph more easily in your head.

Answer 4

They are both true:P