Is this statement true or false?

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.  If you think the statement is true, then prove it. On the other hand, if you think the statement is false, then give an example that disproves the statement.

Answer 1

So the slope of our line is -3/4. Slope is rise / run, so the line goes down 3 units for every 4 units to the right. So if the line passed through the origin (0,0), the next two easiest points on the line to graph would be (4, -3) and (-4, 3.) Drawing such a line might make the mental math easier for you.

 

So if we start at a point P on our line L, and we want to move around but stay on the line, then for every four units we move to the right, we must move three units down. So if we move eight units to the right, we would have to move six units down, if we moved 40 units to the right, we would need to move 30 units down, etc. This math is no different for when we move a fraction of 4 units to the right. In particular, if we go 2 units to the right (which is half of the "run" of 4,) then we would need to go 3/2 units down (which is half of the "rise" of -3.)

 

The way to "prove" this, arithmetically:

• Slope = rise / run = (y1 - y0) / (x1 - x0) where P - (x0, y0) and Q = (x1, y1)
• (-3 / 4) = (y1 - y0) / (x1 - x0)  ... keep in mind the problem statement tells us that (x1 - x0) = 2.
• (-3 / 4) = (y1 - y0) / 2
• (-3 / 4) * 2 = (y1 - y0)
• (-3 / 2) = y1 - y0 ... which means that Q is situated 3/2 units below P.

 

Hope this helps!

 

 

Answer 2

P is {x,y}

Q is {x',y'}

Since Q lies 2 units to the right of P, x' = x+2

And since Q lines on L, (y' - y)/(x'-x) = -3/4
(y'-y)/(x+2 - x) = -3/4
(4y'-4y)/2 = -3

y'-y = -3/2

y' = y-3/2

Therefore Q lies 3/2 units below P