# Math Help - Slope problems

1. ## Slope problems

Hey guys,

I'm stumped on these two problems, and was wondering if someone could shed some light on them for me.

1. Parallelogram ABCD has vertices at A(-1, x - 1), B(x, x + 1), C(3, 1), and D(x - 2, -1). Use the slopes of AB and CD to find "x".

I came up with these answers:

Slope of AB = 0
Slope of CD = -2/x - 5

So now I can't figure out what to do. How am I supposed to find "x" with these answers? Did I do something wrong?

2. According to the Americans With Disabilities Act, a ramp is a route with a slope greater than 1/20. The maximum allowable slope of a ramp is 1/12, and the maximum rise is 30 in. What are the minimum and maximum runs for a ramp with a rise of 30 in.?

I honestly can not figure out what to do...period. Any help, or advice would be awesome! I really appreciate it!
Thanks.

2. Originally Posted by FS777
Hey guys,

I'm stumped on these two problems, and was wondering if someone could shed some light on them for me.

1. Parallelogram ABCD has vertices at A(-1, x - 1), B(x, x + 1), C(3, 1), and D(x - 2, -1). Use the slopes of AB and CD to find "x".

I came up with these answers:

Slope of AB = 0
Slope of CD = -2/x - 5

So now I can't figure out what to do. How am I supposed to find "x" with these answers? Did I do something wrong?

...
1. I fyou have 2 points $P(x_1, y_1)$ and $Q(x_2, y_2)$ then the slope

$m_{PQ} = \dfrac{y_2 - y_1}{x_2 - x_1}$

2. Therefore

$m_{AB} = \dfrac{(x+1) -(x-1)}{x-(-1)} = \dfrac2{x+1}$

$m_{CD} = \dfrac{-1 - 1}{(x-2)-3} = \dfrac{-2}{x-5}$

3. Since $AB \parallel CD$ the 2 slopes must be equal. Therefore you have to solve for x

$\dfrac2{x+1} = \dfrac{-2}{x-5}$

I'll leave this part for you.

3. Originally Posted by FS777
I'm stumped on these two problems, and was wondering if someone could shed some light on them for me.

1. Parallelogram ABCD has vertices at A(-1, x - 1), B(x, x + 1), C(3, 1), and D(x - 2, -1). Use the slopes of AB and CD to find "x".

I came up with these answers:

Slope of AB = 0.
That is incorrect, Slope of AB is $\frac{2}{x+1}$.

4. Oh Ok! Thank you so much! I see.

x = 2