The coordinates of the mid point of the line AB (1,2).The length of AB is 10 units.

Question

1. If the gradient of AB is 0 ,find the coordinated of AB.

2. If the gradient of AB is 3\4 ,find the coordinates of AB.

Answer 1

1. Since the slope AB is 0, points A, B, and the midpoint of AB have the same y-coordinate, and AB is a horizontal line segment.  Since AB has length 10, A and B are 5 units left and right of the midpoint.  

 

      A = (-4, 2) and B = (6,2)

 

2. Let M=(1,2).  Line segments AM and BM both have length 5 and, since the slope of AB is 3/4, each of them is the hypotenuse of a right triangle with vertical leg 3 and horizontal leg 4.  

 

     To find A, start at M then move 4 units to the left and 3 units down.      To find B, start at M then move 4 units to the right and 3 units up.

 

   So, A = (-3, -1) and B = (5, 5)

 

Answer 2

Hi Shaina,

 

Midpoint Formula: ( (x1+x2)/2 , (y1+y2)/2 )

Distance Formula: d = √[(x1-x2)^2 + (y1-y2)^2]

Gradient (Slope) Formula: (y2-y1) / (x2-x1)

 

From the information given, we know that (x1+x2)/2 = 1, (y1+y2)/2 = 2, and 100 = (x1-x2)^2 + (y1-y2)^2. I simplified the distance formula by squaring both sides of the equation. To simplify the other equations:

 

(x1+x2) = 2   and (y1+y2) = 4

 

1. If the gradient of AB is 0, that must mean that (y2-y1) = 0, because the only way to get 0 as the result of a fraction is to have the numerator equal 0. With this equation, y2 = y1. Now we can use the distance formula:

 

100 = (x1-x2)^2 + 0

 

10 = x1-x2 or x1 = x2 + 10

 

Plug that into (x1+x2) = 2 gives us 2x2 + 10 = 2 → x2 = -4 and x1 = 6

 

Now that we found Bx (x2) and Ax (x1), we need to find y1 or y2 (remember that they're equal). We can use the (y1+y2) = 4 equation. Since they're equal, 2y2 = 4 → y2 = 2 = y1.

 

This gives us A (6, 2) and B (-4, 2).

 

 

2. If the gradient is 3/4, then 4 * (y2-y1) = 3 * (x2-x1). I used fractions and cross multiplication to find this relationship. We can use the distance equation, where I substitute 3/4 * (x2-x1) into (y2-y1):

 

100 = (x2-x1)^2 + (3/4 * (x2-x1))^2

 

100 = (x2-x1)^2 + 9/16 * (x2-x1)^2

 

100 = 25/16 * (x2-x1)^2

 

64 = (x2-x1)^2

 

8 = (x2-x1) → x2 = x1 + 8

 

Plug this back into (x1+x2) = 2:

 

2x1 + 8 = 2 → x1 = -3 and x2 = 5

 

Now we can use the relationship that we first found to find y2 and y1:

 

4 * (y2-y1) = 3 * (5 - -3)

 

4 * (y2-y1) = 24

 

(y2-y1) = 6

 

Using the relationship (y1+y2) = 4:

 

y2 = 4 - y1

 

(4 - y1 - y1) = 6

 

y1 = -1 and y2 = 5

 

This gives us A (-3, -1) and B (5, 5).

 

Hope this helps!