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Rectangle ABCD has an area of 50 units squared. If the coordinates of A are (1,2) and the coordinates of B are (5,-1), what are possible coordinates of points C and D, respectively?

A) (15,-1) and (11,2)
B) (11,7) and (7,10)
C) (5,9) and (1,12)
D) (2,-5) and (-2,-2)

Sagot :

JcAlmighty
The answer is B) (11,7) and (7,10)

First calculate the distance (d) between the points, using the formula:
d = √((y2 - y1)² + (x2 - x1)²)
A distance between points A (1,2) and B (5,-1) is:
d1 = √((-1 - 2)² + (5 - 1)²) = √((-3)² + (4)²) = √(9 + 16) = √25 = 5

So, we know that one side of the rectangle is 5 units. Using the known area of the rectangle, let's calculate the other side:
A = d1 * d2
A = 50
d1 = 5

50 = 5 * d2
d2 = 50/5
d2 = 10

We need to find slope between two points now:
m = (y2 - y1) / (x2 - x1) = (-1 - 2) / (5 - 1) = -3 / 4

Find the negative reciprocal slope:
m1 = -1/m = 4 / 3

Now, you want the points C and D 10 units (d2) away going at the slope m1:

Use the Pythagorean triple 3 : 4 : 5 to calculate sides, if hypotenuse is 10:
3 : 4 : 5 = a : b : 10
3*2 : 4*2 : 5*2 = 6 : 8 : 10

So, sides have length of 6 and 8.
Now, find possible points
A(1,2)     ⇒     D(1 + 6, 2 + 8)            or        D(1 - 6, 2 - 8)
              ⇒     D(7, 10)                       or        D(-5, -6)

B(5,-1)    ⇒     C(5 + 6, -1 + 8)            or        C(5 - 6, -1 - 8)
              ⇒     C(11,7)                          or        C(-1, -9)

Among all choices, the correct one is B) (11,7) and (7,10)
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