Answer:
B (1, 8)
C (13, 4)
D (11, -2)
Explanation:
ABCD is a rectangle, so it has four right angles. The equation of BC is 3y + x = 25, or in slope-intercept form, y = -⅓ x + ²⁵/₃.
That means the slope of AB is 3. So the equation of AB in point-slope form is:
y − 2 = 3 (x − (-1))
Or in slope-intercept form:
y − 2 = 3 (x + 1)
y − 2 = 3x + 3
y = 3x + 5
B is the intersection of these two lines.
3x + 5 = -⅓ x + ²⁵/₃
9x + 15 = -x + 25
10x = 10
x = 1
y = 8
The coordinates of B are (1, 8).
The distance between A and B is:
d = √((x₂ − x₁)² + (y₂ − y₁)²)
d = √((1 − (-1))² + (8 − 2)²)
d = √(2² + 6²)
d = √40
The area of the rectangle is 80 square units, so the distance between B and C is:
A = wh
80 = w√40
w = 80 / √40
w = 80√40 / 40
w = 2√40
In other words, the distance between B and C is double the distance between A and B. We can use distance formula again to find the coordinates of C, or we can use geometry.
If the right triangle formed by hypotenuse AB is a 2×6 triangle, then the right triangle formed by hypotenuse BC is a 4×12 triangle.
So x = 1 + 12 = 13, and y = 8 − 4 = 4.
The coordinates of C are (13, 4).
Similarly, the coordinates of D are:
x = -1 + 12 = 11
y = 2 − 4 = -2
D (11, -2)