the three vertices of a parallelogram ABCD taken in order are a (1, -2) b (3,6)and C (5, 10) find the coordinates of the fourth vertex d
Answers
Answer:
d = (3 , 2)
Step-by-step explanation:
the three vertices of a parallelogram ABCD taken in order are a (1, -2) b (3,6)and C (5, 10) find the coordinates of the fourth vertex d
ABCD is a parallelogram
=> AD ║ BC
=> Slope of AD = Slope of BC & AD = BC
Let say D coordinates -= (x , y)
Slope of BC = (10-6)/(5-3) = 4/2 = 2
Slope of AD = (y - (-2))/(x - 1) = 2
=> y + 2 = 2x - 2
=> y = 2x - 4
BC = √(5-3)² + (10-6)² = √20
AD = √(y + 2)² + (x -1)² = √20
Squaring both sides
(y + 2)² + (x -1)² = 20
putting y = 2x - 4
=> (2x - 2)² + (x -1)² = 20
=> 4x² + 4 - 8x + x² + 1 - 2x = 20
=> 5x² - 10x -15 = 0
=> 5x² + 5x - 15x - 15 = 0
=> 5x(x + 1) - 15(x + 1) =0
=> (5x - 15)(x + 1) = 0
=> x = 3 , x = -1
y = 2 , y = -6
Now CD = AB
AB = √(3-1)² +(6-(-2))² = √68
When x = - 1 & y = -6
CD = √(5-(-1))² + (10-(-6))² = √132
=> AB ≠ CD
When x =3 & y = 2
CD = √(5-3)² + (10-2)² = √68
=> AB = CD
Hence the coordinates of the fourth vertex d = (3 , 2)