The base BC on an equilateral triangle ABC lies on y-axis . The co -ordinates of the point C are (0,-3).If origin is the mid point of BC , Find the co-ordinates of point A and B
Answers
Answer:
A (3√3, 0), B (0, 3)
Step-by-step explanation:
O is the midpoint of base BC.
So OC = OB
=> Since OC = (0, -3), OB = (0,3)
Since ABC is an Equilateral Triangle.
=> AB = BC and AB = AC
We know that to find length the formula is √ (x₂ - x₁)² + (y₂ - y₁)²
AB = √ (x - 0)² + (y - 3)² = √[x² + y² - 6y + 9]
BC = √(0 - 0)² + (3+3)² = √36 = 6.
AC = √ (x - 0)² + (y + 3)² = √[x² + y² + 6y + 9
Firstly, AB = BC
Squaring on both sides,
AB² = BC²
[√x² + y² - 6y + 9]² = [6]²
x² + y² - 6y + 9 = 36
x² + y² - 6y = 27 --------------- [1]
Secondly AB = AC
Squaring on both sides,
AB² = AC²
[√x² + y² - 6y + 9]² = [√x² + y² + 6y + 9]²
x² + y² - 6y + 9 = x² + y² + 6y + 9
12y = 0
=> y = 0 ------------------ [2]
Put [2] in [1]
x² + y² - 6y = 27
x² = 27
x = 3√3.
Thus A = (3√3, 0)
