if (-4,3)&(4,3) are the two vertices of an equilateral triangle find the third vertex?
Answers
hope it help u
Solution :---
let the Third vertices be (x,y)
then Distance between (x,y) & (4,3) is :--
→ √(x-4)² + (y-3)² ---------------- Equation (1)
and Distance between (x,y) & (-4,3) is :-----
→ √(x+4)² + (y-3)² ---------------- Equation (2)
Distance between (4,3) &(-4,3) is :-------
→ √(4+4)² + (3-3)² = 8 units. ---------------- Equation (3)
Now, since, Distance Between them all is Equal , as it is Equaliteral ∆.
so, Equation (1) = Equation (2)
→ √(x-4)² + (y-3)² = √(x+4)² + (y-3)²
→ (x-4)² = (x+4)²
→ x² - 8x + 16 = x² +8x +16
→ 16x = 0
→ x = 0
And, also , Equation (1) = Equation (3)
→ √(x-4)² + (y-3)² = 8
Squaring both sides
→ (x-4)² + (y-3)² = 64
Putting value of x = 0, now,
→ (y-3)² = 64-16
→ (y-3)² = 48
Square - root both sides now,
→ (y-3) = ±4√3
→ y = ±4√3 + 3
Now, as origin lies in the interior of the triangle,
y ≠ 3+4√3 .
∴ Third vertex = (x, y) = (0, 3 - 4√3).