Question

⏩Heya,⏪

50 points ✌✌✌

Please solve this Q25. step by step ASAP:
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Q25. If (-4, 3) and (4, 3) are the two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the

(i.) interior

(ii.) exterior of the triangle.
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Answer for convenience:

(i.)(0,3-4√3)

(ii.)(0, 4+√3)
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➡ Note:- PLEASE GIVE ONLY RELEVANT ANSWERS:)

Answer 1

Here is your answer.
Given, P(0, 2) is equidistant from A(3, k) and B(k, 5).

∴ AP = PB

⇒ AP2 = PB2

⇒ (3 – 0)2 + (k – 2)2 = (k – 0)2 + (5 – 2)2    [ Using Distance formula]

⇒ 9 + k2 – 4k + 4 = k2 + 9

⇒ – 4k + 4 = 0

⇒ 4k = 4

⇒ k = 1

Thus, the value of k is 1.

Hope it helps you.
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Answer 2

i)Let the co-ordinate of third vertex be (x, y) Now Using Distance formula BC = [4 - (- 4)] 2 + (3 - 3) 2       = (4 + 4) 2 + 0 BC = 8 2= 8 Now , AB = [x - (- 4)] 2 + (y - 3) 2          AB = (x + 4) 2 + (y - 3) 2 and AC = (x - 4) 2+ (y - 3) 2 Given, ΔABC is equilateral triangle
∴ AB = AC = BC
Now, AB = AC ⇒ (x + 4) 2 + (y - 3) 2   = (x - 4) 2 + (y - 3) 2
On Squaring both sides, we get
(x + 4)2 + (y – 3)2 = (x – 4)2 + (y – 3)2
(x + 4)2 = (x – 4)2
or x 2 + 16 + 8x = x 2 + 16 – 8x
⇒ 16x = 0
x = 0  ....(1)
AC = BC implies that (x - 4) 2 + (y - 3) 2 = 8(0 - 4) 2 + (y - 3) 2 = 8                [from (1)]
On squaring both sides, we get
16 + y 2 + 9 – 6y = 64
y 2 – 6y – 39 = 0 y = -(-6) ± (- 6) 2 - 4(1)(-39) 2(1) y = 6 ± 36 + 156 2 = 6 ± 192 2 y =  6 ± 8 3 2 = 3 ± 4 3 ∴ y = 3 + 4√3 and 3 - 4√3 y ≠ 3 + 4 √3 , as origin lies in the interior of the triangle. Third vertex = (x, y) = (0, 3 - 4√3).