Question:
1) Find the third vertices of an equilateral triangle if two vertices are (1, 2) and (1, 6).
2) Find the third vertices of an equilateral triangle if two vertices are (3,4) and (-2,3).
3) Find the third vertices of an equilateral triangle if two vertices are(0,0) and (root3, root3).
Answer:
2. Two vertices of an equilateral triangle are (3, 4) and (-2, 3)
Let the third vertex of the triangle be (x, y)
Distance between (3, 4) and (-2, 3)
=√[(-2 - 3)2 + (3 -4)2 ]
= (-5)2 + (-1)2
= 26
Distance between (3, 4) and (x, y)
= √[(x - 3)2 + (y - 4)2 ]
= [(x - 3)2 + (y - 4)2 ]
= [(x - 3)2 + (y - 4)2 ]
= x2 - 6x + 9 + y2 - 8y + 16
= x2 - 6x + y2 - 8y + 25 .....................1
Distance between (-2, 3) and (x, y)
= √[(x + 2)2 + (y - 3)2 ]
= [(x + 2)2 + (y - 3)2 ] = 26
= x2 + 4x + 4 + y2 - 6y + 9
= x2 + 4x + y2 - 6y + 13 .................2
Equating the distances we get,
x2 - 6x + y2 - 8y + 25 = x2 + 4x + y2 - 6y + 13
10x + 2y - 12 = 0
5x + y - 6 = 0
y = (6 - 5x)
Substituting the value of y in equation 1 and equating it to 26, we get
x2 - 6x + y2 - 8y + 25 = 26
=> x2 - 6x + (6 - 5x)2 - 8(6 - 5x) + 25 = 26
=> x2 - 6x + 36 + 25x2 - 60x - 48 + 40x + 25 = 26
=> 26x2 - 26x - 13 = 0
=> 2x2 - 2x - 1 = 0
Solving the quadratic equation using the quadratic formula, [-b ± √(b2 - 4ac)]/2a
x = [2 ± √(4+8)]/4
x = [2 ± √(12)]/4
x = [2 ± 2√(3)]/4
x = [1 ± √(3)]/2
y = (6 - 5x)
= 6 - 5 [1 ± √(3)]/2
= [12 - 5 ± 5√(3)]/2
= [7 ± 5√(3)]/2
Hence, the coordinates of the third vertex of the equilateral triangle are:
([1 ± √(3)]/2, [7 ± 5√(3)]/2
In the same way, we can solve part 1 and 3 also.
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