If( 3, 4),( - 2, 3 )and (x, y) are the vertices of the equilateral triangle find X and Y
Answers
Answer:
Step-by-step explanation:
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
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