Question : An equilateral triangle has one vertex at the point (3,4) and another at the point (-2,3). Find the co-ordinates of the third vertex.
Doubt by Muskan
Solution :
Let A(3,4) & B(-2,3)
Let the coordinates of the third vertex be C(x,y)
AB = BC = AC [ABC is an equilateral Triangle]
By Distance Formula
AB = √[(x2-x1)2 + (y2-y1)2]
AB = √[(-5)2 + (-1)2]
AB = √[25+1]
AB = √26
Now
BC = AC
S.B.S
BC2 = AC2
(x+2)2 + (y-3)2 = (x-3)2 + (y-4)2
x2 + 4 + 4x + y2 + 9 - 6y = x2 + 9 - 6x + y2 + 16 - 8y
4x - 6y + 6x + 8y = 9 + 16 - 4 - 9
10x + 2y = 12
5x + y = 6
y = 6 - 5x — (1)
Also
AB = BC
SBS
26 = (x+2)2 + (y-3)2
26 = (x+2)2 + (6-5x-3)2 [Using (1)]
26 = (x+2)2 + (3-5x)2
26 = x2 + 4 + 4x + 9 + 25x2 -30x
26 = 26x2 - 26x + 13
0 = 26x2 - 26x + 13 - 26
0 = 26x2 - 26x -13
26x2 - 26x -13 = 0
2x2 - 2x - 1 = 0
Solving by using Quadratic Equation
a = 2, b= -2, c = -1
D = b2-4ac
D = (-2)2 - 4(2)(-1)
D = 4 + 8
D = 12
x = [-b ± √D]/2a
x = [-(-2)±√12]/2(2)
x = [2±2√3]/4
x = 2[1±√3]/4
x = [1± √3]/2
Putting this value of x in equation (1)
y = 6 - 5x
y = 6 - 5[1± √3]/2
y = [12 - 5 ∓5√3]/2
y = [7∓5√3]/2
∴ Coordinates of the third vertex will be
{ [1 + √3]/2, [7 - 5√3]/2 } or { [1 - √3]/2, [7 + 5√3]/2 }