Question

An equilateral triangle has two vertices at the point 3,4 and -2,3 find the coordinates of the third vertex?

Answer 1

let 3rd vertex of an equilatral triangle is (x, y)
we know,
all side of equilatral triangle are same
so,
(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2

now,
(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2
-6x-8y+9+16=4x-6y+4+9
10x+2y=12
5x+y=6 -----------------(1)

again ,
(x+2)^2+(y-3)^2=25+1
x^2+y^2+4x-6y=13--------------(2)
solve equation (1) and (2)
x= {1+root3}/2 and {1-root3 }/2
and y=6-5x

Answer 2

Given two vertices of equilateral triangle are
A(3,4) & B(-2,3)

Let the third vertex of equilateral triangle be P(x,y)
Distance between points A & B
= √[(3+2)^2 + (4-3)^2
= √(25+1)
= √26

Distance between P & A = Distance between P & B = Distance between A & B

√[(x-3)^2 + (y-4)^2] = √[(x+2)^2 + (y-3)^2] = √26

squaring on both sides ;

(x-3)^2 + (y-4)^2 = (x+2)^2 + (y-3)^2 = 26
-------------------------------------
x^2+9-6x +y^2+16-8y = 26
x^2+y^2-6x-8y = 1
x^2 + y^2 = 6x+8y+1 -------------(1)
------------------------------------
26 = x^2+y^2+4x-6y+11
x^2+y^2+4x-6y = 15
x^2 + y^2 = -4x+6y+15 ----------(2)
-----------------------------------
x^2+9-6x+y^2+16-8y = x^2+y^2+4x-6y+11
10x+3y = 12 ----------------(3)
-----------------------------------
In equation (1) & (2) RHS is equal , therefore ;
6x+8y+1 = -4x+6y+15
10x+2y = 14 --------------(4)

____________________
Solving equation (3) & (4)

y = -2

x = 18
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Hence the third vertex of the triangle is (-2,18)