How to solve this?We have #O_(((0,0))),A_(((3,4))),B_(((x,y))).#Determine real numbers #x,y# such that #Delta_(OAB)# be equilateral.

2 Answers
Vinícius Ferraz
Apr 14, 2017

#B_1 = (x_1, y_1)#

#B_2 = (x_2, y_2)#

Explanation:

#|OA| = sqrt{3^2 + 4^2} = 5#

#|OB|^2 = x^2 + y^2 = 25#

#|AB|^2 = (x-3)^2 + (y-4)^2 = 25#

#(x^2) - 6x + 9 + (y^2) - 8y + 16 = (25)#

#25 - 6x = 8y#

#y = frac{25 - 6x}{8}#

#x^2 + frac{(25 - 6x)^2}{64} = 25#

#64x^2 + 625 - 300x + 36x^2 = 1600#

#100x^2 - 300x - 975 = 0#

#4x^2 - 12x - 39 = 0#

#Delta = 144 + 4 * 4 * 39 = 768 = 2^8 * 3#

#x = 12/8 ± 16/8sqrt 3#

#x = 3/2 + 2jsqrt 3, j = ±1#

#y = frac{25 - 6(3/2 + 2jsqrt 3)}{8} = frac{25 - 9 - 12jsqrt 3}{8}#

#y = 2 - 3/2jsqrt 3#

Ratnaker Mehta
Apr 14, 2017

#(x,y)=(3/2+2sqrt3,2-3/2sqrt3), or, #

#(x,y)=(3/2-2sqrt3,2+3/2sqrt3).#

Explanation:

In an equilateral #DeltaOAB,# we know that,

#OA=AB=OB rArr OA^2=AB^2=OB^2....(ast)#

With #O(0,0), A(3,4), and B(x,y),# using the Distance Formula,

#(ast) rArr 3^2+4^2=(x-3)^2+(y-4)^2=x^2+y^2, i.e., #

#25=x^2+y^2-6x-8y+25=x^2+y^2.#

# OA^2=OB^2 rArr x^2+y^2=25...(1).#

8y#AB^2=OB^2 rArr 6x+8y=25, or, y=(25-6x)/8.........(2).#

#(2), and, (1) rArr x^2+(25-6x)^2/64=25.#

#:. 64x^2+(625-300x+36x^2)=1600#

#:. 100x^2-300x=975,# completing square, we get,

# 100x^2-300x+15^2=975+15^2=1200.#

#:. (10x-15)^2=(20sqrt3)^2#

#:. 10x-15=+-20sqrt3, or, 10x=15+-20sqrt3#

#:. x=3/2+-2sqrt3#

#x=3/2+2sqrt3, and (2) rArr y=1/8{25-6(3/2+2sqrt3)}=1/8(25-9-12sqrt3)=1/8(16-12sqrt3)=2-3/2sqrt3.#

Similarly, #x=3/2-2sqrt3 rArr y=2+3/2sqrt3.#

Therefore, #(x,y)=(3/2+2sqrt3,2-3/2sqrt3), or, #

#(x,y)=(3/2-2sqrt3,2+3/2sqrt3).#

Enjoy Maths.!

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