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).

8yAB^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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