Question

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.

Answer 1

`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`

Answer 2

`(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.!