Question

15. If (0,0) and (3, 13) are two vertices of an
equilateral triangle, then find third vertex.​

Answer 1

Let ABC is an equilateral triangle in which A(0, 0) and B(3, 13) are given then we have to find third vertex C(x, y) (let)

here, AB = BC = CA

from distance formula,

AB = √(3² + 13²) = √(178)

BC = √{(x - 3)² + (y - 13)²}

CA = √{x² + y²}

so, √178 = √{(x - 3)² + (y - 13)²} = √(x² + y²)

from √(178) = √(x² + y²)

or, x² + y² = 178 .......(1)

from √{(x - 3)² + (y - 13)²} = √{x² + y²}

or, (x - 3)² + (y - 13)² = x² + y²

or, - 6x + 9 - 26y + 169 = 0

or, -3x - 13y + 89 = 0

or, 3x + 13y = 89 ......(2)

from equations (1) and (2),

x² + (89 - 3x)²/169 = 178

or, 169x² + 89² + 9x² - 6x × 89 = 178 × 169

or, 178x² + 89² - 6x × 89 = 178 × 169

or, 2x² + 89 - 6x = 338

or, 2x² - 6x - 249 = 0

or, x = {6 ± √(36 + 1992)}/4

= (3± 13√3)/2

= (3 + 13√3)/2 , (3 - 13√3)/2

and y = (89 - 3x)/13 = (13 $\mp$ 3√3)/2

hence, if x = (3 + 13√3)/2 , y = (13 - 3√3)/2

and if x = (3 - 13√3)/2 , y = (13 + 3√3)/2