Question

Dois dos vertices de um triangulo equilatero ABC sao A (2,-5) e B(3,-4). Determine as cordenadas do vertice C

Resposta com explicação. Obrigada!

Answer 1

Resposta:

dAB²=(2-3)²+(-5+4)²=1+1  =2

C=(a,b)

dAC²=(2-a)²+(-5-b)²=2

dBC²=(3-a)²+(-4-b)²=2

(2-a)²+(-5-b)²=2

(3-a)²+(-4-b)²=2

Desenvolvendo

4-4a+a²+25+10b+b²=2 (i)

9-6a+a²+16+8b+b²=2   (ii)

(i)-(ii)

-5+2a+9+2b=0  ==> 2a+2b=-4 ==> a=-2-b   (iii)

(iii) em (i)

4-4*(-2-b)+(-2-b)²+25+10b+b²=2

4+8+4b+4+4b+b²+25+10b+b²=2

2b²+18b+39=0

b'=[-18+√(324-312)]/4 =-9/2+√3/2  ==>a'=-2+9/2-√3/2 =5/2-√3/2

b''=[-18-√(324-312)]/4 =-9/2-√3/2 ==>-2 +9/2+√3/2 =5/2+√3/2

C é  (5/2-√3/2 ; -9/2+√3/2)  ou (5/2+√3/2 ; -9/2-√3/2)