Question

If the points of vertices of an equilateral triangle ABC are A(3,4) and B(-2,3) then find the point C of the third vertex​

Answer 1

Answer:

Step-by-step explanation:

let missing coordinate be (a,b)

so,

AB²=(-2-3)²+(3-4)²                 {distance formula}

AB²= 26 units²

AC²=(3-a)²+(4-b)²

AC²=a²+b²-6a-8b+25    -------------(1)

BC²=(-2-a)²+(3-b)²

BC²=a²+b²+4a-6b+13     --------------(2)

since ΔABC is equilateral

∴AB²=BC²=AC²

a²+b²-6a-8b+25=a²+b²+4a-6b+13

5a+b=6 ---------------(A)

area of ΔABC=1/2[3(3-b)-2(b-4)+a(4-3)]

         √3×26/4=1/2[a-5b+5]

          a-5b=13√3-5

          a-5{6-4}=13√3-5                      [using eq A]

         by solving

         a=(13√3+25)/26

b=6-5a

b=6-5{(13√3+25)/26}

b=(131-65√3)/26

so the third vertex is [(13√3+25)/26 , (131-65√3)/26]

i hope that i solved it right. if not then i am really sorry

Answer 2

Answer:

Step-by-step explanation:

let missing coordinate be (a,b)

so,

AB²=(-2-3)²+(3-4)²                 {distance formula}

AB²= 26 units²

AC²=(3-a)²+(4-b)²

AC²=a²+b²-6a-8b+25    -------------(1)

BC²=(-2-a)²+(3-b)²

BC²=a²+b²+4a-6b+13     --------------(2)

since ΔABC is equilateral

∴AB²=BC²=AC²

a²+b²-6a-8b+25=a²+b²+4a-6b+13

5a+b=6 ---------------(A)

area of ΔABC=1/2[3(3-b)-2(b-4)+a(4-3)]

        √3×26/4=1/2[a-5b+5]

         a-5b=13√3-5

         a-5{6-4}=13√3-5                      [using eq A]

        by solving

        a=(13√3+25)/26

b=6-5a

b=6-5{(13√3+25)/26}

b=(131-65√3)/26

so the third vertex is [(13√3+25)/26 , (131-65√3)/26]

i hope that i solved it right. if not then i am really sorry