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
Answers
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:
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