6. Show that the following points form a equilateral
triangle A(a,0), B(-a, 0), C(0, a (3)
Answers
Answer:
Vertices of Triangle ABC :-
A ( a , 0 )
B ( -a , 0 )
C ( 0 , a sqrt 3 )
To Show :-
These Points form an Equilateral Triangle
Formula Used :-
\begin{gathered}distance \: formula \: = \sqrt{(a _{2} - a_{1}} )^{2} + \sqrt{ ({b_{2} - b_{1}} )^{2} } \\\end{gathered}
distanceformula=
(a
2
−a
1
)
2
+
(b
2
−b
1
)
2
Solution :-
An Equilateral Triangle has all sides equal
so AB = BC = CA
AB :-
\begin{gathered}\sqrt{(a + a} )^{2} + \sqrt{ ( 0 + 0 )^{2} } \\ \\ \\ \sqrt{ ( 2a)^{2} } \\ \\ \\ 2a\end{gathered}
(a+a
)
2
+
(0+0)
2
(2a)
2
2a
BC :-
\begin{gathered}\sqrt{ ( - a - 0 )^{2} } + \sqrt{ ( a \sqrt{3} - 0 )^{2} } \\ \\ \\ \sqrt{ ( a ^{2} + 3a^{2} ) } \\ \\ \\ \sqrt{ ( 4a^{2} )} \\ \\ \\ 2a\end{gathered}
(−a−0)
2
+
(a
3
−0)
2
(a
2
+3a
2
)
(4a
2
)
2a
CA :-
\begin{gathered}\sqrt{ ( a - 0 )^{2} } + \sqrt{ ( a \sqrt{3} - 0 )^{2} } \\ \\ \\ \sqrt{ ( a ^{2} + 3a^{2} ) } \\ \\ \\ \sqrt{ ( 4a^{2} )} \\ \\ \\ 2a\end{gathered}
(a−0)
2
+
(a
3
−0)
2
(a
2
+3a
2
)
(4a
2
)
2a
As AB = BC = CA
It is an equilateral triangle