A(-3,0) and B(3,0) are the vertices of an equilateral triangle ABC. Find the coordinates of C.
Answers
Answer:
(0,5.2)
Step-by-step explanation:
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Step-by-step explanation:
Given :-
A(-3,0) and B(3,0) are the vertices of an equilateral triangle ABC.
To find:-
Find the coordinates of C?
Solution :-
Given that
A(-3,0) and B(3,0) are the vertices of an equilateral triangle ABC.
Let the coordinates of C be (x,y)
Since ∆ABC is an equilateral trianagle, all the three sides are equal .
AB = BC = CA
Finding AB:-
Let (x1, y1) = (-3,0)=>x1 = -3 and y1 = 0
Let (x2, y2) = (3,0) => x2 = 3 and y2 = 0
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> AB =√[(3-(-3))²+(0-0)²]
=> AB = √[(3+3)²+0²]
=> AB = √(6²+0)
=> AB =√6²
=> AB = 6 units -----------(1)
Finding BC:-
Let (x1, y1) = (3,0)=>x1 = 3 and y1 = 0
Let (x2, y2) = (x,y) => x2 = xand y2 = y
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> BC =√[(x-3)²+(y-0)²]
=> BC = √[(x-3)²+y²]
=> BC= √(x²-6x+9+y²) units -----------(2)
Finding AC:-
Let (x1, y1) = (-3,0)=>x1 = -3 and y1 = 0
Let (x2, y2) = (x,y) => x2 = x and y2 = y
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> AC =√[(x-(-3))²+(y-0)²]
=> AC = √[(x+3)²+y²]
=> AC = √(x²+6x+9+y²) units -----------(3)
On taking (1) &(2)
AB = BC
=> 6 = √(x²-6x+9+y²)
On squaring both sides then
=> 6² = [√(x²-6x+9+y²)]²
=> 36 = x²-6x+9+y²
=> x²-6x+y² = 36-9
=> x²-6x+y² = 27 --------------(4)
On taking (2) &(3)
=>BC = AC
√(x²-6x+9+y²) = √(x²+6x+9+y²)
On squaring both sides then
[√(x²-6x+9+y²)]² = [√(x²+6x+9+y²)]²
=> x²-6x+9+y² = x²+6x+9+y²
=> -6x = 6x
=> 6x+6x = 0
=> 12x = 0
=> x = 0/12
=> x = 0
On Substituting the value of x = 0 in (4) then
=> 0²-6(0)+y² = 27
=> 0-0+y² = 27
=> y² = 27
=> y = ±√27
=>y = ±√(3×3×3)
=> y = ±3√3
Therefore, y = 3√3 or -3√3
We know that
√3 = 1.732
3√3 = 3×1.732 = 5.196=5.2(Correct it one decimal)
Therefore, (x,y) = (0,3√3) or (0,-3√3)
or (0,5.2) or (0,-5.2)
Answer:-
The coordinates of the third vertex of the equilateral triangle ABC is (0,3√3 ) or (0,-3√3) or (0,5.2) or (0,-5.2)
Used Concept:-
All the lengths of the sides of a triangle are equal then they are the vertices of an equilateral triangle.
Used formulae:-
Distance formula:-
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
- √3 = 1.732
