Question

The base AB of the two equilateral triangles ABC and ABC' with side 2a lies along the X-axis such that the mid-point of AB is at the origin.  Find the coordinates of the vertices C and C' of the triangles. 


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Answer 1

Answer:

Since the mid-point of AB is at the origin O and AB=2a

∴ OA=OB=a.

Thus, the coordinates of A and B are (a,0) and (−a,0) respectively.

Since triangles ABC. and ABC' are equilateral. Therefore, their third vertices C and C'

lie on the perpendicular bisector of base AB. Clearly, YOY is the perpendicular bisector

of AB. Thus, C and C' lie on Y-axis. Consequently, their x-coordinates are equal to Zero.

In △AOC,wehave

OA

2

+OC

2

=AC

2

⇒a

2

+OC

2

=(2a)

2

⇒OC

2

=4a

2

−a

2

⇒OC

2

=3a

2

OC=

3a

Similarly, by applying Pythagoras theorem in △AOC; we have OC

′

=

3a

Thus, the coordinates of C and C' are (0,

3

a)and(0,−

3

a)respectively.

Answer 2

$\large\bold{\underline{\underline{Question:-}}}$

The base AB of the two equilateral triangles ABC and ABC' with side 2a lies along the X-axis such that the mid-point of AB is at the origin.  Find the coordinates of the vertices C and C' of the triangles. 

$\large\bold{\underline{\underline{Answer:-}}}$

∆ABC and ∆ABC’ are equilateral triangles with side 2a and side AB is on x axis with it's midpoint at the origin O.

AO =a (AB = 2a , and O is midpoint)

∆AOC is a right angle triangle. AC is hypotenuse. Using Pythogoras theorem

${O}^{2} = {AC}^{2} - {AO}^{2}$

${O}^{2} = ( {2a)}^{2} - {a}^{2} = 4 {a}^{2} - {a}^{2} = 3$

$OC = \sqrt{(3a ^{2}) } - \sqrt{3a}$

Coordinates of x is Zero on y axis.

Coordinates of C is$(0 ,\sqrt{3a} )$ and

Coordinates of C' is $(0, - \sqrt{3a} )$