Q. 31

The base BC of an

Given: The base (BC) of the equilateral triangle ABC lies on y - axis, where, C has the coordinates: (0, - 3).The origin is the midpoint of the base.

To find: The coordinates of the points A and B. Also, the coordinates of another point D such that ABCD is a rhombus.

Solution:

Now, Δ ABC is an equilateral triangle

AB = AC = BC …(1)

By symmetry the coordinate A lies on x axis.

Also D is another point such that ABCD is rhombus and every side of rhombus is equal to each other.

So For this condition to be possible D will also lie on x axis.

Now,

Let coordinates of A be (x,0),B be (0,y) and D be ( -x,0).

or coordinates of A be ( - x,0),B be (0,y) and D be (x,0).

The figures are shown below: ...... (a) ....... (b)
From figure (a)

BC = √(0 - 0)2 + ( - 3 - y)2

⇒ BC =  √0 + 9 + y2 + 6y

⇒ BC =  √9 + y2 + 6y

Now, AC = √(0 - x)2 + ( - 3 - 0)2

AC = √x2 + ( - 3)2

⇒ AC = √(x2 + 9)

And

AB = √(0 - x)2 + (y - 0)2

AB = √x2 + y2

From (1)

AB = AC

⇒ √x2 + y2 =  √x2 + 9

Taking square on both sides we get,

x2 + y2 = x2 + 9

⇒ y2 =  9

⇒ y  =   ± 3

Since B lies in positive y direction.

∴ The coordinates of B are (0,3)

Now from (1) AB = BC

⇒√ x2 + y2 =   √9 + y2 + 6y

Take square on both sides

⇒ x2 + y2 =   9 + y2 + 6y

⇒ x2 =   9 + 6y

Put the value of y to get,

⇒ x2 =   9 + 6(3)

⇒ x2 =   9 + 18

⇒ x2 =   27

⇒ x  =   ± 3√3

Hence the coordinates of A can be ( 3√3,0) or ( - 3√3,0)

Also, ABCD is a rhombus.

⇒ AB = BC = DC = BD

So coordinates of D will be  ( -3√3,0) or ( 3√3,0)

Hence coordinates are A( 3√3,0) , B(0,3) , D( - 3√3,0)

Or coordinates are A( -3√3,0) , B(0,3) , D( 3√3,0)

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