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Let the coordinates of the third vertex be (x,y).

So, A(3,0), B(6,0) and C(x,y) are the three points of an equilateral triangle.

We know that in an equilateral triangle all the three sides are equal.

So, AB=BC=CA

Length of AB =√ (6-3)2 + (0-0)2 =3

Length of BC =√ (x-6)2 + (y-0)2

Length of CA =√ (x-3)2 + (y-0)2

BC=CA

√ (x-6)2 + y2 =√ (x-3)2 + y2

(x-6)2 + y2 =(x-3)2 + y2

(x-6)2 = (x-3)2

x-6= -(x-3)

x-6= -x + 3

2x=9

.

BC=AB

√ (x-6)2 + y2 = 3

(x-6)2 + y2 = 9

.

.

The possible coordinates of the third vertex are . a.

Or

Given the points P(5, 4), Q(7, k) and R(9, -2) are collinear.

Condition for collinearity is:

x1(y2-y3) + x2(y3-y1) + x3(y1-y2) = 0 .

+ 2) + 7(-2-4) + 9(4-k) = 0.

5k + 10 -42 + 36 -9k =0

- 4k + 4 =0

4k=4

k=1

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