Q24 of 49 Page 1

If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex.

Or


Find the value of k, if the points P(5, 4), Q(7, k) and R(9, -2) are collinear.

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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