Question

If the points (k,3k),(3k,3k) and (3,1) are collinear,then the value of k is ______​

Answer 1

Answer:

Correct option is

D

3

−1

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero

The area of the triangle with vertices (x

1

,y

1

) ; (x

2

,y

2

) and (x

3

,y

3

) is

∣

∣

∣

∣

2

x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)

∣

∣

∣

∣

Hence, substituting the points (x

1

,y

1

)=(k,2k) ; (x

2

,y

2

)=(3k,3k) and (x

3

,y

3

)=(3,1) in the area formula, we get

∣

∣

∣

∣

2

k(3k−1)+3k(1−2k)+3(2k−3k)

∣

∣

∣

∣

=0

=>3k

2

−k+3k−6k

2

+6k−9k=0

=>−3k

2

−k=0

−k(3k+1)=0

k=0 or k=−

3

1

Step-by-step explanation:

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