Question

2. Find the value of ‘K'for which the points are collinear. (i) (7,-2) (5, 1) (3, K) (ii) (8, 1), (K, -4), (2,-5) (iii) (K, K) (2, 3) and (4, -1).

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​

Answer 1

Answer:

Answer

Correct option is

A

(i) k=4

C

(ii) k=3

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

Area of a 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

)

∣

∣

∣

∣

∣

1) Substituting the points (x

1

,y

1

)=(7,−2) ; (x

2

,y

2

)=(5,1) and (x

3

,y

3

)=(3,k)

In the area formula, we get

∣

∣

∣

∣

∣

2

7(1−k)+5(k+2)+3(−2−1)

∣

∣

∣

∣

∣

=0

∣

∣

∣

∣

∣

2

7−7k+5k+10−9

∣

∣

∣

∣

∣

=0

∣

∣

∣

∣

∣

2

8−2k

∣

∣

∣

∣

∣

=0

⇒8−2k=0

⇒k=4

2) Substituting the points (x

1

,y

1

)=(8,1) ; (x

2

,y

2

)=(k,−4) and (x

3

,y

3

)=(2,−5) in the area formula, we get

∣

∣

∣

∣

∣

2

8(−4+5)+k(−5−1)+2(1+4)

∣

∣

∣

∣

∣

=0

∣

∣

∣

∣

∣

2

8−6k+10

∣

∣

∣

∣

∣

=0

∣

∣

∣

∣

∣

2

18−6k

∣

∣

∣

∣

∣

=0

⇒18−6k=0

⇒k=3