Question

find the value of k if the points A(k+1,2k) B (3k,2k+3) C(5k-1,5k)​

Answer 1

Step-by-step explanation:

Let us assume that the points A(x

1

,y

1

)=(k+1,2k),B(x

2

,y

2

)=(3k,2k+3) & C(x

3

,y

3

)=(5k+1,5k) form a triangle.

Now, A(ΔABC) with vertices A(x

1

,y

1

),B(x

2

,y

2

) and C(x

3

,y

3

) is given by

A(ΔABC)=

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

∴A(ΔABC)=

2

1

[(k+1)(2k+3−5k)+3k(5k−2k)+(5k+1)(2k−2k−3)]

∴A(ΔABC)=2k

2

−5k+2

Since, A,B,C are collinear ⟹A(ΔABC)=0

⟹2k

2

−5k+2=0

⟹(k−2)(2k−1)=0

⟹k=2,

2

1

Thank me for the answer