Question

answer this question with answer only​

Answer 1

Please mark me as brainliest

Answer 2

Answer:

k=3

Step-by-step explanation:

In this question, let

$x_{1}=2 \\x_{2}=4 \\x_{3}=6$

and

$y_{1}=3\\ y_{2} =k \\y_{3} =3$

Since points A, B and C are collinear, therefore by section formula

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

⇒   2(k - 3) + 4 (3 -3) + 6(3 - k)  =   0

⇒   2k - 6 + 4 (0) + 18 - 6k   =  0

⇒   2k - 6k - 6 + 18 = 0

⇒   -4k + 12  = 0

⇒   -4k = -12

⇒      k = 3

Hope this helps you!!!