Question

The points (k+1 1) ( 2k+1 3) and ( 2k+2 2k) are collinear if

Answer 1

Answer:

k = 2 or -1/2

Step-by-step explanation:

Δ = 1/2| x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃ ( y₁ - y₂) |

   => 1/2| (k+1)(3-2k) + (2k+1)(2k-1) + (2k+2)(1-3)| = 0

   => 1/2| 3k + 3 - 2k² - 2k + 4k² - 1 - 4k - 4|

   => 1/2| 2k² - 3k -2| = 0

   => 2k² - 3k - 2 = 0

   => 2k² - 4k + k - 2 = 0

   => 2k(k - 2) + 1(k - 2) = 0

   => (k - 2)(2k + 1) = 0

   => k = 2 or -1/2  

Answer 2

Answer:

Answer:

k = 2 or -1/2

Step-by-step explanation:

Δ = 1/2| x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃ ( y₁ - y₂) |

   => 1/2| (k+1)(3-2k) + (2k+1)(2k-1) + (2k+2)(1-3)| = 0

   => 1/2| 3k + 3 - 2k² - 2k + 4k² - 1 - 4k - 4|

   => 1/2| 2k² - 3k -2| = 0

   => 2k² - 3k - 2 = 0

   => 2k² - 4k + k - 2 = 0

   => 2k(k - 2) + 1(k - 2) = 0

   => (k - 2)(2k + 1) = 0

   => k = 2 or -1/2