Question

Find the values of k if the points a (k + 1, 2k), b (3k, 2k + 3) and c (5k 1, 5k) are collinear

Answer 1

GIVEN:

Points : a(k + 1, 2k) , b(3k, 2k + 3) and c(5k - 1 , 5k)

We know that,

Ar(∆) when points are collinear = 0

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

0 = 1/2 | (k + 1)(2k + 3 - 5k) + 3k(5k - 2k) + (5k - 1)(2k - 2k - 3)

0 = 1/2 | k(-3k + 3) + ( -3k + 3) + 3k(3k) + 5k(-3) - 1(-3) |

0 = 1/2 | -3k² + 3k - 3k + 3 + 9k² - 15k + 3 |

0 = 1/2 |9k² - 3k² - 15k + 6 |

0 = 1/2 |6k² - 15k + 6 |

Divide with three

0 = 1/2 | 2k² - 5k + 2 |

0 = 1/2 | 2k² - 4k - k + 2 |

0 = 1/2 |2k(k - 2) - 1(k - 2)|

0 = 1/2 |(k - 2) (2k - 1)|

(k - 2) (2k - 1) = 0

k - 2 = 2 or 2k - 1 = 0

k = 2 or k = 1/2

Therefore, the value of k is 2 or 1/2.

Answer 2

$\huge{\boxed{\boxed{\red{\mathbb{ANSWER}}}}}$

Step-by-step explanation:

½ [(k + 1)(2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1)(2k – 2k – 3)] = 0

(k + 1)(3 – 3k) + 9k2 – 3(5k – 1) = 0

2k2 – 5k + 2 = 0

(k – 2)(2k – 1) = 0

k = 2, 1/2

may be it's helpful for you ✴️