If the points (k, 2-2 k),(1-k, 2 k) and (-k-4,6-2 k) are collinear, then possible values of k are

Question

If the points (k, 2-2 k),(1-k, 2 k) and (-k-4,6-2 k) are collinear, then possible values of k are (A) -1 / 2,1 (B) 1 / 2,-1 (C) 1,2 (D) 1,3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/c648d88576aa5fa89ba6c28d3ea1c0ed-.png" /> Since A, B, C are collinear Slope of A B= Slope of B C (2-2 k -2 k / k -1+ k )=(2 k -6+2 k /1- k + k +4) ⇒ (2-4 k /2 k -1)=(4 k -6/5) ⇒ 10-20 k =(4 k -6)(2 k -1) ⇒ (4 k -6)(2 k -1)+10(2 k -1)=0 ⇒ k =1 / 2,-1

Q. If the points $(k, 2 - 2 k), (1 - k, 2 k)$ and $(- k - 4, 6 - 2 k)$ are collinear, then possible values of $k$ are

$- 1/2, 1$

$1/2, - 1$

1,2

1,3

Q. If the points (k,2−2k),(1−k,2k) and (−k−4,6−2k) are collinear, then possible values of k are

−1/2,1

1/2,−1

1,2

1,3

Since A, B, C are collinear Slope of AB= Slope of BC k−1+k2−2k−2k​=1−k+k+42k−6+2k​ ⇒2k−12−4k​=54k−6​ ⇒10−20k=(4k−6)(2k−1) ⇒(4k−6)(2k−1)+10(2k−1)=0 ⇒k=1/2,−1

Q. If the points $(k, 2 - 2 k), (1 - k, 2 k)$ and $(- k - 4, 6 - 2 k)$ are collinear, then possible values of $k$ are

$- 1/2, 1$

$1/2, - 1$