Question

if points (3k-1,k-2), (k, k-7)and(k-1, -k-2) are colonist find K. ​

Answer 1

Answer:

let A(3k-1,k-2)

B(k,k-7)

C(k-1,-k-2)

since they are collinear

area of triABC =0

$0 = \frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) | \\ 0 \times 2 = |(3k - 1)(k - 7 + k + 2) + k( - k - 2 - k + 2) + (k - 1)(k - 2 - k + 7)| \\ 0 = |(3k - 1)(2k - 5) + k( - 2k) + (k - 1)(5)| \\ 0 = |6 {k}^{2} - 15k - 2k + 5 - 2 {k}^{2} + 5k - 5 | \\ 0 = |4 {k}^{2} - 12k| \\ 0 = 4 {k}^{2} - 12k \\ 0 = 4k(k - 3) \\ 4k = 0 \: \: \: \: \: \: \: \: \: k - 3 = 0 \\ k = 0 \: \: \: \: \: k = 3$