Question

find the value of k for the points [3k-1,k-2] [k,k-7] [k-1,-k-2] are collinear

Answer 1

k-7-(k-2)/k-(3k-1)=(-k-2)-(k-7)/k-1-k
k-7-k+2/k-3k+1=-k-1-k+7/k-1-k
-5/-2k+1=6/-1
(-5)(-1)=(-2k+1)(6)
5=-12k+6
12k=6-5
12k=1
:.k=1/12

Answer 2

Answer:

The value of k is either 0 or 3.

Step-by-step explanation:

The given points are A[3k-1,k-2], B[k,k-7], C[k-1,-k-2].

These points are collinear if slope of AB is equal to slope of BC.

Slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

$m_{AB}=\frac{k-7-(k-2)}{k-(3k-1)}$

$m_{AB}=\frac{-5}{1-2k}$

$m_{BC}=\frac{-k-2-(k-7)}{k-1-k}$

$m_{BC}=\frac{-2k+5}{-1}$

All point are collinear, so

$m_{AB}=m_{BC}$

$\frac{-5}{1-2k}=\frac{(-2k+5)}{-1}$

$5=(1-2k)(-2k+5)$

$4k^2-12k=0$

$4k(k-3)=0$

$k=0,3$

Therefore the value of k is either 0 or 3.