Question

If the points A(1, 2), B(5, 4), and C(-1, 1) lie on a straight line, find the value of k.​

Answer 1

Answer:

Consider the given points.

(k,−1),(2,1) and (4,5)

Since, these points are collinear means that the area of triangle must me zero.

So,

21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣=0

where (x1,y1),(x2,y2),(x3,y3) are the points

Therefore,

k(1−5)+2(5+1)+4(−1−1)=0

k(−4)+2(6)+4(−2)=0

−4k+12−8=0

−4k+4=0

4k=4

k=1

Hence, this is the answer.