We know that, if three points are collinear, then the area of triangle formed by these points is zero.
Since, the points A(k+1,2k), B(3k,2k+3) and C(5k-1,5k), are collinear.
Then, area of Δ ABC=0.
⇒12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=0Here, x1=k+1,x2=3k,x3=5k−1and y1−2k,y2−2k+3,y3=5k⇒12[(k+1)(2k+3−5k)+3k(5k−2k)+(5k−1)(2k−(2k+3))]=0⇒12[(k+1)(3−3k)+3k(3k)+(5k−1)(2k−2k−3)]=0⇒12[−3k2+3k−3k+3+9k2−15k+3]=0⇒12(6k2−15k+6)=0 [multiply by 2]⇒6k2−15k+6=0 [by factorization method]⇒2k2−5k+2=0⇒2k2−4k−k+2=0⇒2k(k−2)−1(k−2)=0⇒(k−2)(2k−1)=0
If k – 2 = 0, then k = 2
If 2k – 1, then k = 12
∴k=2,12
Hence, the required values of k are 2 and 12.