UPSC NDA Math Model Paper 9

Here, we have to find the value of t for which the points (2, 4), (2t, 6t) and (3, 8) are collinear
Let A = (2, 4), B = (2t, 6t) and C = (3, 8)
Let x1=2,y1=4,x2=2t,y2=6t,x3=3 and y3=8
As we know that, if A(x1,y1),B(x2,y2) and C(x3,y3) are the vertices of a Δ ABC then area of Δ ABC = |A| where A=

1

2

.|

x1	y1	1
x2	y2	1
x3	y3	1

|
⇒A=

1

2

.|

2	4	1
2t	6t	1
3	8	1

|
⇒ |A| = t - 2
∵ The given points are collinear.
As we know that, if the points A(x1,y1),B(x2,y2) and C(x3,y3) are collinear then area of ΔABC = 0.
⇒ A = t - 2 = 0
⇒ k = 2
Hence, option B is the correct answer.