Latika Leekha
Last Activity: 10 Years ago
The given points are (2,6), (1,2) and (p,10).
for the points to be collinear, the detreminant should be zero.
Hence, we must have the value of the determinant as zero.
This gives 2(2-10) - 1(6-10) + p (6-2) = 0.
Hence, -16 + 4+ 4p = 0.
This yields 4p = 12.
Hence, p = 3.
Hence, for p = 3, the following points will be collinear.
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Latika Leekha
askIITians Faculty