if the point a[k+1,2k],b[3k,2k+3]and c[5k-1,5k]are collinear then find the value of k
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a, b,and c are colinear so,
slope of ab = slope of bc = slope of ca
or
area of ∆ form by joining the point abc =0
we use, 2nd concept .
ar ∆ = 1/2 { (K +1)(2K +3 - 5K) + 3K(5K - 2K) + (5K -1)( 2K -2K -3)} =0
(K +1)(3 -3K) + 3K × 3K+ (5K -1)(-3) = 0
3( 1 - K²) + 9K² -15K + 3 = 0
3 + 6K² -15K +3 = 0
6K² -15K + 6 = 0
2K² - 5K + 2 = 0
2K² -4K - K + 2 = 0
2K( K -2) -(K -2) = 0
K = 2 , 1/2
slope of ab = slope of bc = slope of ca
or
area of ∆ form by joining the point abc =0
we use, 2nd concept .
ar ∆ = 1/2 { (K +1)(2K +3 - 5K) + 3K(5K - 2K) + (5K -1)( 2K -2K -3)} =0
(K +1)(3 -3K) + 3K × 3K+ (5K -1)(-3) = 0
3( 1 - K²) + 9K² -15K + 3 = 0
3 + 6K² -15K +3 = 0
6K² -15K + 6 = 0
2K² - 5K + 2 = 0
2K² -4K - K + 2 = 0
2K( K -2) -(K -2) = 0
K = 2 , 1/2
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