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got it
since F, E, and D are collinear
FE=hFD, where h is a constant
q-p = h(p(3-4k) + 2kp)
-p+q = hp (3-4k) + 2hkq
comparing the coefficients of p and q
-1=h(3-4k) which implies -1 = 3h-4hk ... (1)
1=2hk which implies k=1/2h ... (2)
substitute (2) in to (1)
-1= 3h-4h (1/2h)
h= 1/3
substitute h=1/3 in to (2)
k=1/2(1/3)
k=1.5
since F, E, and D are collinear
FE=hFD, where h is a constant
q-p = h(p(3-4k) + 2kp)
-p+q = hp (3-4k) + 2hkq
comparing the coefficients of p and q
-1=h(3-4k) which implies -1 = 3h-4hk ... (1)
1=2hk which implies k=1/2h ... (2)
substitute (2) in to (1)
-1= 3h-4h (1/2h)
h= 1/3
substitute h=1/3 in to (2)
k=1/2(1/3)
k=1.5