Question

find the value of k for which the points A(k+1,2k)B(3k,2k+3),C5k-1,5k) are collinear

Answer 1

if A (x1, y1) , B(x2, y2), C(x2,y2) are three collinear points then

area of triangleABC = 0

1/2 l (x1[ y2-y1] +x2[y3-y1]+x3[y1-y2]l =0

given A(k+1,2k) ,B(3k,2k+3) ,C(5k-1,5k)

therefore

1/2l(k+1)[2k+3-5k]+3k[5k-2k]+(5k-1)[2k-(2k+3)]l=0

(k+1)[3-3k]+3k*3k+(5k-1)(-3)=0

3k-3k²+3-3k+9k²-15k+3=0

6k²-15k+6=0

divide each term with 2

2k²-5k+2=0

2k²-4k-k+2=0

2k(k-2)-(k-2)=0

(k-2)(2k-1)=0

k-2 =- or 2k-1 =0

k=2 or k= 1/2

Hope it helps you

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