find the value of k if A(k+1,2k) ,B (3k,2k+3), C(5k-1,5k) are collinear
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Answered by
1231
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
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
mysticd:
ur welcome
Answered by
341
Answer:plzzgo through the image below.
Step-by-step explanation:First use formula then try to solve.
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