find the value of k for which A(1,2),(k,4)and C(7,-1) are collinear
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Answer:
Step-by-step explanation:
A(1,2) B(k,4) C(7,-1)
Area of triangle will be equal to zero bcoz the points are coolinear
1/2(1*5+K*-3+7*-2)=0
1/2(5-3k-14)=0
-9-3k/2=0
-9-3k=0
-3k=9
K=9/-3
K=-3
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since the points are collinear ,area of triangle is zero
area of ∆ = 1\2[x1y2+x2y3+x3y1]-[x2y1+x3y2+x1y3]
on substituting x1,y1=1,2&x2,y1=k,4;x3,y3=7,-1
1\2[x1y2+x2y3+x3y1]-[x2y1+x3y2+x1y3]=0
we get
k=-3
area of ∆ = 1\2[x1y2+x2y3+x3y1]-[x2y1+x3y2+x1y3]
on substituting x1,y1=1,2&x2,y1=k,4;x3,y3=7,-1
1\2[x1y2+x2y3+x3y1]-[x2y1+x3y2+x1y3]=0
we get
k=-3
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