Find the value of K if the points A(2,3),B(4,k) and C(6,-5) collinear
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Step-by-step explanation:
A(2,3)[ x1,y1]
B(4,k)[ x2, y2]
C(6, -5)[ x3, y3]
GN.that ar(∆ABC)= 0{collinear}
ar(∆ABC)= 1/2[x1(y2-y3)+ x2(y3-y1) + x3(y1-y2)]
0= 3(k+5)+4(-5-3)+6(3-k)
0= 3k+15+4(-8)+18k-6k
0= 3k-6k+18k+15-32
17= 15k
K= 15/17
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