Question

plz solve it fast plz plz​

Answer 1

Step-by-step explanation:

when these points r collinear,then

area of ∆=0

x¹(y²-y³)+x²(y³-y¹)+x³(y¹-y³)=0

k+1(2k+3-5k)+3k(5k-2k)+5k-1(2k-2k-3)=0

k+1(3-3k)+3k(3k)+5k-1(-3)=0

3k+3-3k-3k²+9k²-15k+3=0

6k²-15k+6=0

by spitting the middle term,

6k²-12k-3k+6=0

6k(k-2)-3(k-2)=0

(6k-3)(k-2)=0

k=2 and k=1/2

hope it helps you.....

Answer 2

Answer:

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