find the value of K for which the point (1,-1),(2,-1) and (k,-1) lie on the same straight line
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1
Answer:
Consider the given points.
(k,−1),(2,1) and (4,5)
Since, these points are collinear means that the area of triangle must me zero.
So,
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)∣=0
where (x
1
,y
1
),(x
2
,y
2
),(x
3
,y
3
) are the points
Therefore,
k(1−5)+2(5+1)+4(−1−1)=0
k(−4)+2(6)+4(−2)=0
−4k+12−8=0
−4k+4=0
4k=4
k=1
Hence, this is the answer.
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