Find the value of k for which the points P(k,-1) Q(2,1) & (4,5) are collinear
Answers
Answered by
4
Answer:
collinear formula : x1 ( y2-y3) + x2 (y3-y1) + x3( y1-y2) = 0
here, x1= k y1 = -1 x2 = 2 y2 = 1 x3 = 4 y3 = 5
k ( 1 - 5 ) + 2 ( 5 - {- 1}) + 4 ( - 1 - 1 ) = 0
= k ( - 4) + 12 - 8 = 0
= - 4 k + 4 = 0
k = - 4 / - 4 = 1
so, the value of k is 1.
Answered by
0
Answer:
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Step-by-step explanation:
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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