find the value of k for which the points are collinear if k, k ;2, 3 ;4, -1;
Answers
Answered by
1
Step-by-step explanation:
We know that,
Area of ΔABC=0
=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]=0
Here,
x
1
=k,y
1
=k
x
2
=2,y
2
=3
x
3
=4,y
3
=−1
Putting values,
=
2
1
[k(3−(−1))+2((−1)−k)+4(k−3)]=0
=
2
1
[3k+k−2−2k+4k−12]=0
=
2
1
[6k−14]=0
2
6k−14
=0
6k−14=0
6k=14
k=
6
14
=
3
7
k=
3
7
Hence, this is the answer.
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