Find the value of K, if the points A(2,3) B(4, K) & C(6.3) are collinear.
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Answer:
Points A(2,3),B(4,k) and C(6,−3) are collinear.
Area of triangle having vertices A, B and C=0
Area of a triangle =
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Area of given ΔABC=0
⇒
2
1
[2(k−(−3))+4(−3−3)+6(3−k))]=0
⇒2k+6−24+18−6k=0
⇒−4k=0
or k=0
The value of k is zero.
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