Find the value of ‘k’, for which the points are collinear A(2,3),B(4,k) and C(6,−3)

Given 
Points A(2,3),B(4,k) and C(6,−3) are collinear.
Find out
We have to determine the value of k
Solution
Area of triangle having vertices A, B and C=0
We know that
Area of a triangle is given by =1/2 [x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
When points are collinear the area of triangle is zero
Area of given ΔABC=0
On substituting the values of coordiantes we get
⇒1/2[2(k−(−3))+4(−3−3)+6(3−k))]=0
⇒2k+6−24+18−6k=0
⇒−4k=0
 k=0

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