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