Find the values of k, if the points A(k+1,2k),B(3k,2k+3) and C(5k+1,5k) are collinear.
হানি নাটস
Let us assume that the points A(x1,y1)=(k+1,2k),B(x2,y2)=(3k,2k+3) & C(x3,y3)=(5k+1,5k) form a triangle.
Now,
area(ΔABC) with vertices
A(x1,y1),B(x2,y2) and
C(x3,y3) is given by
Area(ΔABC)=21[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] ∴Area(ΔABC)=21[(k+1)(2k+3−5k)+3k(5k−2k)+(5k+1)(2k−2k−3)]
∴Area(ΔABC)=2k2−5k+2 Since,
A,B,C are collinear
⟹Area(ΔABC)=0⟹2k2−5k+2=0 ⟹(k−2)(2k−1)=0 ⟹k=2,21