Find the value of k in k + - 2/5 = 1/3
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Since points are collinear,
so, x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0
Take (7,−2) as (x1,y1), (5,1) as (x2,y2) and (3,k) as (x3,y3), we have
7(1−k)+5(k+2)+3(−2−1)=0
⇒ 7−7k+5k+10−9=0
⇒ 2k=8
⇒ k=4
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