यदि बिंदु `A(k+1,2k), B(3k,2k+3)` तथा `C(5k-1,5k)` सरीख हों, तो k का मान ज्ञात कीजिए।

Explanation

यहाँ `x_(1)=k+1,x_(2)=3k, x_(3)=5k-1`
`y_(1)=2k,y_(2)=2k+3, y_(3)=5k`
सरीख होने का प्रतिबन्ध
`x_(1)(y_(2)-y_(3)+x_(2)y_(3)-y_(1))+x_(3)(y_(1)-y_(2))=0`
`x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))=0`
`(k+1)(2k+3-5k)+3k(5k-2k)+(5k-1)(2k-2k-3)=0`
`(k+1)(3-3k)+3k(3k)+(5k-1)(-3)=0`
`(k+1)(3-3k)+3k(3k)+(5k-1)(-3)=0`
`6k^(2)-15k+6=0`
`2k^(2)-5k+2=0`
`2k^(2)-4k-k+2=0`
`2k(k-2)-1(k-2)=0`
`(k-2)(2k-1)=0`
`(k-2)(2k-1)=0`
जब `(k-2)(2k-1)=0`
jab `k-2=0` हो , तब
`k=2`
और जब `2k-1=0` हो, तब `k=1/2`
अतः `k=2` या `1/2`