UPSC NDA Math Model Paper 4

Given, the points A=(2,4),B=(2,6) and C=(2+√3,k) are the vertices of an equilateral triangle.

AB = AC
By distance formula, we have
√(2−2)2+(4−6)2 =√(2−2−√3)2+(4−k)2
√(−2)2=√(−√3)2+(4−k)2
Squaring both side, we get
(−2)2=(−√3)2+(4−k)2
⇒4=3+16−8k+k2
⇒k2−8k+15=0
⇒(k−5)(k−3)=0
⇒k=5,3
Hence,the points (2,4),(2,6) and (2+√3,k) are the vertices of an equilateral triangle, then the value of k is 5