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