UPSC NDA Math Model Paper 10

Let ABC is equilateral triangle where,
A = (0, 2) B = (0, 4) and C = (x, y)
AB = BC = AC (sides of equilateral triangle)
AB=√(0)+(4−2)2=2units (Using distance formula)
BC=√(x−0)2+(y−4)2 =√(x)2+(y−4)2 units
AC=√(x−0)2+(y−2)2 =√(x)2+(y−2)2 units
Now, BC = AC (sides of equilateral triangle)
√(x)2+(y−4)2=√(x)2+(y−2)2
⇒(x)2+(y−4)2=(x)2+(y−2)2 (Squaring both sides)
⇒(y−4)2=(y−2)2
⇒y2−8y+16=y2−4y+4
⇒4y=12
⇒y=3
And AB=AC
√(x)2+(y−2)2=2
⇒x2+(y−2)2=4 (Squaring both sides)
⇒x2+(3−2)2=4
⇒x2=4−1=3⇒x=√3
∴C=(√3,3)
Hence, option (2) is correct.