Answer:
Let A (2a, 4a)
B(2a, 6a)
C (2a + sqrt (3a), 5a)
Side AC:
[tex]\sqrt{(2a+\sqrt{3a}-2a)^2+(5a-4a)^2 }=\sqrt{3a+a^2}[/tex]
Side BC:
[tex]\sqrt{(2a+\sqrt{3a}-2a)^2+(5a-6a)^2 }=\sqrt{3a+a^2}[/tex]
Hence AC = BC
It is an isoceles triangle.
Side AB:
[tex]\sqrt{(2a-2a)^2+(6a-4a)^2 }=\sqrt{2a^2}=2a[/tex]
It proves that the triangle is NOT an equilateral triangle.
