If A $(2 a, 4 a)$ and B $(2 a, 6 a)$ are two vertices of a triangle ABC and the vertex C is given by

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Solution

Length of $A B = 2 a$
(a) $C (4 a, 5 a)$
$B C = \sqrt{(2 a)^{2} + a^{2}} = \sqrt{5} a$
$C A = \sqrt{(2 a)^{2} + a^{2}} = \sqrt{5} a \Rightarrow B C = C A \neq A B$
$Δ A B C$ is isosceles
(b) $C ((2 + \sqrt{3}) a, 5 a)$
$B C = \sqrt{3 a^{2} + a^{2}} = 2 a$
$C A = \sqrt{3 a^{2} + a^{2}} = 2 a, Δ A B C$ is equilateral
(c) $C (6 a, 4 a)$
$B C = \sqrt{(4 a)^{2} + (2 a)^{2}} = 2 \sqrt{5} a$
$C A = \sqrt{(4 a)^{2} + 0} = 4 a$
So $(B C)^{2} = 20 a^{2} = (A B)^{2} + (A C)^{2}, Δ A B C$ is right-angled.
(d) $C (a, 3 a)$
$B C = \sqrt{a^{2} + (3 a)^{2}} = \sqrt{10} a$
$C A = \sqrt{a^{2} + a^{2}} = \sqrt{2} a$
so $cos A = \frac{(A B)^{2} + (A C)^{2} - (B C)^{2}}{2 A B \times A C}$
$= \frac{4 a^{2} + 2 a^{2} - 10 a^{2}}{2 \times 2 a \times \sqrt{2} a} < 0$
$\Rightarrow$ A is obtuse and thus $Δ A B C$ is obtuse angled.

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