Correct Answer: Option B

Explanation

 Let A = (1, -3), B = (6, 2) and C = (0,4)

\(d =\sqrt{ (y_2 - y_1)^2 + (x_2 - x_1)^2}\)
|AB| = \(\sqrt{(2 - (-3))^2 + (6 - 1)^2}\) 
= \(\sqrt{(2 + 3)^2 + (6 - 1)^2}\) 
= \(\sqrt{5^2 + 5^2}\) 
= \(\sqrt{25 + 25}\)
= \(\sqrt50\)
= 5\(\sqrt2\) units

|BC| = \(\sqrt{(4 - 2)^2 + (0 - 6)^2}\)
= \(\sqrt{2^2 + (-6)^2}\)
= \(\sqrt{4 + 36}\)
= \(\sqrt40\) 
= 2\(\sqrt10\)  units

|AC| = \(\sqrt{(4 - (-3)^2 + (0 - 1)^2}\)
= \(\sqrt{(4 + 3)^2 + (0 - 1)^2}\)
= \(\sqrt{7^2 + (-1)^2}\) 
= \(\sqrt{49 + 1}\) 
= \(\sqrt50\)
= 5\(\sqrt{2}\)  units

Since two sides are equal (|AB| = |AC|), then the triangle is Isosceles.

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