A(2, 3, k), B(-1, k, -1), and C(4, -3, 2) are | vedclass

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(B) Given that $AB = AC$. Using the distance formula, $AB^2 = AC^2$. $AB^2 = (2 - (-1))^2 + (3 - k)^2 + (k - (-1))^2 = 3^2 + (3 - k)^2 + (k + 1)^2 = 9

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a right-angled isosceles triangle

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(B) Given that $AB = AC$. Using the distance formula, $AB^2 = AC^2$. $AB^2 = (2 - (-1))^2 + (3 - k)^2 + (k - (-1))^2 = 3^2 + (3 - k)^2 + (k + 1)^2 = 9 + 9 - 6k + k^2 + k^2 + 2k + 1 = 2k^2 - 4k + 19$. $AC^2 = (2 - 4)^2 + (3 - (-3))^2 + (k - 2)^2 = (-2)^2 + 6^2 + (k - 2)^2 = 4 + 36 + k^2 - 4k + 4 = k^2 - 4k + 44$. Equating $AB^2 = AC^2$: $2k^2 - 4k + 19 = k^2 - 4k + 44$. $k^2 = 25$. Since $k > 0$, we have $k = 5$. Now, calculate the side lengths: $AB^2 = 2(5)^2 - 4(5) + 19 = 50 - 20 + 19 = 49 \Rightarrow AB = 7$. $AC^2 = 49 \Rightarrow AC = 7$. $BC^2 = (-1 - 4)^2 + (5 - (-3))^2 + (-1 - 2)^2 = (-5)^2 + 8^2 + (-3)^2 = 25 + 64 + 9 = 98$. Since $AB^2 + AC^2 = 49 + 49 = 98 = BC^2$, the triangle satisfies the Pythagorean theorem. Thus, $	riangle ABC$ is a right-angled isosceles triangle.

$A(2, 3, k)$, $B(-1, k, -1)$, and $C(4, -3, 2)$ are the vertices of $\triangle ABC$. If $AB = AC$ and $k > 0$, then $\triangle ABC$ is:

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