The points (-4,0), (4,0), (0,3) are the verti | vedclass

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(C) Let the vertices be $A(-4,0), B(4,0),$ and $C(0,3)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(4 - (-4))^2

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isosceles triangle

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(C) Let the vertices be $A(-4,0), B(4,0),$ and $C(0,3)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2 + 0^2} = 8$$BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$AC = \sqrt{(0 - (-4))^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$Since $BC = AC = 5$,two sides of the triangle are equal.Therefore,$	riangle ABC$ is an isosceles triangle.

The points $(-4,0), (4,0), (0,3)$ are the vertices of a

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