In Delta ABC, if A(0, 0), B(4, 0) and C(0, 3) | vedclass

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

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right angled

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(A) The coordinates of the vertices are $A(0, 0)$,$B(4, 0)$,and $C(0, 3)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4$ units.$BC = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$ units.$AC = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{3^2} = 3$ units.Since $AB^2 + AC^2 = 4^2 + 3^2 = 16 + 9 = 25$ and $BC^2 = 5^2 = 25$,we have $AB^2 + AC^2 = BC^2$.This satisfies the Pythagorean theorem,indicating that $\Delta ABC$ is a right-angled triangle at vertex $A$. Since $AB 
eq AC$,it is not isosceles.

In $\Delta ABC,$ if $A(0, 0), B(4, 0)$ and $C(0, 3)$ then $\Delta ABC$ is a $\ldots \ldots \ldots$ triangle.

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