A triangle with vertices (4, 0), (-1, -1), (3 | vedclass

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Question

(A) Let the vertices be $A(4, 0)$,$B(-1, -1)$,and $C(3, 5)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(-1 - 4)

Vedclass · Accepted Answer

Isosceles and right angled

Vedclass · Answer

(A) Let the vertices be $A(4, 0)$,$B(-1, -1)$,and $C(3, 5)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(-1 - 4)^2 + (-1 - 0)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$.$AC = \sqrt{(3 - 4)^2 + (5 - 0)^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$.$BC = \sqrt{(3 - (-1))^2 + (5 - (-1))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$.Since $AB = AC = \sqrt{26}$,the triangle is isosceles.Also,$AB^2 + AC^2 = 26 + 26 = 52$ and $BC^2 = 52$.Since $AB^2 + AC^2 = BC^2$,the triangle is right-angled at $A$ by the converse of the Pythagorean theorem.Thus,the triangle is isosceles and right-angled.

$A$ triangle with vertices $(4, 0), (-1, -1), (3, 5)$ is

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