What is the distance of the orthocenter from | vedclass

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Question

(A) Let the vertices be $A(2, 3)$,$B(4, 5)$,and $C(-1, 10)$.First,check the slopes of the sides:Slope of $AB = \frac{5-3}{4-2} = \frac{2}{2} = 1$.Slop

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(A) Let the vertices be $A(2, 3)$,$B(4, 5)$,and $C(-1, 10)$.First,check the slopes of the sides:Slope of $AB = \frac{5-3}{4-2} = \frac{2}{2} = 1$.Slope of $BC = \frac{10-5}{-1-4} = \frac{5}{-5} = -1$.Since (Slope of $AB$) $	imes$ (Slope of $BC$) = $1 	imes (-1) = -1$,the triangle is a right-angled triangle at vertex $B(4, 5)$.In a right-angled triangle,the orthocenter is the vertex at which the right angle is formed.Therefore,the orthocenter $H$ is at $B(4, 5)$.The distance of the orthocenter from vertex $B$ is the distance from $B$ to $B$,which is $0$. However,the question asks for the distance from the vertex (usually implying the distance from the orthocenter to the vertices of the triangle).If the orthocenter is $B(4, 5)$,the distance from $B$ to $A$ is $\sqrt{(4-2)^2 + (5-3)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.The distance from $B$ to $C$ is $\sqrt{(4 - (-1))^2 + (5-10)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2}$.Given the options,the distance from the orthocenter to vertex $A$ is $2\sqrt{2}$.

What is the distance of the orthocenter from the vertex of a triangle with vertices $(2, 3), (4, 5),$ and $(-1, 10)$?

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