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Question

(B) Let the vertices be $A(2, \frac{\sqrt{3}-1}{2})$,$B(\frac{1}{2}, -\frac{1}{2})$,and $C(2, -\frac{1}{2})$.Calculate the lengths of the sides:$AB^2

Vedclass · Accepted Answer

$(2, -\frac{1}{2})$

Vedclass · Answer

(B) Let the vertices be $A(2, \frac{\sqrt{3}-1}{2})$,$B(\frac{1}{2}, -\frac{1}{2})$,and $C(2, -\frac{1}{2})$.Calculate the lengths of the sides:$AB^2 = (2 - \frac{1}{2})^2 + (\frac{\sqrt{3}-1}{2} - (-\frac{1}{2}))^2 = (\frac{3}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{9}{4} + \frac{3}{4} = 3$.$BC^2 = (2 - \frac{1}{2})^2 + (-\frac{1}{2} - (-\frac{1}{2}))^2 = (\frac{3}{2})^2 + 0 = \frac{9}{4}$.$CA^2 = (2 - 2)^2 + (\frac{\sqrt{3}-1}{2} - (-\frac{1}{2}))^2 = 0 + (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$.Since $BC^2 + CA^2 = \frac{9}{4} + \frac{3}{4} = \frac{12}{4} = 3 = AB^2$,the triangle is a right-angled triangle with the right angle at vertex $C(2, -\frac{1}{2})$.In a right-angled triangle,the orthocenter is the vertex where the right angle is located.Therefore,the orthocenter is $C(2, -\frac{1}{2})$.

What is the orthocenter of the triangle with vertices $(2, \frac{\sqrt{3}-1}{2})$,$(\frac{1}{2}, -\frac{1}{2})$,and $(2, -\frac{1}{2})$?

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