The orthocentre of the triangle with vertices | vedclass

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Question

(B) Let the vertices be $A = \left( 2, \frac{\sqrt{3} - 1}{2} \right)$,$B = \left( \frac{1}{2}, -\frac{1}{2} \right)$,and $C = \left( 2, -\frac{1}{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the vertices be $A = \left( 2, \frac{\sqrt{3} - 1}{2} \right)$,$B = \left( \frac{1}{2}, -\frac{1}{2} \right)$,and $C = \left( 2, -\frac{1}{2} \right)$.Observe that the side $AC$ is vertical (both $x$-coordinates are $2$) and the side $BC$ is horizontal (both $y$-coordinates are $-\frac{1}{2}$).Since $AC$ is perpendicular to $BC$,the triangle is a right-angled triangle with the right angle at vertex $C = \left( 2, -\frac{1}{2} \right)$.In a right-angled triangle,the orthocentre is the vertex where the right angle is located.Therefore,the orthocentre is $\left( 2, -\frac{1}{2} \right)$.

The orthocentre of the triangle with vertices $\left( 2, \frac{\sqrt{3} - 1}{2} \right)$,$\left( \frac{1}{2}, -\frac{1}{2} \right)$,and $\left( 2, -\frac{1}{2} \right)$ is:

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