For a triangle formed by the vertices (0,0),( | vedclass

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(A) Let the vertices of the triangle be $O(0,0)$,$B(4,0)$,and $C(3,4)$.To find the orthocenter,we need the intersection of the altitudes.$1$. The alti

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$\left(3, \frac{3}{4}\right)$

Vedclass · Answer

(A) Let the vertices of the triangle be $O(0,0)$,$B(4,0)$,and $C(3,4)$.To find the orthocenter,we need the intersection of the altitudes.$1$. The altitude from $C(3,4)$ to the side $OB$ (which lies on the $x$-axis) is a vertical line passing through $x=3$. Thus,the equation of this altitude is $x=3$.$2$. The slope of side $BC$ is $m_{BC} = \frac{4-0}{3-4} = \frac{4}{-1} = -4$.$3$. The altitude from $O(0,0)$ to $BC$ is perpendicular to $BC$. Its slope is $m_{\perp} = -\frac{1}{m_{BC}} = -\frac{1}{-4} = \frac{1}{4}$.$4$. The equation of the altitude from $O$ is $y - 0 = \frac{1}{4}(x - 0)$,which simplifies to $y = \frac{1}{4}x$.$5$. Substituting $x=3$ into the equation $y = \frac{1}{4}x$,we get $y = \frac{1}{4}(3) = \frac{3}{4}$.Therefore,the orthocenter is $\left(3, \frac{3}{4}\right)$.

For a triangle formed by the vertices $(0,0)$,$(4,0)$,and $(3,4)$,the orthocenter is:

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