Find the orthocenter of the triangle with ver | vedclass

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Question

(A) Let the vertices be $A(0, 0)$,$B(3, 4)$,and $C(4, 0)$.Since $AC$ lies on the $x$-axis $(y=0)$,the altitude from $B$ to $AC$ is a vertical line pas

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the vertices be $A(0, 0)$,$B(3, 4)$,and $C(4, 0)$.Since $AC$ lies on the $x$-axis $(y=0)$,the altitude from $B$ to $AC$ is a vertical line passing through $B(3, 4)$. Thus,the equation of this altitude is $x = 3$.Now,consider the altitude from $A(0, 0)$ to $BC$. The slope of $BC$ is $m_{BC} = \frac{0 - 4}{4 - 3} = -4$.The slope of the altitude from $A$ to $BC$ is $m_{\perp} = -\frac{1}{m_{BC}} = \frac{1}{4}$.The equation of this altitude is $y - 0 = \frac{1}{4}(x - 0)$,which simplifies to $x = 4y$.To find the orthocenter,we solve the system of equations $x = 3$ and $x = 4y$.Substituting $x = 3$ into $x = 4y$,we get $3 = 4y$,so $y = \frac{3}{4}$.Therefore,the orthocenter is $\left( 3, \frac{3}{4} \right)$.

Find the orthocenter of the triangle with vertices $A(0, 0)$,$B(3, 4)$,and $C(4, 0)$.

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