What is the orthocenter of a triangle with corners at $(7, 3)$ $(7 ,3 )$ , $(4, 8)$ $(4 ,8 )$ , and (6 ,8 )#?

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Answer 1 · 2017-06-19T12:52:56

The orthocenter is $= (7, \frac{42}{5})$ $=(7,42/5)$

Let the triangle $DeltaABC$ be

$A=(7,3)$

$B=(4,8)$

$C=(6,8)$

The slope of the line $BC$ is $=(8-8)/(6-4)=0/2=0$

The slope of the line perpendicular to $BC$ is $=-1/0=-oo$

The equation of the line through $A$ and perpendicular to $BC$ is

$x=7$ ................... $(1)$

The slope of the line $AB$ is $=(8-3)/(4-7)=5/-2=-5/2$

The slope of the line perpendicular to $AB$ is $=2/5$

The equation of the line through $C$ and perpendicular to $AB$ is

$y-8=2/5(x-6)$

$y-8=2/5x-12/5$

$y-2/5x=28/5$ ................... $(2)$

Solving for $x$ and $y$ in equations $(1)$ and $(2)$

$y-2/5*7=28/5$

$y-14/5=28/5$

$y=28/5-14/5=42/5$

The orthocenter of the triangle is $=(7,42/5)$

What is the orthocenter of a triangle with corners at (7 ,3 )(7,3)(7 ,3 ), (4 ,8 )(4,8)(4 ,8 ), and (6 ,8 )#?