SOLUTION: 1. An altitude of a triangle is a line segment from a vertex perpendicular to the opposite side. Find the equations of the altitudes of the triangle with vertices (4, 5),(-4, 1) an

Algebra ->  Graphs -> SOLUTION: 1. An altitude of a triangle is a line segment from a vertex perpendicular to the opposite side. Find the equations of the altitudes of the triangle with vertices (4, 5),(-4, 1) an      Log On


   

Question 127639: 1. An altitude of a triangle is a line segment from a vertex perpendicular to the opposite side. Find the equations of the altitudes of the triangle with vertices (4, 5),(-4, 1) and (2, -5). Do this by solving a system of two of two of the altitude equations and showing that the intersection point also belongs to the third line.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If we plot the points and connect them, we get this triangle:



Let point
A=(xA,yA)
B=(xB,yB)
C=(xC,yC)



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Let's find the equation of the segment AB


Start with the general formula

y=%28%28yA-yB%29%2F%28xA-xB%29%29%28x-xA%29%2ByA


Plug in the given points

y=%28%28+5-+1%29%2F%284%2B4%29%29%28x-4%29%2B+5


Simplify and combine like terms

y=%281%2F2%29x%2B3


So the equation of the line through AB is y=%281%2F2%29x%2B3


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Let's find the equation of the segment BC


Start with the general formula

y=%28%28yB-yC%29%2F%28xB-xC%29%29%28x-xB%29%2ByB


Plug in the given points

y=%28%28+1%2B5%29%2F%28-4-2%29%29%28x%2B4%29%2B+1


Simplify and combine like terms

y=-x-3


So the equation of the line through BC is y=-x-3




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Let's find the equation of the segment CA


Start with the general formula

y=%28%28yC-yA%29%2F%28xC-xA%29%29%28x-xC%29%2ByC


Plug in the given points

y=%28%28+-5-+5%29%2F%282-4%29%29%28x-2%29%2B+-5


Simplify and combine like terms

y=5x-15


So the equation of the line through CA is y=5x-15




So we have these equations of the lines that make up the triangle





So to find the equation of the line that is perpendicular to y=%281%2F2%29x%2B3 that goes through the point C(2,-5), simply negate and invert the slope 1%2F2 to get -2%2F1=-2

Now plug the slope and the point (2,-5) into y-y%5B1%5D=m%28x-x%5B1%5D%29


y-%28-5%29=-2%28x-2%29

y=-2x-1 Solve for y and simplify

So the altitude for vertex C is y=-2x-1



Now to find the equation of the line that is perpendicular to y=-x-3 that goes through the point A(4,5), simply negate and invert the slope -1%2F1 to get -%28-1%2F1%29=1

Now plug the slope and the point (2,-5) into y-y%5B1%5D=m%28x-x%5B1%5D%29


y-5=1%28x-4%29

y=x%2B1 Solve for y and simplify

So the altitude for vertex A is y=x%2B1




Now to find the equation of the line that is perpendicular to y=5x-15 that goes through the point B(-4,1), simply negate and invert the slope 5%2F1 to get -%281%2F5%29=-1%2F5

Now plug the slope and the point (-4,1) into y-y%5B1%5D=m%28x-x%5B1%5D%29


y-1=-%281%2F5%29%28x%2B4%29

y=-%281%2F5%29x%2B1%2F5 Solve for y and simplify

So the altitude for vertex B is y=-%281%2F5%29x%2B1%2F5



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Now let's solve the system

y=-2x-1
y=x%2B1

x%2B1=-2x-1 Plug in y=x%2B1 into the first equation

3x=-2 Add 2x to both sides and subtract 2 from both sides

x=-2%2F3 Divide both sides by 3 to isolate x


Now plug this into y=x%2B1

y=-2%2F3%2B1

y=1%2F3



So the orthocenter is (-2/3,1/3)

So if we plug in x=-2%2F3 into the third equation y=-%281%2F5%29x%2B1%2F5, we get


y=-%281%2F5%29%28-2%2F3%29%2B1%2F5


y=2%2F15%2B1%2F5


y=3%2F15

y=1%2F3

So the orthocenter lies on the third altitude




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Summary:

So the equations of the altidudes are:
y=-2x-1, y=x%2B1 and y=-%281%2F5%29%2B1%2F5