5 : 5, 10, 15, 20, 25
6 : 6, 12, 18, 24, 30
<span><span><span>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. </span>
(Scroll Down for Answer!)</span><span>Answer by </span>jim_thompson5910(34047) (Show Source):You can put this solution on YOUR website!
<span>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
Plug in the given points
Simplify and combine like terms
So the equation of the line through AB is
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Let's find the equation of the segment BC
Start with the general formula
Plug in the given points
Simplify and combine like terms
So the equation of the line through BC is
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Let's find the equation of the segment CA
Start with the general formula
Plug in the given points
Simplify and combine like terms
So the equation of the line through CA is
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 that goes through the point C(2,-5), simply negate and invert the slope to get
Now plug the slope and the point (2,-5) into
Solve for y and simplify
So the altitude for vertex C is
Now to find the equation of the line that is perpendicular to that goes through the point A(4,5), simply negate and invert the slope to get
Now plug the slope and the point (2,-5) into
Solve for y and simplify
So the altitude for vertex A is
Now to find the equation of the line that is perpendicular to that goes through the point B(-4,1), simply negate and invert the slope to get
Now plug the slope and the point (-4,1) into
Solve for y and simplify
So the altitude for vertex B is
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Now let's solve the system
Plug in into the first equation
Add 2x to both sides and subtract 2 from both sides
Divide both sides by 3 to isolate x
Now plug this into
So the orthocenter is (-2/3,1/3)
So if we plug in into the third equation , we get
So the orthocenter lies on the third altitude
</span><span>
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Answer:
The answer is 250 in interest.
Step-by-step explanation:
I=prt, I=(500)(0.05)(10)
The scale is that 3 the force is going on a colicky of 350
Shortest side (a) = 58
middle side (b) = 64
longest side (c) = 77
the 3 sides a + b + c = 199
b = a + 6
c = a + 19
substitute your new values for b & c into your original formula, so:
a + (a+6) + (a+19) = 199
3a + 25 = 199
3a = 174
a = 58
then substitute 58 into your b & c formulas to figure out the rest
b = a + 6 = 58 + 6 = 64
c = a + 19 = 58 + 19 = 77
