<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
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Answer:
1. 35 - 176 = -141
2. 0.55 × 1.79 = 0.9845
3. 11.38 × 5.46 = 62.1348
Hope it helps and have a great day! =D
~sunshine~
Use slope formula
y^2-y^1/x^2-x^1
the answer is -4/3
Answer:
First one: function and linear
Second one: function and non-linear
Third one: not a function and linear
Step-by-step explanation:
• The first one is set in the linear function formula y=mx+b, so it is a linear function.
• The second one is a non-linear because functions with exponents have different shapes and can intersect more than once unline linear functions.
• The third one isn't a function but it is linear because it isn't set equal to a variable and it would become a linear if you simply it to isolate a variable.
I hope this helps!!!
Answer:
45 mph
Step-by-step explanation:
This is a really good question to know the answer to. It is tricky and a bit indirect (which means you have to find something else before you can find the speed of the car.)
Let's keep track of what he does in the time allotted.
How far does Joe go in 5 minutes? That's the amount of time he's on the road before she is.
convert 5 minutes into hours. 5 minutes * 1 hour / 60 minutes = 1/12 of an hour
d = r*t
r = 30 km/hour
t = 1/12 hour
d = 30 km/hr * 1/12 hour = 2.5 km
Now she's about to start. She wants to catch him in 10 minutes
d = r*t
r = x mph
t = 10 minutes = 10 minutes * 1 hour * 60 minutes = 1/6 of an hour.
How far does he go in the 10 minute time?
d = 30 * 1/6 = 5 km
What is his total distance
5 km + 2.5 km = 7.5 km
Finally how fast does she need to go to catch him
d = 7.5 km
r = ? This is what you are trying to find
t = 1/6 of an hour
d = r*t
7.5 km = r * (1/6)hour divide by 1/6 hour
7.5 km // 1/6 hour = r
r = 7.5 * 6 = 45 mph
