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. 
(Scroll Down for Answer!)Answer by jim_thompson5910(34047) (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

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

3 0

3 years ago

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

$BC^2=BD\cdot AB.$