Absolutely clueless on how to do this Geometry Help

[Frogeyedpeas]

Homework Statement

There are three problems based on the following diagram.

You have a triangle on the cartesian plane with each corner having the coordinates (-a, 0) (a,0) (b,c)

Find the coordinates (in terms of a,b,c) of the point where all three perpendicular bisectors of the triangle intersect.

Find the coordinates (in terms of a,b,c) of the point where all three medians of the triangle intersect.

Find the coordinates (in terms of a,b,c) of the point where all three altitudes of the triangle intersect.

Prove that all three points lie on one line.

Homework Equations

Well as far as I know you can't use any calculus on this problem (as it came in the section before calculus started) although I don't believe you should have to.

The Attempt at a Solution

I basically was trying to use the slopes, distance formula, pythagorean theorem, and law of sines to play around with this one but I'm still pretty clueless.

I believe the line that they lie on is the euler line although I know nothing about that or any of the other central points of a triangle.

 

[SammyS]

Homework Statement

There are three problems based on the following diagram.

You have a triangle on the cartesian plane with each corner having the coordinates (-a, 0) (a,0) (b,c)

Find the coordinates (in terms of a,b,c) of the point where all three perpendicular bisectors of the triangle intersect.

Find the coordinates (in terms of a,b,c) of the point where all three medians of the triangle intersect.

Find the coordinates (in terms of a,b,c) of the point where all three altitudes of the triangle intersect.

Prove that all three points lie on one line.

Homework Equations

Well as far as I know you can't use any calculus on this problem (as it came in the section before calculus started) although I don't believe you should have to.

The Attempt at a Solution

I basically was trying to use the slopes, distance formula, pythagorean theorem, and law of sines to play around with this one but I'm still pretty clueless.

I believe the line that they lie on is the euler line although I know nothing about that or any of the other central points of a triangle.

What are the coordinates of the midpoints of each side?

What is the x-coordinate of the perpendicular bisector of the side whose vertices are (-a,0) and (a,0) ?

 

[Frogeyedpeas]

None of that information is given... we have to derive using slope and distance formula so the answer will be symbolic as opposed to numerical

 

[Frogeyedpeas]

If it helps I realized the answer has to be of the form (0, h) because two sides are (-a,0) and (a,0) so their bisector would be through the point (0,0) but we still are left with the difficult part of deriving h

 

[SammyS]

None of that information is given... we have to derive using slope and distance formula so the answer will be symbolic as opposed to numerical

All of that information can be easily obtained from the coordinates of the vertices, and perhaps a small amount of thinking.

 

[Frogeyedpeas]

Okay so here is the information:

midpoint 1 (0,0) midpoint 2 (a/2+b/2 , c/2) midpoint 3 (b/2-a/2 , c/2) in terms where the lines once again intersect sides (not at midpoints) I do not know...

 

[SammyS]

What is the equation of the perpendicular bisector of the side whose vertices are (-a,0) and (a,0) ?

 

[Frogeyedpeas]

it will be x = 0 because it has to be vertical since (-a,0) and (a,0) denote a line that is horizontal at y = 0

 

[Frogeyedpeas]

but we still need to find the equation of another perpendicular bisector which has undefined sloped and y intercept (at least the slope can be expressed through symbols) but how do you solve for y intercept? because the y intercept can only be defined by solving for zero which ironically is the point that we are dealing with :(

 

[SammyS]

What is the slope of the line passing through points (-a,0) & (b,c) ?

 

[NascentOxygen]

Okay so here is the information:

midpoint 1 (0,0) midpoint 2 (a/2+b/2 , c/2) midpoint 3 (b/2-a/2 , c/2)

That's right.

Next, determine the slopes of each side of the triangle. We know the base has slope 0, so that leaves just 2 slopes to be determined.

What is the slope of the line joining points (-a,0) and (b,c)?

 

[Frogeyedpeas]

Well the slope is what it is...

Slope formula: (y₂ - y₁)/(x₂-x₁)

So for the side (b,c) and (-a,0)

we get a slope of c/(b+a)

and for the sdie (a,0) and (b,c)

we get a slope of c/(b-a)

The perpendiculars (respectively) therefore have slopes

-(b+a)/c and -(b-a)/c

So where do we move from there?

 

[Frogeyedpeas]

I'll keep giving you guys info but you do realize that all I'm looking for is a proof? This isn't a homework problem, its an independent study and I quite honestly would benefit more from having a single proof through the problem so I could understand it, (then ask questions about what you did), and get a more thorough grasp that way.

 

[micromass]

I'll keep giving you guys info but you do realize that all I'm looking for is a proof? This isn't a homework problem, its an independent study and I quite honestly would benefit more from having a single proof through the problem so I could understand it, (then ask questions about what you did), and get a more thorough grasp that way.

You do realize that at this forum we don't just give people the answer right?? We give you hints and nudges, but eventually, it is you who needs to do the work.

 

[NascentOxygen]

The perpendiculars (respectively) therefore have slopes

-(b+a)/c and -(b-a)/c

So where do we move from there?

Right. Now you have all the information you need to write the equation for every perpendicular bisector. For each you know its slope, and you know the co-ordinates of a point it goes through.

Considering one perpendicular at a time, substitute for what you know into the straight line equation: y=mx+c

 

[Frogeyedpeas]

No i do totally understand that you want me to derive it with nudge and hints but I've already gone down this same path that you've given me. Give it a shot yourself because here is what happens (I wanted to avoid writing down the entire failed attempt as I have a tendency to write excessively but here it anyways)

Your given a triangle with vertices (-a,0) (a,0) (b,c)

The perpendicular bisector to (-a,0) and (a,0) is met by the equation x = 0

The next side we address has the endpoints (-a,0) and (b,c) it's slope is c/(b+a) so the perpendicular bisector's slope will be -(b+a)/c

The third side has the endpoints (b,c) and (a,0) with a slope of c/(b-a) therefore giving the perpendicular bisector's slope as (b-a)/-c

So now we know that our x coordinate in the intersection point is most definitely going to be equal to 0.

So we now have to solve for the Y coordinate of the intersection point.

The two other perpendicular bisectors take on the form:

y₁ = -(b+a)/c + K
y₂ = -(b-a)/c + K

So their point of intersection is (0,K)

So how do we algebraically derive K?

Well we attempt to solve one of our original side equations as a starting place. So given a slope of c/(b-a) for the side whose endpoints are (b,c) and (a,0) we know it has the equation: c/(b-a)x + i associated with where i is the y intercept, we also know that the zero fo this equation is at x=a.

Therefore (using some basic algebra) -i/(c/(b-a)) = a So, -i(b-a)/c = a or i = -ac/(b-a)

Doing the same thing for the other sidelength we find it's x intercept is as follows:

y = c/(b+a)x + j
-j/(c/(b+a)) = -a
j/(c/(b+a)) = a
j(b+a)/c = a
j = ac/(b+a)

So there are our two y intercepts for the sidelengths, now time to see how we can use those to find the value of K. Using the second equation we derived:

y = c/(b+a)x + ac/(b+a) through the use of the distance formula from (-a,0) to (0, ac/(b+a)) tells us that the length of this segment is (time for the tex...)

\sqrt{((0-(-a))^{2} + (\frac{ac}{b+a}-0)^{2}}

Simplifying to: \sqrt{a^{2} +(\frac{ac}{b+a})^{2}}
Now distributing the exponential results in:

\sqrt{a^{2} +(\frac{a^{2}c^{2}}{(b+a)^{2}})}

and combining fractional terms gets us to:

\sqrt{\frac{a^{2}(b+a)^{2}}{(b+a)^{2}} +(\frac{a^{2}c^{2}}{(b+a)^{2}})}

simplifying to

\sqrt{(\frac{a^{2}c^{2}+a^{2}(b+a)^{2}}{(b+a)^{2}})}

factoring out the a term:

a\sqrt{(\frac{c^{2}+(b+a)^{2}}{(b+a)^{2}})}

Pushing onwards from there... Where do we go?

 

[Frogeyedpeas]

There is no information right now introduced that allows us to connect the midpoints to the actual equations of the perpendicular bisectors...

 

[NascentOxygen]

The two other perpendicular bisectors take on the form:

y₁ = -(b+a)/c + K
y₂ = -(b-a)/c + K

The general equation for a straight line is y=mx+c
You seem to be overlooking that x.

As I pointed out, you now have all the information you need to write the equation for every perpendicular bisector.

For each you know its slope, and you know the co-ordinates of a point on the side of the triangle that it passes through. So, considering one perpendicular at a time, substitute for what you know into the straight line equation y=mx+c and hence determine its equation.

 

[SammyS]

...

Your given a triangle with vertices (-a,0) (a,0) (b,c)

The perpendicular bisector to (-a,0) and (a,0) is met by the equation x = 0

The next side we address has the endpoints (-a,0) and (b,c) it's slope is c/(b+a) so the perpendicular bisector's slope will be -(b+a)/c

The third side has the endpoints (b,c) and (a,0) with a slope of c/(b-a) therefore giving the perpendicular bisector's slope as (b-a)/-c

So now we know that our x coordinate in the intersection point is most definitely going to be equal to 0.

So we now have to solve for the Y coordinate of the intersection point.

The two other perpendicular bisectors take on the form:

y₁ = -(b+a)/c + K
y₂ = -(b-a)/c + K
...

Find the equation for each of the lines corresponding to these bisectors.

For the perpendicular bisector to the segment with vertices (-a,0) and (b,c):

\displaystyle y=-\frac{b+a}{c}\,x+k_1

Put in the coordinates for the midpoint to find k₁​

k₁ is the y-intercept for this line, so that's where this line intersects the line, x=0.

Do similar for the other bisector.

 

[Frogeyedpeas]

Okay so let's use the first perpendicular bisector: and the corresponding midpoint

((b-a)/2, c/2)

We know that y = -(b+a)/c*x + K1 So to substitute the x value we get

c/2 = -(b²-a²)/c + K1

multplying the c we recieve:

c²/2 = k -b² + a²

Which leaves us finally with

K = 1/2(c²+b²-a² = k

And since that is the Y coordinate of the point of intersection we get the intersection of three perpendicular bisectors as:

(0, 1/2c² + b²-a²) which unfortunately doesn't match up with my book's answer (I wanted to do this from scratch without working backwards) which is:

(0, (-a² + b²+²)/2c)

The two expressions are not equal either, yet this one that I have attained seems quite logically deduced

 

[SammyS]

Okay so let's use the first perpendicular bisector: and the corresponding midpoint

((b-a)/2, c/2)

We know that y = -(b+a)/c*x + K1 So to substitute the x value we get

c/2 = -(b²-a²)/c + K1

multplying the c we recieve:

c²/2 = k -b² + a²

...

Not quite right.

You failed to multiply k by c. Anyway there should have been a 2 in both denominator.

So, after multiplying by 2c you should have:

\displaystyle c^2=2ck-b^2+a^2​

Solving for k gives:

\displaystyle k=\frac{b^2+c^2-a^2}{2c}​