MHB How Do You Solve This Cartesian Geometry Problem?

AI Thread Summary
The discussion revolves around solving a Cartesian geometry problem involving the equations of medians in a triangle. The initial equation proposed for median [AQ] is incorrect; the correct equation is cx + by = 2ac, derived from the coordinates of points A and Q. Participants suggest finding the coordinates of the centroid G and proving that the third median [BR] passes through G by determining its equation. The conversation emphasizes using the midpoint formula and the relationships between the coordinates of the triangle's vertices. Overall, the thread provides guidance on deriving the necessary equations and verifying the intersection points.
abrk
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Hello

I've got a problem with Cartesian Geometry and cannot find a solution.

A will appretiate any help I can get.
View attachment 1893
b) Show that $$[AQ]$$ has equation $$cx + by = -2ac$$

c) Prove that the third median $$[BR]$$ passes through the point of intersection $$G$$ of medians $$[OP]$$ and $$[AQ]$$

Cheers!
 

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Hi,
The coordinates of the midpoint Q are (b,c). Unfortunately, x=b and y=c does not satisfy cx+by=-2ac unless b=-a (c is not zero if ABC is to be a triangle). So the equation of AQ is wrong. It should be:
cx+(2a-b)y=2ac
See if you can't get this. Similarly find the equations of the other medians OP and BR. You can "cheat" by knowing the coordinates of the centroid, namely (2(a+b)/3,2c/3) and showing this point satisfies all 3 equations.
 
abrk said:
Hello

I've got a problem with Cartesian Geometry and cannot find a solution.

A will appretiate any help I can get.
View attachment 1893
b) Show that $$[AQ]$$ has equation $$cx + by = -2ac$$
This is clearly NOT true. For example, point A has x= 2a, y= 0 which in this equation give 2ac not -2ac.
What is true is that AQ has equation cx+ by= 2ac.
To find that equation of use the fact that the coordinates of Q are (b, c) and find the equation of the line through (b, c) and (2a, 0).
(To show that this cx+ by= 2ac is the correct equation, it is enough to show that A and Q both satisfy that equation.)

c) Prove that the third median $$[BR]$$ passes through the point of intersection $$G$$ of medians $$[OP]$$ and $$[AQ]$$

Cheers!
P has coordinates (a+ b, c). The equation of the line through the origin and P is y= cx/(a+ b). Use that and the equation of the line AQ to find the coordinates of point G.

The third median is through points B with coordinates (2b, 2c) and R with coordinates (a, 0). Find the equation of the line through those two points and show that G also satisfies that equation.

I have used a couple of times that the midpoint of a line through points (x0, y0) and (x1, y1) is ((x0+x1)/2, (y0+y1)/2).
 
My teacher told me I have to do the following in C:
first find the coordinates of the point G.
Then write the equation for [BR].
Finally, check that G belong to the line [BR].

Can someone help me, thank you for your time :D
 
Line GP passes through (0, 0) and (a+ b, c). Any line can be written as y= px+ q. Since the line passes through (0, 0), 0= p(0)+ q so q= 0. Since the line passes through (a= b, c) c= p(a+ b) so p= c/(a+ b). That line is y= [c/(a+ b)]x.

Line GQ passes through (2a, 0) and (b, c). Writing y= px+ q, since the line passes through (2a, 0), 0= p(2a)+ q. Since the line passes through (b, c), c= bp+ q. Subtracting the first equation from the second, c= (b- 2a)p so p= c/(b- 2a). Putting that into the first equation, 0= 2ac/(b- 2a)+ q so q= -2a/(b- 2a). That line is y= [c/(b- 2a)]x- 2a/(b- 2a).

G is the point where those two lines intersect: [c/(a+ b)]x= [c/(b- 2a)]x- 2a/(b- 2a). [c/(a+ b)- c/(b- 2a)]x= -2a/(b- 2a). x= \frac{3ac}{(a+ b)(b- 2a)}.
 
I told you what x is and gave 2 equivalent formulas for y as a function of x. Do some of the work yourself!
 
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