How Do You Solve This Cartesian Geometry Problem?

[abrk]

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 \(\displaystyle [AQ]\) has equation \(\displaystyle cx + by = -2ac\)

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

Cheers!

 

[johng1]

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.

 

[HallsofIvy]

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 \(\displaystyle [AQ]\) has equation \(\displaystyle cx + by = -2ac\)

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 \(\displaystyle [BR]\) passes through the point of intersection \(\displaystyle G\) of medians \(\displaystyle [OP]\) and \(\displaystyle [AQ]\)

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).

 

[charlottekriste]

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

 

[HallsofIvy]

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). $$x= \frac{3ac}{(a+ b)(b- 2a)}$$.

 

[HallsofIvy]

I told you what x is and gave 2 equivalent formulas for y as a function of x. Do some of the work yourself!