Medians of a triangle

deltaxray · #1 $Old$ November 1st, 2009, 02:44 AM

Can someone help me please? This is very confusing. =S

Prove that the three medians of a triangle are concurrent by choosing as vertices A(6a,6b) B(-6a-6b) and C(0,6c) and following these steps:

a) Find the Midpoints P,Q and R of BC, CA and AB respectively. Show that the median through C is x=0 and find the equation of the other two medians.

b) Find where the median through C meets another median, and show that the point lies on the third median.

Thanks

Grandad · #2 $Old$ November 1st, 2009, 03:12 AM

Hello deltaxray

Quote:

Originally Posted by deltaxray $View Post$

Can someone help me please? This is very confusing. =S

Prove that the three medians of a triangle are concurrent by choosing as vertices A(6a,6b) B(-6a-6b) and C(0,6c) and following these steps:

a) Find the Midpoints P,Q and R of BC, CA and AB respectively. Show that the median through C is x=0 and find the equation of the other two medians.

b) Find where the median through C meets another median, and show that the point lies on the third median.

Thanks

You need to know:

the mid-point of the line joining $(x_1,y_1)$ to $(x_2,y_2)$ is $\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)$

the equation of the line joining $(x_1,y_1)$ to $(x_2,y_2)$ is $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Use these results to find the mid-points and the equations of the three medians. Then solve these equations simultaneously to find their point(s) of intersection.

Let us know if you can't complete it now.

Grandad

ukorov · #3 $Old$ November 1st, 2009, 04:11 AM

If my calculations are correct....

R = (0, 0)
P = (-3a, 3c - 3b)
Q = (3a, 3c + 3b)

I did not try to work out the equations for AB, BC, AC in terms of a, b, c, x, y.......they will look too messy.

median RC is x = 0 so lies on y-axis. if any other medians, PA or BQ, is to intersect with RC, the x coordinate of the intersection point must be 0.

let AP intersects with RC at (0, m1).
(6b - m1)/(6a - 0) = (3c - 3b - m1)/(-3a - 0)
in the end i get m1 = 2c

by the same method for the interection between BQ and RC....say they intersect at (0, m2).....i also get m2 = 2c in the end.

Hence the three medians PA, BQ, RC all intersect at (0, 2c)