B(-3, -3) and C(8, -5) are two vertices of triangle ABC. The centroid of the triangle is G(3, -1).
A centroid of a triangle is the point where the three medians of the triangle meet. A median of a triangle is a line segment from one vertex to the mid point on the opposite side of the triangle. The centroid is also called the center of gravity of the triangle.
It is best to do a rough sketch to get the idea of what you are doing.
Find the equation of the median of vertex B.
This is the line through B and G
I think of these lines a bit differently to how they are normally taught. normally there are about three different formulas that you are expected to learn. I just have 1 and it is very simple
\(gradient = gradient\\ \frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}\\ \frac{-1--3}{3--3}=\frac{y--3}{x--3}\\ \frac{2}{6}=\frac{y+3}{x+3}\\ \frac{1}{3}=\frac{y+3}{x+3}\\ \frac{1}{3}=\frac{y+3}{x+3}\\ x+3=3(y+3)\\ x+3=3y+9\\ x-3y-6=0 \)
Find the equation of the median of vertex C.
\(gradient = gradient\\ \frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}\\ \frac{-1--5}{3-8}=\frac{y--5}{x-8}\\ \frac{4}{-5}=\frac{y+5}{x-8}\\ 4(x-8)=-5(y+5)\\ 4x-32=-5y-25\\ 4x+5y-32+25=0\\ 4x+5y-7=0\)
If E(0, 6) is the midpoint of AC, find the coordinates of vertex A.
\(Let\;\; A(x_1,y_1)\;\;then\\ \frac{x_1+8}{2}=0 \qquad \frac{y_1+-5}{2}=6\\ \frac{x_1+8}{2}=0 \qquad \frac{y_1-5}{2}=6\\ x_1+8=0 \qquad y_1-5=12\\ x_1=-8 \qquad y_1=17\\ A(-8,17)\)
I will now draw the picture to see if it all makes sense.
The picture does not make sense - that is because if G is the centroid then E(0,6) cannot be the midpoint of AC.
So it is the question that is incorrect!
suck it
