Question

The steps shown can be used to prove that the medians of a triangle meet at a point. Triangle A B C is shown on an x and y-axis. Lines are drawn from each point to the opposite side and intersect at point G. Point A is at (0, 0), point D is at (c, 0), point C is at (2 c, 0), point B is at (2 a, 2 b), and point E is at (a, b). Define segments BD and CE as medians of triangle ABC.

Answer 1

The medians of triangle ABC intersect at a point G.

Step-by-step explanation:

The medians of a triangle are the line segments drawn from each vertex to the midpoint of the opposite side. Let's denote the vertices of triangle ABC as A(a, b), B(2a, 2b), and C(2c, 0). The midpoints of the sides are D(c, 0) for side AC and E(a, b) for side AB.

Now, let's find the equations of the medians BD and CE. The equation of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by y - y₁ = (y₂ - y₁)/(x₂ - x₁) * (x - x₁).

For median BD, connecting B and D, the equation is y - 2b = (0 - 2b)/(2c - 2a) * (x - 2a), which simplifies to y = -x/c + 3a/c + 2b.

For median CE, connecting C and E, the equation is y - b = (0 - b)/(2c - a) * (x - a), which simplifies to y = -x/c + 2a/c + b.

The intersection point G is the solution to the system of equations formed by setting the two median equations equal to each other. Solving for x and y yields x = 3c/4 + a/4 and y = b/2 + a/4. Thus, point G has coordinates (3c/4 + a/4, b/2 + a/4). The intersection point G is independent of the specific triangle dimensions, demonstrating that the medians of any triangle indeed intersect at a common point.

Answer 2

The solution tot he centroid of the triangles are:

a) y = ((c - a)/b) * x

b) Linear equations:

BD: y = (2b/(2a - c))(x - c)

CE: y = [b/(a - 2c)] (x - 2c)

How to solve the centroid of the triangle?

a) The median part is solved as:

(y/x) * 2b/(2a - 2c) = -1

(y/x) * b/(a - c) = -1

Make y the subject of the formula to get:

y = ((c - a)/b) * x

b) Linear equations:

BD: y = (2b/(2a - c))(x - c)

CE: y = [b/(a - 2c)] (x - 2c)