Properties of Triangle Medians in Coordinate Geometry
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A median of a triangle is a line segment from a vertex to the midpoint of the opposite side.
a Show that the median OP has equation $cx - (a + b)y = 0$.
b Show that the median AQ has equation $cx - (b - 2a)y = 2ac$.
c Prove that the third median BR passes through the point of intersection G of medians OP and AQ.
This question includes visual content: The image shows a triangle OBA on a Cartesian plane with vertices at O(0,0), A(2a,0), and B(2b,2c). Three medians are drawn: OP (to side AB), AQ (to side OB), and BR (to side OA). The intersection of the medians is labeled G. The sides are marked with hash marks indicating equal segments (midpoints P, Q, R). A faint vertical dashed line drops from P to the x-axis.
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Step by Step Written Solution
Hi Melek, let's look at this coordinate geometry problem exploring the properties of medians in a triangle.
Triangle Median Proof
First, let's identify our vertices from the diagram. Vertex O is at the origin, A is at two a, zero, and B is at two b, two c.
Vertices
- $O(0, 0)$
- $A(2a, 0)$
- $B(2b, 2c)$
For part a, we need the equation for median O P. P is the midpoint of side A B.
Part (a): Median OP
Substituting the coordinates of A and B, we find that P is at a plus b, c.
Since median O P passes through the origin 0, 0 and P, its gradient m is c over a plus b.
Using the line equation y equals m x, we get y equals c over a plus b times x.
Cross multiplying and rearranging, we get c x minus, bracket, a plus b, bracket, y equals zero. This proves part a.
Now for part b, we find median A Q. Q is the midpoint of side O B.
Part (b): Median AQ
The gradient of A Q, from A at two a, zero to Q at b, c, is c minus zero over b minus two a.
Using the point-slope form with point A, we have y minus zero equals the gradient times x minus two a.
Multiply both sides by b minus two a to clear the fraction.
Expanding the right side gives c x minus two a c. Moving the terms, we get c x minus, bracket, b minus two a, bracket, y equals two a c. That's our second equation.
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