Question

The medians of a triangle are the line segments from each vertex to the midpoint of the opposite side. Find the lengths of the medians of the triangle with vertices at A=(0,0), B=(6,0), C=(4,4) ...?

Answer 1

Draw the triangle, find the midpoints of each side and find the distance from the midpoints to the opposite vertex to get ur result
sqrt[(y2-y1)^2 +(x2-x1)^2]

Answer 2

`√(29) , √(20) , √(17)`

Explanation:

Consider ΔABC with vertices
`A\left ( 0,0 \right )\,,\,B\left ( 6,0 \right )\,,\,C\left ( 4,4 \right )` such that P , Q , R are midpoints of sides BC , AC and AB .

We know that midpoint of line segment joining points
`\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )` is equal to
`\left ( (x_1+x_2)/(2)\,,\,(y_1+y_2)/(2) \right )`

Midpoints P , Q , R :

`P\left ( (6+4)/(2)\,,\,(0+4)/(2) \right )=P\left ( 5\,,\,2 \right )\\Q\left ( (0+4)/(2)\,,\,(0+4)/(2) \right )=Q\left ( 2\,,\,2 \right )\\R\left ( (6+0)/(2)\,,\,(0+0)/(2) \right )=R\left ( 3\,,\,0 \right )`

We know that distance between points
`\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )` is given by
`√(\left ( x_2-x_1 \right )^2+\left ( y_2-y_1 \right )^2)`

Length of AP :

AP =
`√(\left ( 5-0 \right )^2+\left ( 2-0\right )^2)=√(25+4)=√(29)`

Length of BQ :

BQ =
`√(\left (2-6 \right )^2+\left ( 2-0 \right )^2)=√(16+4)=√(20)`

Length of CR :

`√(\left (3-4\right )^2+\left ( 0-4 \right )^2)=√(1+16)=√(17)`