Question

Triangle ABC has vertices ofA(–6, 7), B(4, –1), and C(–2, –9).Find the length of the median from

Answer 1

We know that in geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. So, in a triangle there are three medians. We will find them.

As shown in the figure below, we have the medians:

P1C
P2A
P3B

We need to find P1, P2 and P3. The midpoint of the segment (x1, y1) to (x2, y2) is:


`((x_(1)+x_(2) )/(2), (y_(1)+y_(2) )/(2))`

Therefore:

For the segment AB:


`P_(1) = ( (4-6)/(2), (7-1)/(2))`

`P_(1) = (-1,3)`

For the segment BC:

`P_(2) = ( (4-2)/(2), (-1-9)/(2))`

`P_(2) = (1,-5)`


For the segment CA:

`P_(3) = ( (-6-2)/(2), (7-9)/(2))`

`P_(3) = (-4,-1)`

We know that the distance d between two points P1(x1,y1) and P2(x2,y2) is given by the formula:


`d = \sqrt{(x_(2)- x_(1))^(2)+(y_(2)- y_(1))^(2) }`

Then of each median is:

Median P1C:


`d_(1) = \sqrt{(-9-3)^(2)+(-2-(-1))^(2)} = √(145)`

Median P2A:


`d_(2) = \sqrt{(7-(-5))^(2)+(-6-1)^(2)} = √(193)`

Median P3B:


`d_(3) = \sqrt{(-1-(-1))^(2)+(4-(-4))^(2) } = 8`