# Find the equation of the median which starts from the vertex A, of the triangle ABC. A(5,2), B(-3,7), C(1,-5)

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Definition of a median in a triangle: Join the vertex with the middle of the opposite side to the vertex of the triangle.

In our case, the opposite side of the A vertex is the BC side.

Let's denote the middle of the side BC, as being the point M.

To write the equation of the line (AM), we have to know the coordinates of the points A and M.

For the moment, we know, from the enunciation, just the coordinates o fthe A point.

To find the coordinates of the middle point M, we'll use the formula:

xM= (xB+xC)/2

xM=(-3+1)/2=-2/2

**xM=-1**

yM=(yB+yC)/2

yM=(7-5)/2

yM=2/2

**yM=1**

**So, the coordinates for the point M are: M(-1,1).**

We can find now the equation of the line which is passing through the points A and M.

**(xM-xA)/(x-xA)=(yM-yA)/(y-yA)**

(-1-5)/(x-5)=(1-2)/(y-2)

-6/(x-5)=-1/(y-2)

6(y-2)=(x-5)

6y-12-x+5=0

6y-x-7=0

6y=x+7

y=x/6 + 7/6

**(AM):6y-x-7=0 or y=x/6 + 7/6**

To find the equation of median from A(5,2) to the other side BC, where B(-3,7) and C(1,-5).

Solution Le D be the mid point of BC whose coordinates are {(-3+1)/2 , 7-5)/2} = (-1,1).

Therefore the equation of the median is the equation of the line AD:

y-y1 = (y2-y1)(x2-x1){x-x1) is theline joining (x1,y1) and (x2,y2).

y-2= (1-2)/(-1-5)(x-5). Or

3(y-2) = (x-5) . Or

x-3y -5+6 = 0. Or

x-3y+1 = 0

In the second answer is missing (-1-5)=-6. Why?