# Determine the median which goes from A in triangle ABC if A(5,2), B(-3,7), C(1,-5).

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The mid point of B(-3, 7) and C(1, -5) is M(-1, 1)

The median from A is the line that joins A and the mid point of BC.

This line is between A(5, 2) and M(-1, 1)

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

=> (y - 1)/(x + 1) = 1/6

=> 6(y - 1) = (x + 1)

=> 6y - 6 = x + 1

=> x - 6y + 7 = 0

**The equation of the median is x - 6y + 7 = 0**

We'll recall that the median of a triangle is the segment that joins the vertex of triangle to the midpoint of the opposite side.

If the median is drawn from vertex A, then the opposite side is BC.

We'll determine the equation of the median, and for this reason, we'll determine the midpoint of the segment [BC].

xM = (xB+xC)/2 => xM = (-3+1)/2 => xM = -1

yM = (yB+yC)/2 => yM = (7-5)/2 => yM = 1

Since we know two points, we'll write the equation of the line that passes through the points:

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

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

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

We'll cross multiply:

-x + 5 = -6y + 12

We'll shift all the terms to one side:

-x + 6y + 5 - 12 = 0

x - 6y + 7 = 0

**The equation of the median that joins the vertex A to the midpoint M of the side BC, is x - 6y + 7 = 0.**