Question 175133
the median of the triangle from A to BC intersects line BC at its midpoint.
we'll call that point D.
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you need to find the midpoint of the line BC and then you need to find the equation of the line AD.
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the coordinates for the midpoint of a line is ((x1+x2)/2,(y1+y2)/2)
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line BC is given by the points:
(x1,y1) = (-7,4)
(x2,y2) = (1,12)
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the midpoint of line BD is:
((-7+1)/2,(4+12)/2)
this equals:
(-3,8)
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the coordinates for point D are (-3,8) which is the midpoint of line BC.
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line AD is the median of line BC.
the length of line AD is given by the equation:
{{{sqrt((x2-x1)^2 + (y2-y1)^2)}}}
where:
(x1,y1) = (2,-6)
(x2,y2) = (-3,8)
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this comes out to be:
{{{sqrt((-5)^2 + 14^2) = sqrt(25+196) = sqrt(221) = 14.86606875}}}
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the equation for this line would be:
y = m*x + b
m = slope = {{{(y2-y1)/(x2-x1) = (8-(-6))/(-3-2) = 14/-5 = -14/5}}}
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take one of the points (either one will do) and substitute in the equation to get the y intercept.
y = m*x + b
m = -14/5
point used = (-3,8)
8 = (-14/5)*(-3) + b
8 = 42/5 + b
8 - 42/5 = b
b = -2/5
equation for the median line AD is:
y = -14/5*x - 2/5
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in the graph below:
point A (2,-6) is the intersection of 3 lines on the lower right (AB,AD,AC).
point B (-7,4) is the intersection of 2 lines on the middle left (AB,BC).
point C (1,12) is the intersection of 2 lines on the upper right (AC,BC).
point D (-3,8) is the intersection of 2 lines in between points B and C (AD,BC). 
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if you print the graph and label them, it should be easier to see.
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{{{graph(800,800,-16,4,-8,14,-10/9*x - 34/9,-18*x+30,x+11,-14/5*x - 2/5)}}}