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Question

Find the equations of the medians of a triangle, the equations of whose sides are :
3x+2y+6=0,2x5y+4=0 and x3y6=0.

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Solution


The equations of the sides of a triangle are given by the following equations
3x+2y+6=0(1)
2x5y+4=0(2)
x3y6=0(3)
Solving the equations taking two at a time we can find the vertices of the triangle whose sides are given by (1), (2) and (3)
Solving (1) (2)$
3x+2y+6=0 and 2x5y+4=0
we get, x=2 and y=0
Solving (2) & (3)
x3y6=0
2x5y+4=0 } x=42 and y=16
Solving (3) & (1)
x3y6=0
3x+2y+6=0 } x=611 and y=2411
So, the vertices are
A(2,0) B=(42,16) C=(611,2411)
let the midpoint of the side AB be P, which has coordinates (22,8)
let the midpoint of the side BA be Q, which has coordinates (46811,20011)
let the midpoint of the side CA be R, which has coordinates (2811,2411)
Now the equation of a line in the two-point form is given by
yy1=y2y1x2x1(xx1)
So the equation of the median PC is given by : y+2411=8+241122+611(x+611)
11y+24=64236(11x+6)
11y+24=1659(11x+6)
176x649y1032=0
the equation of the median RB is given by
y+2411=16+24/1142+28/11(x+2811)2387y+5208=836x+2128
836x2387y3080=0
the equation of the median AQ is given by
y0=200/1146811+2(x+2)
100x223y+200=0

1217966_1304416_ans_50fd325d0b964d56b19f08f1102ba0ef.png

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