Question

Find the equations of the medians of a triangle, the equations of whose sides are:

$3x+2y+6=0,2x-5y+4=0andx-3y-6=0$

Answer 1

$\text{The given equationa re as follow}$

$3x+2y+6=0$

$2x-5y+4=0$

$x-3y-6=0$

$\text{Solving the equations}3x+2y+6=0$

$and2x-5y+4=0$

$\text{we get}x=-2andy=0$

$\text{Solving the equation}x-3y-6=0and$

$2x-5y+4=0$

$\text{we get}x=-42andy=-16$

$\text{Solving the equations}3x+2y+6=0$

$andx-3y-6=0$

$\text{we get}x=-6/11andy=-24/11$

$\text{sol let the intersection points be A, B and C o.e., the triangle be ABC}$

$\text{Coordinate of A, B and C will be}$

$A(-2,0),B(-42,-16)andC(-6/11,-24/11)$

$\text{By mid-point formula the mid-point of}$

$\text{AB will be D}=(\frac{2-42}{2},\frac{0-10}{2})$

$=(-22,-8)$

$E=\frac{\frac{-6}{11}-42}{2},\frac{\frac{-24}{11}-16}{2}=(\frac{-234}{11},\frac{-100}{11})$

$F=(\frac{\frac{-6}{11}-2}{2},\frac{\frac{-24}{11}+0}{2})(\frac{-14}{11},\frac{-12}{11})$

$\text{Equation of line passing through this mid-point and the opposite vertex}C(-6/11,-24/11)\text{will be the equation of the median from C. The equation will be}$

$\frac{y+8}{x+22}=\frac{-8+\frac{24}{11}}{-22+\frac{6}{11}}$

$\frac{y+8}{x+22}=\frac{-88+24}{=242+6}=\frac{16}{59}$

$16x-59y+352-472=0$

$16x-59y-120=0\text{Median through C Similar procedure has to be used for getting other medians as well}$

$\text{For getting median through B find midpoint of AC and apply the two point form of line equation. Similarly for median through A}$

$\text{Final median equations are}$

$41x-112y-70=0$

$25x-53y+50=0$

$16x-59y-120=0$