Question

Find the equations of the medians of a triangle, the coordinates of whose vertices are (-1, 6), (-3, -9) and (5, -8).

Answer 1

$LetA(-1,6)be\left(x_1{y}_1\right)B(-3,-9)be\left(x_2{y}_2\right)C(5,-8)be\left(x_3{y}_3\right)$

$\text{Median is a line segement which join a vertex to the mid-point of the side opposite to it. Let D, E and F be the mid points of sides AB, BC, and CA.}\text{Then, using mid point formula}(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\text{we can find the coordinates of D, E and F as :}D=(\frac{-3+5}{2},\frac{-9-8}{2})=(-1,\frac{-17}{2})E=(\frac{-1+5}{2},\frac{6-8}{2})=(2,-1)F=(\frac{-1-3}{2},\frac{6-9}{2})=(-2,\frac{-3}{2})\text{Equation of median AD is}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)y-6=\frac{\frac{-17}{2}-6}{1-(-1)}(x+1)=\frac{-29}{4}(x+1)\left[A\right(-1,6),D(1,\frac{-17}{2})]29x+4y+5=0\text{Equation of median BE is}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)y-(-9)=\frac{-1-(-9)}{2-(-3)}\{x-(-3\left)\right\}\left[B\right(-3,-9),E(2,-1)]y+9=\frac{8}{5}(x+3)5y+45=8x+248x-5y-21=0\text{Equation of median CF is}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)y-(-8)=\frac{\frac{-3}{2}-(-8)}{-2-5}(x-5)\left[C\right(5,-8),F(-2\frac{-3}{2})]y+8=\frac{-3+16}{2\times (-7)}(x-5)y+8=\frac{-13}{14}(x-5)13x+14y+47=0$