the midpoints of the sides BC, CA and AB of a ΔABC are D(2, 1), B

Solution:

Let us consider the coordinates of vertices of triangle A, B, C be (a, b), (c, d) and (e, f). Now using mid - point formula

$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)$

For side BC (midpoint D): $(2,1)=\frac{c+e}{2}, \frac{d+f}{2}$

For side $\mathrm{AC}($ midpoint $\mathrm{E}):(-5,7)=\frac{\mathrm{a}+\mathrm{e}}{2}, \frac{\mathrm{b}+\mathrm{f}}{2}$

For side $\mathrm{AB}($ midpoint $\mathrm{F}):(-5,-5)=\frac{\mathrm{a}+\mathrm{c}}{2}, \frac{\mathrm{b}+\mathrm{d}}{2}$

Now from above equations, we have

$c+e=4, d+f=2$ (i)

$a+e=-10, b+f=14$ (ii)

$a+c=-10, b+d=-10$ (iii)

From subtract (i) from (ii),we get

$a-c=-14, b-d=12$ (iv)

Adding (iii) and (iv)

$2 \mathrm{a}=-24 \Rightarrow \mathrm{a}=-12,2 \mathrm{~b}=2 \Rightarrow \mathrm{b}=1$

Putting values of a, b in equation (iii)

$c=-10-(-12) \Rightarrow c=2, d=-10-1 \Rightarrow d=-11$

Again putting values in (i)

$\mathrm{e}=4-2 \Rightarrow \mathrm{e}=2, \mathrm{f}=2-(-11) \Rightarrow \mathrm{f}=13$

So coordinates of $A(-12,1), B(2,-11)$ and $C(2,13)$.

Using two point form of the equation

Equation of side AB:

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

$y-1=\frac{-11-1}{2-(-12)}(x-(-12)) \Rightarrow y-1=\frac{-12}{14}(x+12)$

$14(y-1)=-12(x+12)$

$14 y-14+12 x+144=0$

$12 x+14 y+130=0$

$6 x+7 y+65=0$

Equation of side BC:

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

$y-(-11)=\frac{13-(-11)}{2-2}(x-2) \Rightarrow y+11=\frac{24}{0}(x-2)$

y = - 11(slope is not defined i.e. line is vertical)

Equation of side CA:

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

$y-13=\frac{1-13}{-12-2}(x-2) \Rightarrow y-13=\frac{-12}{-14}(x-2)$

$14(y-13)=12(x-2)$

$12 x-24-14 y+182=0$

$12 x-14 y+158=0$

$6 x-7 y+79=0$

So, the required equations of sides for $A B: 6 x+7 y+65=0$

For BC: $y=-11$

For CA: $6 x-7 y+79=0$