Question

let d(3,-2),e(-3,1),f(4,-3) be the midpoints of the sides bc,ac,ab respectively of triangle abc.find the coordinates of a,b,c​

Answer 1

Please see the attachment

Answer 2

Hence  the coordinates of a,b,c  are a(-2,0), b(-10,-6), c(-4,2).

Step-by-step explanation:

Given:

Here d(3,-2),e(-3,1),f(4,-3) be the midpoints of the sides bc, ac, ab respectively of triangle abc.

Let the co-ordinates of the triangle abc be $a(x_1,y_1), b(x_2,y_2), c(x_3,y_3)$

Now, d(3.-2) is the midpoint of side bc, therefore

$(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}) =d(3,-2)$

Also, e(-3,1)  is the midpoint of side ac, therefore

$\frac{x_1+x_3}{2},\frac{y_1+y_3}{2} = e(-3,1)$

Also, f(4,-3) is the midpoint of side ab, therefore

$\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} = f(4,-3)$

On comparing the co-ordinates in the above 3, we get

$\frac{x_1+x_2}{2}=4$ and $\frac{y_1+y_2}{2}=-3$

${x_1+x_2}=4 \times 2$ and ${y_1+y_2}=-3\times2$  

${x_1+x_2}=8$ and ${y_1+y_2}=-6$   [1]

$\frac{x_1+x_3}{2}=-3$ and $\frac{y_1+y_3}{2}=1$

${x_1+x_3}=-3\times 2$ and ${y_1+y_3}=2$  

${x_1+x_3}=-6$ and ${y_1+y_3}=2$   [2]

$\frac{x_2+x_3}{2}=3$ and $\frac{y_2+y_3}{2}=-2$

${x_2+x_3}=3\times 2$ and ${y_1+y_3}=-2\times2$  

${x_2+x_3}=6$ and ${y_1+y_3}=-4$   [3]

Adding the equations in x's and y's separately, we get

$2x_1+2x_2+2x_3=8-6+6$ and $2y_1+2y_2+2y_3=-6+2-4$

$2(x_1+x_2+x_3)=8$ and $2(y_1+y_2+y_3)=-8$

$(x_1+x_2+x_3)=\frac{8}{2}$ and $(y_1+y_2+y_3)=\frac{-8}{2}$

$(x_1+x_2+x_3)=4$ and $(y_1+y_2+y_3)=-4$    [4]

Subtracting 1 from 4, we get

$x_1+x_2+x_3-(x_1+x_2)=4-8$ and $y_1+y_2+y_3-(y_1+y_2)=-4-(-6)$

$x_1+x_2+x_3-x_1-x_2=-4$ and $y_1+y_2+y_3-y_1-y_2=2$

$x_3=-4$ and $y_3=2$

Subtracting 2 from 4, we get

$x_1+x_2+x_3-(x_1+x_3)=4-(-6)$ and $y_1+y_2+y_3-(y_1+y_3)=-4-2$

$x_1+x_2+x_3-x_1-x_3=-10$ and $y_1+y_2+y_3-y_1-y_3=-6$

$x_2=-10$ and $y_2=-6$

Subtracting 3 from 4, we get

$x_1+x_2+x_3-(x_2+x_3)=4-6$ and $y_1+y_2+y_3-(y_2+y_3)=-4-(-4)$

$x_1+x_2+x_3-x_2-x_3=-2$ and $y_1+y_2+y_3-y_2-y_3=0$

$x_1=-2$ and $y_1=0$

Hence the points are a(-2,0), b(-10,-6), c(-4,2).