Answer :

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of the given triangle


and D (1, 2), E (2, –1) and F (0, –1) be the mid points of the sides BC, CA and AB respectively.



Using the mid–point formula,


=(0, –1)


x1 +x2 = 0 and y1 +y2 = –2 ……(i)


=(2, –1)


x1 +x3 = 4 and y1 +y3 = –2 ……(ii)


=(1, 2)


x2 +x3 = 2 and y2 +y3 = 4 ……(iii)


Adding all the equations obtained for x coordinates and y coordinates respectively, we get


2(x1 + x2 + x3) = 6


and 2(y1 + y2 + y3) = 0


x1 + x2 + x3 = 3 and y1 + y2 + y3 = 0 –––(iv)


From eq (i) and (iv), we get x3 = 3 , y3 = 2 , i.e. C (3, 2)


From eq (ii) and (iv), we get x2 = –1 , y2 = 2, i.e. B (–1, 2)


From eq (iii) and (iv), we get x1 = 1 , y3 = –4, i.e. A (1, –4)


The coordinates of the vertices of the triangle are A (1, –4), B (–1, 2) and C (3, 2).


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