Question

If (2, 1), (4, 5), (-1, - 3) are the mid points of the sides of a triangle, then the co-ordinates of its vertices are

• A. $(-3,-7),(17,9),(1,1)$
• B. $(-3.7),(7,9),(-1,1)$
• C. $(-3,-7),(7,9),(1,1)$
• D. none

Answer 1

Answer: (C)

$(-3,-7),(7,9),(1,1)$

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ be the vertices of $△ABC$
Let D(2,1) , F(-1,-3) and E(4,5) be the mid points of AB,AC and BC.

D and F are mid points pf ABC and AC

$\therefore DF\parallel BE$

E and F are mid points pf BC and Ac

$\therefore EF\parallel BD$

$\therefore$ DBEF is a parallelogram.

The diagonals of a parallelogram bisect each other i.e, both diagonals have same mid point

i.e, Midpoint BF=Midpoint of DE

$(\frac{x_2+(-1)}{2},\frac{y_2+(-3)}{2})=(\frac{2+4}{2},\frac{1+5}{2})\therefore \frac{x_2+(-1)}{2}=\frac{2+4}{2}\therefore x_2=7$

Similarly $\frac{y_2+(-3)}{2}=\frac{1+5}{2}\therefore y_2=9i.e,(x_2,y_2)=(7,9)$

D is the mid point of AB

$D=(2,1)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\frac{x_1+7}{2}=2\therefore x_1=-3i.e,(x_1,y_1)=(-3,-7)\frac{y_1+9}{2}=1\therefore y_1=-7$

F is the midpoint of AC

$F=(-1,-3)=(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})-1=\frac{-3+x_3}{2}\therefore x_3=1i.e,(x_3,y_3)=(1,1)-3=\frac{-7+y_3}{2}\therefore y_3=1$

$\therefore$ The vertices of triangle are $=(-3,-7),(7,9),(1,1)$