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  • Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

3.   Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

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From the figure:

The coordinates of the point P, Q, and R are:

Point P is the midpoint of side AB, hence the coordinates of P are :

P(x_{1},y_{1}) = \left (\frac{0+2}{2}, \frac{3+1}{2} \right ) = \left (1, 2 \right )

Point Q is the midpoint of side AC, hence the coordinates of Q are :

Q(x_{2},y_{2}) = \left (\frac{2+0}{2}, \frac{1-1}{2} \right ) = \left (1, 0 \right )

Point R is the midpoint of side BC, hence the coordinates of R are :

R(x_{3},y_{3}) = \left (\frac{0+0}{2}, \frac{-1+3}{2} \right ) = \left (0, 1 \right )

Hence, the area of the triangle formed by the midpoints PQR will be,

Area_{(PQR)} = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]

                       = \frac{1}{2}\left [ (2-1)+1(1-0)+0(0-2) \right ]

                       =\frac{1}{2}(1+1) = 1\ square\ units.

And the area formed by the triangle ABC will be:

Area_{(ABC)} = \frac{1}{2}\left [ 0(1-3)+2(3-(-1))+0(-1-1) \right ]

                       = \frac{1}{2}\left [ 8 \right ] = 4\ square\ units.

Thus, the ratio of Area of \triangle PQR to the Area of \triangle ABC  will be 1:4.

Posted by

Divya Prakash Singh

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