Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are $(0. - 1), (2, 1) a n d (0, 3)$

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Solution

The correct option is A $8$
Area of triangle having vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by
Area $= \frac{1}{2} \times [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
Also, we know $A r e a o f t r i a n g l e = 4 \times o f t r i a n g l e f o r m e d u s i n g m i d - p o i n t = 4 \times (\frac{1}{2}) [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} + y_{1}) + x_{3} (y_{1} - y_{2})] = 2 [0 + 2 (3 - 1) + 0] = 8 s q u n i t$

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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)

The correct option is A 8Area of triangle having vertices (x1,y1),(x2,y2) and (x3,y3) is given byArea =12×[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]Also, we know Areaoftriangle=4×oftriangleformedusingmid−point=4×(12)[x1(y2−y3)+x2(y3+y1)+x3(y1−y2)]=2[0+2(3−1)+0]=8squnit

Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are $(0. - 1), (2, 1) a n d (0, 3)$