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Find the area of the triangle whose vertices are :
(i) (2, 3), (–1, 0), (2, – 4)
(ii) (–5, –1), (3, –5), (5, 2)
(i) Area of a triangle is given by
(ii) 
In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, –5)
(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, −2) (5, 1), and (3, k), area = 0
(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, −4), and (2, −5), area = 0
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.
Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by
Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).
Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.
You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for รข?? ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).
Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
However, area cannot be negative. Therefore, area of ΔADC is 3 square units.
Clearly, median AD has divided ΔABC in two triangles of equal areas.
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