**Find the area of the triangle whose vertices are: **

**(i) (2, 3), (− 1, 0), (2, − 4) **

**(ii) (− 5, − 1), (3, − 5), (5, 2)**

Solution:

(i) Area of a triangle is given by

(ii)

**2. 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) **

Solution:

(i)For collinear points, area of triangle formed by them is zero. Therefore, for points (7, −2) (5, 1), and (3, k), area = 0

½[7{1-k} + 5{k – (-2) + 3{(-2) – 1}] = 0

7 – 7k + 5k + 10 – 9 = 0

-2k + 8 = 0

k = 4

(ii) For collinear points, area of triangle formed by them is zero. Therefore, for points (8, 1), (k, −4), and (2, −5), area = 0

½[8{-4 – (-5)} + k{(-5) – (1)} + 2{1 – (-4)}] = 0

8 – 6k + 10 = 0

6k = 18

k = 3

**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. **

Solution:

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

**4. Find the area of the quadrilateral whose vertices, taken in order, are (− 4, − 2), (− 3, − 5), (3, − 2) and (2, 3) **

Solution:

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 ∆ABC and ∆ACD.

**5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4, − 6), B (3, − 2) and C (5, 2) **

Solution:

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 ∆ABC. Therefore, AD is the median in ∆ABC.

However, area cannot be negative. Therefore, area of ∆ADC is 3 square units. Clearly, median AD has divided ∆ABC in two triangles of equal areas.

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