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(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

Answer :

We know that formula for area of a triangle whose vertices are \( (x_1,y_1) , (x_2,y_2) , (x_3,y_3) \) is,

= \( \frac{1}{2} \ | \ [x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \ | \)

(i) So, here \(x_1 \ = \ 2 \ , \ y_1 \ = \ 3 \ , \ x_2 \ = \ -1 \ , \ y_2 \ = \ 0 \ x_3 \ = \ 2 \ , \ y_3 \ = \ -4 \)

So, area of triangle = \( \frac{1}{2} \ | \ [ 2( 0 + 4 ) \ - \ 1(-4 - 3) \ + \ 2(3 - 0)] \ | \ = \ \frac{1}{2} \ | \ [ 8 + 7 + 6 ] \ | \ = \ \frac{21}{2} \)

∴ Area of triangle is \( \frac{21}{2} \) sq. units.

(ii) Similarly, here \(x_1 \ = \ -5 \ , \ y_1 \ = \ -1 \ , \ x_2 \ = \ 3 \ , \ y_2 \ = \ -5 \ x_3 \ = \ 5 \ , \ y_3 \ = \ 2 \)

So, area of triangle = \( \frac{1}{2} \ | \ [ (-5)( -5 - 2 ) \ - \ 3(2 + 1) \ + \ 5(-1 + 5)] \ | \ = \ \frac{1}{2} \ | \ [ 35 + 9 + 20 ] \ | \ = \ 32 \)

∴ Area of triangle is \( 32 \) sq. units.

- 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)
- 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.
- Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
- You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \(\triangle \) ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

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