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Area of the given triangle = ½[2 {0 − (−4)} – 1 (−4 − 3) + 2 (3 − 0)]
=½ (8 + 7 + 6) = 21/2 sq. units
(ii) (–5, –1), (3, –5), (5, 2)
Area of the given triangle = ½ [−5 (−5 − 2) + 3 {2 − (−1)} + 5 {−1 − (−5)}]
= ½(35 + 9 + 20)
= 32 sq. units
2. In each of the following find the value of ‘k’, for which the points are collinear.
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
D = (0+2/2 , -1+1/2) = (1,0)
E = (0+0/2 , -3-1/2) = (0,1)
Area of a triangle = 1/2 {x
1
(y
2
- y
3
) + x
2
(y
3
- y
1
) + x
3
(y
1
- y
2
)}
Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}
= 1/2 (1+1) = 1 square units
Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]
= 1/2 {8} = 4 square units
Therefore, the required ratio is 1:4.
4. 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.
Area of a triangle = 1/2 {x
1
(y
2
- y
3
) + x
2
(y
3
- y
1
) + x
3
(y
1
- y
2
)}
Area of ΔABC =1/2[−4 (−5 − 2) – 3 {-2 − (−2)} + 3 {−2 − (−5)}]
=1/2 [12 + 0 + 9]
= 21/2 sq. units
Again using formula to find area of triangle:
Area of △ACD = [−4 (−2 − 3) + 3 {3 − (−2)} + 2 {−2 − (-2)}]
= 1/2 [20 + 15 + 0]
= 35/2 sq. units
Area of ☐ABCD = Area of ΔABC + Area of ΔACD
= 21/2 + 35/2 = 28 sq. units
5. We know that 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).
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.
Coordinates of point D = (3+5/2, -2+2/2) = (4,0)
Area of a triangle = 1/2 {x
1
(y
2
- y
3
) + x
2
(y
3
- y
1
) + x
3
(y
1
- y
2
)}
Area of △ABD = 1/2[4 (−2 − 0) + 3 {0 − (−6)} + 4 {−6 − (−2)}]
=1/2(−8 + 18 −16)
=1/2 (−6) = −3 sq units
Area cannot be in negative.
Therefore, we just consider its numerical value.
Therefore, area of △ABD = 3 sq units
Again using formula to find area of triangle:
Area of △ABD = 1/2 [4 (0 − 2) + 4{2 − (−6)} + 5 {−6 −0 )}]
= 1/2 (- 8 + 32 −30) = ½ (-6) = -3 sq units.
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas
Hence Proved.
