(2) Vertices of the triangle
taken in order and their areas are given below. In each of the following find
the value of a.
(i) (0, 0), (4, a) and (6, 4) and its area is 17 sq.units
Solution :
Area of the triangle = 17 square units
(1/2)[(0 + 16 + 0) – (0 + 6a + 0)] = 17
(1/2)(16 – 6 a) = 17
8 - 3a = 17
-3a = 9
a = -3
(ii) (a, a) , (4 , 5) and ( 6 , -1) and its area is 9 sq.units
Solution :
Area of the triangle = 9 square units
(1/2)[(5a - 4 + 6a) – (4a + 30 - a)]
(1/2)[(11a – 4) – ( 3a+30)] = 9
(1/2)[11a – 4 – 3a - 30] = 9
(1/2)[8a – 34] = 9
4a – 17 = 9
4a = 26
a = 26/4
a = 13/2
(iii) (a, -3),(3, a) and (-1, 5) and its area is 12 sq.units
Solution :
Area of the triangle = 12 square units
(1/2)[(a2 + 15 + 3) – (-9 - a + 5a)] = 12
(1/2)[(a2 + 18 ) – (-9 + 4a)] = 12
[a2 + 18 + 9 - 4a] = 12 x 2
a2 - 4a + 27 = 24
a2 - 4a + 27 - 24 = 0
a2 - 4a + 3 = 0
(a - 1)(a - 3) = 0
a = 1 , 3
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