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Question
Vertices of given triangles are taken in order and their areas are provided aside. Find the value of ‘p’.
| Vertices | Area (sq.units) |
| (p, p), (5, 6), (5, –2) | 32 |
Solution
Let the vertices be A(p, p), B(5, 6) and C(5, –2)
Area of a triangle = 32 sq. units
`1/2[(x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3)]` = 32
`1/2[(6"p" - 10 + 5"p") - (5"p" + 30 - 2"p")]` = 32
`1/2 [11"p" - 10 - 3"p" - 30]` = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40
⇒ 8p = 104
p = `104/8`
= 13
The value of p = 13
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