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प्रश्न
Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.
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उत्तर
The formula for the area ‘A’ encompassed by three points( x1 , y1) , (x2 , y2) and (x3 , y3) is given by the formula,
\[∆ = \frac{1}{2}\left| \left( x_1 y_2 + x_2 y_3 + x_3 y_1 \right) - \left( x_2 y_1 + x_3 y_2 + x_1 y_3 \right) \right|\]
The three given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units. Substituting these values in the above mentioned formula we have,
\[∆ = \frac{1}{2}\left| \left( a \times 6 + \left( - 2 \right) \times 1 + 3 \times 2a \right) - \left( \left( - 2 \right) \times 2a + 3 \times 6 + a \times 1 \right) \right|\]
\[10 = \frac{1}{2}\left| \left( 6a - 2 + 6a \right) - \left( - 4a + 18 + a \right) \right|\]
\[10 = \frac{1}{2}\left| 15a - 20 \right|\]
\[20 = \left| 15a - 20 \right|\]
\[4 = \left| 3a - 4 \right|\]
We have |3a - 4 | = 4. Hence either
3a - 4 = 4
3a = 8
a = `8/3`
Or
-(3a - 4 ) = 4
4 - 3a = 4
a = 0
Hence the values of ‘a’ which satisfies the given conditions are a = 0 `a = 8/3` .
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