Question

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.

Answer 1

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`.