Find the values of the parameter a so that the point (a, 2) is an interior point of the triangle formed by the lines x + y − 4 = 0, 3x − 7y − 8 = 0 and 4x − y − 31 = 0.
Solution
Let ABC be the triangle of sides AB, BC and CA whose equations are x + y − 4 = 0, 3x − 7y − 8 = 0 and 4x − y − 31 = 0, respectively.
On solving them, we get
\[A \left( 7, - 3 \right)\],
\[B \left( \frac{18}{5}, \frac{2}{5} \right)\] and \[C \left( \frac{209}{25}, \frac{61}{25} \right)\] as the coordinates of the vertices.
Let P (a, 2) be the given point.
It is given that point P (a,2) lies inside the triangle. So, we have the following:
(i) A and P must lie on the same side of BC.
(ii) B and P must lie on the same side of AC.
(iii) C and P must lie on the same side of AB.
Thus, if A and P lie on the same side of BC, then \[\left( 21 + 21 - 8 \right)\left( 3a - 14 - 8 \right) > 0\]
\[\Rightarrow a > \frac{22}{3}\] ... (1)
If B and P lie on the same side of AC, then \[\left( \frac{4 \times 18}{5} - \frac{2}{5} - 31 \right)\left( 4a - 2 - 31 \right) > 0\]
\[\Rightarrow a < \frac{33}{4}\] ... (2)
If C and P lie on the same side of AB, then
\[\left( \frac{209}{25} + \frac{61}{25} - 4 \right)\left( a + 2 - 4 \right) > 0\]
\[ \Rightarrow \left( \frac{34}{5} - 4 \right)\left( a + 2 - 4 \right) > 0\]
\[\Rightarrow a > 2\] ... (3)
From (1), (2) and (3), we get:
\[a \in \left( \frac{22}{3}, \frac{33}{4} \right)\]
