Question

Find the values of α so that the point P (α², α) lies inside or on the triangle formed by the lines x − 5y+ 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0.

Answer 1

Let ABC be the triangle of sides AB, BC and CA whose equations are x − 5y + 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0, respectively.

On solving the equations, we get A (9,3), B (4, 2) and C (13, 5) as the coordinates of the vertices.

It is given that point P (α², α) lies either inside or on the triangle. The three conditions are given below.

(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.

If A and P lie on the same side of BC, then

\[\left( 9 - 9 + 2 \right)\left( \alpha^2 - 3\alpha + 2 \right) \geq 0\]

\[ \Rightarrow \left( \alpha - 2 \right)\left( \alpha - 1 \right) \geq 0\]

\[\Rightarrow \alpha \in ( - \infty , 1] \cup [2, \infty )\]      ... (1)
If B and P lie on the same side of AC, then

\[\left( 4 - 4 - 3 \right)\left( \alpha^2 - 2\alpha - 3 \right) \geq 0\]

\[ \Rightarrow \left( \alpha - 3 \right)\left( \alpha + 1 \right) \leq 0\]

\[\Rightarrow \alpha \in \left[ - 1, 3 \right]\]            ... (2)

If C and P lie on the same side of AB, then

\[\left( 13 - 25 + 6 \right)\left( \alpha^2 - 5\alpha + 6 \right) \geq 0\]

\[ \Rightarrow \left( \alpha - 3 \right)\left( \alpha - 2 \right) \leq 0\] 

\[\Rightarrow \alpha \in \left[ 2, 3 \right]\]      ... (3)
From (1), (2) and (3), we get: \[\alpha \in \left[ 2, 3 \right]\]