Q. 14.0( 5 Votes )

# Find the values of α so that the point P(α ^{2}, 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 :

__Given:__ x – 5y + 6 = 0, x – 3y + 2 = 0 and x – 2y – 3 = 0 forming a triangle and point P(α^{2}, α) lies inside or on the triangle

__To find:__ value of α

Explanation:

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.

Diagram:

It is given that point P (α^{2}, 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

(9 – 9 + 2)(α^{2} – 3α + 2) ≥0

⇒ (α – 2)(α – 1) ≥ 0

⇒ α ∈ (- ∞, 1 ] ∪ [ 2, ∞) … (1)

If B and P lie on the same side of AC, then

(4 – 4 – 3) (α^{2} – 2α – 3) ≥ 0

⇒ (α – 3)(α + 1) ≤ 0

⇒ α ∈ [- 1, 3] … (2)

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

(13 – 25 + 6)(α^{2} – 5α + 6) ≥0

⇒ (α – 3)(α – 2) ≤ 0

⇒ α ∈ [ 2, 3] … (3)

From (1), (2) and (3), we get:

α∈ [ 2, 3]

Hence, α∈ [2, 3]

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