# 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.

Given: x – 5y + 6 = 0, x – 3y + 2 = 0 and x – 2y – 3 = 0 forming a triangle and point P(α2, 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 (
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

(9 – 9 + 2)(α2 – 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) (α2 – 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)(α2 – 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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