问题1：找到α的值，使得点P(α2，α)内或由所述线形成的三角形在于X – 5Y + 6 = 0，X – 3Y + 2 = 0和x – 2Y – 3 = 0。
解决方案：
Let the triangle be ABC where sides are AB, BC, and CA with equations as x – 5y + 6 = 0, x – 3y + 2 = 0 and x – 2y – 3 = 0 respectively.
We get, A(9, 3), B(4, 2), and C(13, 5) as the coordinates of the vertices.
Given that point P(α2, α) lies either inside or on the triangle, therefore,
(i) A and P must be on the same side as BC.
(ii) B and P must be on the same side as AC.
(iii) C and P must be on the same side as AB.
Now,
If A and P are on the same side as BC, then
(9(1) + 3(-3) + 2)(α2 – 3α + 2) > 0
(9 – 9 + 2)(α2 – 3α + 2) > 0
α2 – 3α + 2 > 0
(α – 2)(α – 1) >
α ∈ (- ∞, 1 ) ∪ ( 2, ∞) …… (1)
If B and P are on the same side as AC, then
(4(1) + 2(-2) – 3)(α2 – 2α – 3) > 0
(4 – 4 – 3)(α2 – 2α – 3) > 0
(-3)(α2 – 2α – 3) > 0
(α – 3)(α + 1) < 0
α ∈ (- 1, 3) …… (2)
If C and P are on the same side as AB, then
(13(1) + 5(-5) + 6)(α2 – 5α + 6) > 0
(13 – 25 + 6)(α2 – 5α + 6) > 0
α2 – 5α + 6 > 0
(α – 3)(α – 2) < 0
α ∈ ( 2, 3) …… (3)
From equations (1), (2) and (3), we get
α ∈ (2, 3)
Therefore, α ∈ (2, 3)
问题2：找到参数a的值，以使点(a，2)是由x + y – 4 = 0、3x – 7y – 8 = 0和4x – y –线形成的三角形的内部点31 = 0。
解决方案：
Let the triangle be ABC where sides are AB, BC, and CA with equations as x + y – 4 = 0, 3x – 7y – 8 = 0 and 4x – y – 31 = 0 respectively.
We get, A(7, -3), B(18/5, 2/5), and C(209/25, 61/25) as the coordinates of the vertices.
Given that point P(a, 2) is an interior point, therefore,
(i) A and P must be on the same side as BC.
(ii) B and P must be on the same side as AC.
(iii) C and P must be on the same side as AB.
Now,
If A and P are on the same side as BC, then
(7(3) – 7(-3) – 8)(3a – 7(2) – 8) > 0
(21 + 21 – 8)(3a – 7(2) – 8) > 0
3a – 22 > 0
a > 22/3 …… (1)
If B and P are on the same side as AC, then
(4(18/5) – (2/5) – 31)(4a – 2 – 31) > 0
4a – 33 > 0
a > 33/4 …… (2)
If C and P are on the same side as AB, then
(209/25 + 61/25 – 4)(a + 2 – 4) > 0
a + 2 > 0
a > -2 …… (3)
From equations (1), (2) and (3), we get
a ∈ (22/3, 33/4)
Therefore, a ∈ (22/3, 33/4)
问题3：确定点(-3，2)是在三角形的内侧还是外侧，三角形的边由公式x + y – 4 = 0，3x – 7y + 8 = 0，4x – y – 31 = 0给定。
解决方案：
Let the triangle be ABC where sides are AB, BC, and CA with equations as x + y – 4 = 0, 3x – 7y + 8 = 0, 4x – y – 31 = 0 respectively
We get, A(7, -3), B(2, 2), and C(9, 5) as the coordinates of the vertices.
Given that point P(-3, 2) lies either inside or outside the triangle, therefore,
(i) A and P must be on the same side as BC.
(ii) B and P must be on the same side as AC.
(iii) C and P must be on the same side as AB.
Now,
If A and P are on the same side as BC, then
(3(7) – 7(-3) + 8)(3(-3) – 7(2) + 8) > 0
(21 + 21 + 8)(-9 – 14 + 8) > 0
(50)(-15) > 0 which is false
Therefore, point P(-3, 2) lies outside the triangle ABC.