Class 11 RD Sharma Solutions - Chapter 23 The Straight Lines- Exercise 23.14

Question 1: 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.

Solution:

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)

Question 2: 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 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)

Question 3: Determine whether the point (-3, 2) lies inside or outside the triangle whose sides are given by the equations x + y – 4 = 0, 3x – 7y + 8 = 0, 4x – y – 31 = 0.

Solution:

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.