# 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(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.

**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(α

^{2}, 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)(α

^{2}– 3α + 2) > 0(9 – 9 + 2)(α

^{2}– 3α + 2) > 0α

^{2}– 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)(α

^{2}– 2α – 3) > 0(4 – 4 – 3)(α

^{2}– 2α – 3) > 0(-3)(α

^{2}– 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)(α

^{2}– 5α + 6) > 0(13 – 25 + 6)(α

^{2}– 5α + 6) > 0α

^{2}– 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.

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