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

Last Updated : 21 Feb, 2021

### Question 1: Find the values of Î± so that the point P(Î±2, Î±) lies inside or on the triangle formed by the lines x â€“ 5y + 6 = 0, x â€“ 3y + 2 = 0 and x â€“ 2y â€“ 3 = 0.

Solution:

Let the triangle be ABC where sides are AB, BC, and CA with equations as x â€“ 5y + 6 = 0, x â€“ 3y + 2 = 0 and x â€“ 2y â€“ 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(Î±2, Î±) 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)(Î±2 â€“ 3Î± + 2) > 0

(9 â€“ 9 + 2)(Î±2 â€“ 3Î± + 2) > 0

Î±2 â€“ 3Î± + 2 > 0

(Î± â€“ 2)(Î± â€“ 1) >

Î± âˆˆ (- âˆž, 1 ) âˆª ( 2, âˆž) â€¦… (1)

If B and P are on the same side as AC, then

(4(1) + 2(-2) – 3)(Î±2 â€“ 2Î± â€“ 3) > 0

(4 â€“ 4 â€“ 3)(Î±2 â€“ 2Î± â€“ 3) > 0

(-3)(Î±2 â€“ 2Î± â€“ 3) > 0

(Î± â€“ 3)(Î± + 1) < 0

Î± âˆˆ (- 1, 3) â€¦… (2)

If C and P are on the same side as AB, then

(13(1) + 5(-5) + 6)(Î±2 â€“ 5Î± + 6) > 0

(13 â€“ 25 + 6)(Î±2 â€“ 5Î± + 6) > 0

Î±2 â€“ 5Î± + 6 > 0

(Î± â€“ 3)(Î± â€“ 2) < 0

Î± âˆˆ ( 2, 3) â€¦… (3)

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

Î± âˆˆ (2, 3)

Therefore, Î± âˆˆ (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 â€“ 4 = 0, 3x â€“ 7y â€“ 8 = 0 and 4x â€“ y â€“ 31 = 0.

Solution:

Let the triangle be ABC where sides are AB, BC, and CA with equations as x + y â€“ 4 = 0, 3x â€“ 7y â€“ 8 = 0 and 4x â€“ y â€“ 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 âˆˆ (22/3, 33/4)

Therefore, a âˆˆ (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 â€“ 4 = 0, 3x â€“ 7y + 8 = 0, 4x â€“ y â€“ 31 = 0.

Solution:

Let the triangle be ABC where sides are AB, BC, and CA with equations as x + y â€“ 4 = 0, 3x â€“ 7y + 8 = 0, 4x â€“ y â€“ 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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