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Home » 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.
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.
CBSE | Class 11 | Exercise 23.14 | Maths | RD Sharma | Straight Lines
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.

According to ques,:

    \[x\text{ }+\text{ }y\text{ }\text{ }4\text{ }=\text{ }0,\]

    \[3x\text{ }\text{ }7y\text{ }\text{ }8\text{ }=\text{ }0\]

and

    \[4x\text{ }\text{ }y\text{ }\text{ }31\text{ }=\text{ }0\]

forming a triangle and point (a, 2)is an interior point of the triangle

Let ABC be the triangle of sides AB, BC and CA whose equations are:

    \[x\text{ }+\text{ }y~-~4\text{ }=\text{ }0,\]

    \[3x~-~7y~-~8\text{ }=\text{ }0\]

and

    \[4x~-~y~-~31\text{ }=\text{ }0,\]

respectively.

On solving them, we get A (7, – 3), B (18/5, 2/5) and C (209/25, 61/25) as the coordinates of the vertices.

Let P (a, 2) be the given point.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 71

It is given that point P (a, 2) lies inside the triangle. So, we have the following:

(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.

Thus, if A and P lie on the same side of BC, then

    \[21\text{ }+\text{ }21\text{ }\text{ }8\text{ }\text{ }3a\text{ }\text{ }14\text{ }\text{ }8\text{ }>\text{ }0\]

    \[a\text{ }>\text{ }22/3~\ldots \text{ }\left( 1 \right)\]

From (1), (2) and (3), we have

A ∈ (22/3, 33/4)

∴ A ∈ (22/3, 33/4)

i

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