Understanding the Problem: Straight Line and Arithmetic Progression

The problem asks us to find a specific point that a straight line will always pass through. The equation of the line is given as \(Ax + 2By + C = 0\), and we are told that the coefficients A, B, and C are in Arithmetic Progression (AP).

When three numbers A, B, and C are in Arithmetic Progression, it means the difference between consecutive terms is constant. Mathematically, this condition is expressed as:

\(B - A = C - B\)

Rearranging this equation gives us a useful relationship between A, B, and C:

\(2B = A + C\)

This relationship is key to finding the fixed point.

Substituting the AP Condition into the Line Equation

We have the equation of the line: \(Ax + 2By + C = 0\).

Since A, B, and C are in AP, we know that \(2B = A + C\). We can substitute \(A + C\) in place of \(2B\) in the line equation. However, \(2B\) is the coefficient of \(y\). Let's substitute \(2B\) itself using the relation \(2B = A + C\).

So, replace \(2B\) in the equation \(Ax + 2By + C = 0\) with \(A + C\):

\(Ax + (A + C)y + C = 0\)

Finding the Fixed Point Coordinates

Now we have the equation \(Ax + (A + C)y + C = 0\). This equation must hold true for any values of A, B, and C that satisfy the AP condition. We want to find a point \((x, y)\) that lies on this line regardless of the specific values of A and C (since B is determined by A and C through the AP condition).

Let's rearrange the equation to group terms involving A and terms involving C:

\(Ax + Ay + Cy + C = 0\)

Group the terms with A and the terms with C:

\(A(x + y) + C(y + 1) = 0\)

For this equation to be true for any values of A and C (not both zero, as they are coefficients of a line), the coefficients of A and C must both be equal to zero. This is because if the coefficients of A and C were not zero, we could choose specific values for A and C that would make the equation false for a fixed \((x, y)\).

So, we must have the following system of equations:

1. \(x + y = 0\)
2. \(y + 1 = 0\)

Now we solve this simple system of linear equations for \(x\) and \(y\).

From equation (2):

\(y + 1 = 0\)

\(y = -1\)

Substitute the value of \(y\) into equation (1):

\(x + y = 0\)

\(x + (-1) = 0\)

\(x - 1 = 0\)

\(x = 1\)

Thus, the fixed point is \((x, y) = (1, -1)\).

Verification

Let's verify if the point \((1, -1)\) always lies on the line \(Ax + 2By + C = 0\) when A, B, and C are in AP (i.e., \(2B = A + C\)).

Substitute \(x = 1\) and \(y = -1\) into the line equation:

\(A(1) + 2B(-1) + C = 0\)

\(A - 2B + C = 0\)

Rearranging this equation gives:

\(A + C = 2B\)

This is exactly the condition for A, B, and C to be in Arithmetic Progression. Since substituting the point \((1, -1)\) into the line equation results in the AP condition, the line \(Ax + 2By + C = 0\) always passes through the point \((1, -1)\) whenever A, B, and C are in AP.

Summary of Steps

• Recognize the condition for A, B, C to be in AP: \(2B = A + C\).
• Substitute this condition into the equation of the line: \(Ax + 2By + C = 0\).
• Rearrange the resulting equation: \(A(x+y) + C(y+1) = 0\).
• Equate the coefficients of A and C to zero to find the point that satisfies the equation for any A and C: \(x+y=0\) and \(y+1=0\).
• Solve the system of equations to find \(x=1\) and \(y=-1\).
• The fixed point is \((1, -1)\).

Revision Table: Straight Lines and AP

Additional Information: General Approach for Fixed Point Problems

Problems asking for a "fixed point" a line or curve always passes through, often involve parameters (like A, B, C here). A common strategy is to express the equation in terms of a minimal set of independent parameters and then recognize that for the equation to hold for any value of these parameters, the terms multiplying each parameter must individually be zero.

In this case, we used the AP condition \(2B = A+C\) to eliminate B, leaving the equation in terms of A and C: \(A(x+y) + C(y+1) = 0\). For this to hold for any A and C, the expressions \((x+y)\) and \((y+1)\) must both be zero, allowing us to solve for \(x\) and \(y\).

This technique is applicable in various coordinate geometry problems involving parameters.