Problem 88
Question
If non-zero numbers \(a, b, c\) are in H.P., then the straight line \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) always passes through a fixed point. That point is \([2005]\) (A) \((-1,2)\) (B) \((-1,-2)\) (C) \((1,-2)\) (D) \(\left(1,-\frac{1}{2}\right)\)
Step-by-Step Solution
Verified Answer
The fixed point is (1, -2).
1Step 1: Define the Harmonic Progression Condition
Given that the numbers \(a, b, c\) are in Harmonic Progression (H.P.), it means their reciprocals are in Arithmetic Progression (A.P.). Thus, \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) are in A.P. This implies \( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \).
2Step 2: Use the Harmonic Progression Condition
Rewrite the line equation \( \frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0 \). Substitute for \( \frac{1}{c} \) from the H.P. condition as \( \frac{1}{c} = \frac{2}{b} - \frac{1}{a} \). Substitute in the line equation to get \( \frac{x}{a} + \frac{y}{b} + \frac{2}{b} - \frac{1}{a} = 0 \).
3Step 3: Simplify the Equation
Combine the terms in the equation together: \( \frac{x - 1}{a} + \frac{y + 2}{b} = 0 \). This equation describes a family of lines.
4Step 4: Find the Fixed Point
A line that passes through a fixed point does so regardless of \(a\) and \(b\) values. Thus, let both coefficients of the terms with \(\frac{1}{a}\) and \(\frac{1}{b}\) cancel each other out independently: \( x - 1 = 0 \) and \( y + 2 = 0 \) which gives us the fixed point \((1, -2)\).
Key Concepts
Arithmetic ProgressionFixed PointEquation of a Line
Arithmetic Progression
An Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the "common difference". Here is a simple way to understand it:
- If the sequence starts with the number \(a\), then the next term is \(a + d\), where \(d\) is the common difference.
- The sequence continues as \(a, a + d, a + 2d, \ldots\).
Fixed Point
In geometry, a fixed point of a function or transformation is a point that remains unchanged when the function is applied. In context of lines, a fixed point is a specific point that lies on every line within a given family of lines.When dealing with an equation like \( \frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0 \), it describes a family of lines dependent on \(a, b, c\). The sequence condition and equation manipulation, as in the original solution, help identify this invariant point. For instance:
- The substitution and simplification steps lead to finding when the equation holds independently of the specific parameters \(a\) and \(b\).
- Setting the line conditions \(x - 1 = 0\) and \(y + 2 = 0\) results in the point \((1, -2)\), which is fixed for the entire family of lines defined by the given equation.
Equation of a Line
The equation of a line is a critical concept in algebra and geometry, describing a wide range of linear relations between variables. The standard form is typically \( Ax + By + C = 0 \), where \(A, B,\) and \(C\) are constants, describing a straight line with slope and intercept properties.In our specific problem, the equation \( \frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0 \) was used to explore harmonic progression and identify the fixed point for a family of lines. Here are key points about solving line equations:
- The coefficients yourself determine the line's slope and position.
- In problems involving multiple variables, substitution and manipulation changes the line's form to highlight underlying patterns or fixed components.
- Simplifying complex terms can reveal intersections or points common to multiple lines, known as fixed points.
Other exercises in this chapter
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