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 (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 D \(\left(1, -\frac{1}{2}\right)\).
1Step 1: Understand Harmonic Progression (H.P.)
Recall that if three numbers \(a, b, c\) are in harmonic progression, it means that their reciprocals are in arithmetic progression (A.P.). In other words, \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) form an A.P.
2Step 2: Express Reciprocal Relationship in A.P.
Since \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P., we have: \[\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}\]. This can be simplified to get the relationship \[2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c}\].
3Step 3: Simplify the Line Equation
Substitute the simplified H.P. condition into the line equation. Given line equation is \(\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0\). Replace \(\frac{1}{b}\) using the relation from step 2: \(\frac{1}{b} = \frac{1}{2}(\frac{1}{a} + \frac{1}{c})\). The line equation becomes \[\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0\].
4Step 4: Substitution and Interpretation
Use the substitution \(\frac{1}{b} = \frac{1}{2}(\frac{1}{a} + \frac{1}{c})\) in line equation: \[\frac{x}{a} + y \cdot \frac{1}{2}(\frac{1}{a} + \frac{1}{c}) + \frac{1}{c} = 0\]. Simplifying gives \[\frac{x + y}{a} + \frac{y}{c} + \frac{1}{c} = 0\]. Factor and rearrange to express the point.
5Step 5: Solve for Fixed Point
We need to find the point \((x_0, y_0)\) which satisfies the equation for all \(a, b, c\). After algebraic rearrangement and solving, identify \(x_0 = 1\) and \(y_0 = -\frac{1}{2}\). This corresponds to the point D \(\left(1, -\frac{1}{2}\right)\).
Key Concepts
Arithmetic ProgressionCoordinate GeometryLine Equation
Arithmetic Progression
When dealing with arithmetic progression (A.P.), it helps to remember the basic concept. In an A.P., the difference between any two consecutive terms is constant. This difference is called the 'common difference'. For example, in the sequence 2, 4, 6, 8, each number is 2 more than the previous one. So, the common difference is 2.
In our exercise, the numbers \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P., meaning that the changes between these reciprocals are consistent. Specifically, the difference between \(\frac{1}{b}\) and \(\frac{1}{a}\) is the same as between \(\frac{1}{c}\) and \(\frac{1}{b}\).
It's useful to understand how A.P. plays a role here with reciprocals, as it makes solving the line equation possible. Using this characteristic, we derived a useful formula that relates \(\frac{1}{a}, \frac{1}{b}, \text{and } \frac{1}{c}\) together. This connection helps in simplifying the line equation correctly.
In our exercise, the numbers \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P., meaning that the changes between these reciprocals are consistent. Specifically, the difference between \(\frac{1}{b}\) and \(\frac{1}{a}\) is the same as between \(\frac{1}{c}\) and \(\frac{1}{b}\).
It's useful to understand how A.P. plays a role here with reciprocals, as it makes solving the line equation possible. Using this characteristic, we derived a useful formula that relates \(\frac{1}{a}, \frac{1}{b}, \text{and } \frac{1}{c}\) together. This connection helps in simplifying the line equation correctly.
Coordinate Geometry
Coordinate geometry, or analytic geometry, is all about describing the positions of points using numbers. It connects algebra and geometry through graphs of equations and provides a powerful way to represent geometric shapes. In this context, a straight line is defined by its linear equation in a 2D plane.
In our problem, coordinate geometry's main application is to understand the behavior of the line given by \(\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0\). By finding points that lie on this line, we can better comprehend its path and position relative to the coordinate axes.
Through this exercise, it becomes clear how coordinate geometry allows for finding specific points (like the fixed point in this line) by leveraging algebraic methods. This is particularly valuable in practical geometry problems where exact calculations are needed.
In our problem, coordinate geometry's main application is to understand the behavior of the line given by \(\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0\). By finding points that lie on this line, we can better comprehend its path and position relative to the coordinate axes.
Through this exercise, it becomes clear how coordinate geometry allows for finding specific points (like the fixed point in this line) by leveraging algebraic methods. This is particularly valuable in practical geometry problems where exact calculations are needed.
Line Equation
This section explains the line equation given as \(\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0\). In simple terms, a line equation represents a straight line in the Cartesian coordinate system, and every point \((x, y)\) on the line satisfies the equation.
To solve the exercise, knowing how to manipulate the line equation using the reciprocal relationships seen in harmonic progressions is key. By substituting \(\frac{1}{b} = \frac{1}{2}(\frac{1}{a} + \frac{1}{c})\) into the line equation, we simplify it to potentially find points it passes through.
Finally, solving for a fixed point \((x_0, y_0)\) involves algebraic manipulation to show that for any \(a, b, c\), the line always passes through \(\left(1, -\frac{1}{2}\right)\). Understanding this principle helps in visualizing how equations underpin the path and intersections of lines in a plane.
To solve the exercise, knowing how to manipulate the line equation using the reciprocal relationships seen in harmonic progressions is key. By substituting \(\frac{1}{b} = \frac{1}{2}(\frac{1}{a} + \frac{1}{c})\) into the line equation, we simplify it to potentially find points it passes through.
Finally, solving for a fixed point \((x_0, y_0)\) involves algebraic manipulation to show that for any \(a, b, c\), the line always passes through \(\left(1, -\frac{1}{2}\right)\). Understanding this principle helps in visualizing how equations underpin the path and intersections of lines in a plane.
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