Question

If non-zero numbers \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in \(\mathrm{HP}\), then the straight line \((\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})+(1 / \mathrm{c})=0\) always passes through a fixed point that point is (a) \([1,-(1 / 2)]\) (b) \((1,-2)\) (c) \((-1,-2)\) (d) \((-1,2)\)

Short answer

The fixed point through which the straight line (x / a) + (y / b) + (1 / c) = 0 always passes is (b) (1, -2).

Step-by-step solution

Denote the reciprocals and the arithmetic progression of 1/a, 1/b, and 1/c

: Let p = 1/a, q = 1/b, and r = 1/c. Since a, b, and c are in HP, the reciprocals p, q, and r are in an arithmetic progression with a common difference. That implies: q - p = r - q

Express r in terms of p and q and substitute in the given equation

: From step 1, we know that r - q = q - p, which can be written as: r = 2q - p Now replace p, q and r with their respective reciprocals in the given equation of the line: (x / a) + (y / b) + (1 / c) = 0 => (x * p) + (y * q) + (1 * r) = 0 Substitute r = 2q - p in the equation: (x * p) + (y * q) + (1 * (2q - p)) = 0

Rewrite equation in terms of x and y

: Now rearrange and simplify the equation from step 2: xp + yq + 2q - p = 0 => p(x - 1) + q(y + 2) = 0 Since a, b, and c are non-zero, their reciprocals p, q, and r, are also non-zero. So we have a linear equation in x and y with non-zero coefficients.

Find the fixed point

: Now we have the equation p(x - 1) + q(y + 2) = 0. To find the fixed point, let's set values of p and q that satisfy the equation: p = 1 q = -1/2 Now, substituting these values into the equation: (1)(x - 1) + (-1/2)(y + 2) = 0 x - 1 - (y + 2) / 2 = 0 2x - 2 - y - 2 = 0 Simplifying the equation and solving for x and y: 2x - y - 4 = 0 => y = 2x - 4 Now, we can see that the coordinates of the fixed point are (x, y) = (1, -2). Therefore, the correct option is: (b) (1, -2)

Key concepts

Arithmetic progression

An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always the same. This difference is a constant known as the common difference. For example, in the sequence 5, 8, 11, 14... the common difference is 3.

To express the n-th term of an arithmetic progression, we use the formula:

• First term: \(a_1\)
• Common difference: \(d\)
• n-th term: \(a_n = a_1 + (n-1)d\)

Understanding arithmetic progressions is crucial in various fields such as finance, computer science, and even in solving algebraic equations like the one in our exercise, where the reciprocals of the terms form an arithmetic progression.

Reciprocal

The reciprocal of a number is simply the result of dividing 1 by that number. In mathematical terms, if you have a number \(x\), its reciprocal is \(\frac{1}{x}\). Reciprocals are handy in many arithmetic and algebraic manipulations.

When dealing with harmonic progressions (HP), we often work with reciprocals. For example, if numbers \(a, b, c\) are in harmonic progression, it means their reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) form an arithmetic progression.

This concept plays a critical role in solving the given problem, where identifying the pattern in reciprocals helps establish the linear relationship featured in the exercise.

Linear equation

A linear equation is a polynomial equation of the first degree, meaning it has no exponents higher than one. The general form is \(ax + by + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) and \(y\) are variables.

Linear equations graph as straight lines on the coordinate plane. In the context of the exercise, the expression \( (x/a) + (y/b) + (1/c) = 0 \) was rewritten to illustrate the relationship between \(x\) and \(y\) in a linear form. The beauty of linear equations lies in their simplicity and the fact that their solutions can often be visualized directly on a graph.

When simplified to \(p(x - 1) + q(y + 2) = 0\), it becomes evident how the equation describes a linear path, ultimately guiding us to find a particular fixed point on this line.

Fixed point in geometry

In geometry, a fixed point refers to a specific point that remains unchanged under a certain transformation or condition. In our exercise, we sought the fixed point through which the line always passes, regardless of specific values for the coefficients involved.

By setting particular conditions derived from the problem, we were able to simplify the equation to its most fundamental form. This leads to spotting the invariant point (1, -2) on the plane. Geometrically, this means any line conforming to the problem's setup will always intersect at this fixed point.

Understanding fixed points can have numerous applications, from designing stable systems to solving equations where certain values must remain constant despite varying other inputs.