Problem 39 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

1\. The equation of a family of lines is given by \((2+3 t)\) \(x+(1-2 t) y+4=0\), where \(t\) is the parameter. The equation of a straight line, belonging to this family, at the maximum distance from the point \((2,3)\) is (A) \(21 x+14 y=0\) (B) \(21 x-14 y=0\) (C) \(14 x-21 y=0\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (D) none of these.

Step 1: Line Equation Examination

Let's first write down the given family of lines: \((2 + 3t)x + (1 - 2t)y + 4 = 0\). In general, this is the equation of a straight line where the parameters \(a = 2 + 3t\), \(b = 1 - 2t\), and \(c = 4\).

Step 2: Point Distance from a Line

To find which line is at the maximum distance from a given point, we need the distance formula for a point \((x_0, y_0)\) to a line \(ax + by + c = 0\): \[ \text{Distance} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]. Here, \((x_0, y_0) = (2, 3)\).

Step 3: Substitute Point into Line Distance Formula

Now substitute \(x_0 = 2\), \(y_0 = 3\), and \(c = 4\) into the formula: \[ \text{Distance} = \frac{|(2 + 3t)\cdot 2 + (1 - 2t)\cdot 3 + 4|}{\sqrt{(2 + 3t)^2 + (1 - 2t)^2}} \].

Step 4: Simplify Numerator of Distance

Simplify the numerator: \((2 + 3t)\cdot 2 + (1 - 2t)\cdot 3 + 4 = 4 + 6t + 3 - 6t + 4 = 11\).

Step 5: Simplify Denominator of Distance

Simplify the denominator: \(\sqrt{(2 + 3t)^2 + (1 - 2t)^2} = \sqrt{4 + 12t + 9t^2 + 1 - 4t + 4t^2} = \sqrt{13t^2 + 8t + 5}\).

Step 6: Maximize Distance

The distance is maximized when the denominator is minimized, implying each term contributes minimally. Only terms including \(t\) can be affected by choosing \(t\). Set the derivative of \(\sqrt{13t^2 + 8t + 5}\) to zero for \(t\). This specializes our inquiry to specific algebraic valuation of \(t\) that simplifies computation.

Step 7: Determine optimal t

The computed minimum value of the expression \(13t^2 + 8t + 5\) derived from constant evaluation (also motion of contrasting aspects), the solution residual outcome stems \(t = 0\), or derive zero determinant complex constant segmentation.

Step 8: Assess which line satisfies condition for t=0

Plugging \(t = 0\) into the line equation \((2 + 3t)x + (1 - 2t)y + 4 = 0\), results in the line \(2x + y + 4 = 0\), aligning standard forms with various transformations. The congruent response is not directly listed thus review examines these adjustments within residual examination.

Step 9: Evaluate Choices

Translate outcome into provided options. Evaluating lines implicitly determines none among \((A)\), \((B)\), \((C)\) matches directly, signifying mismatched assessed reality, allign with solution investigative nature analyzing evaluated transformations of endpoints.

Key Concepts

Family of Lines

In analytical geometry, a family of lines refers to a set of straight lines governed by an equation containing a parameter. Adjusting this parameter can produce different lines within the family.
In our given exercise, the family of lines is represented by \[(2+3t)x + (1-2t)y + 4 = 0\]where \(t\) is the variable parameter. The values of \(t\) change the coefficients of \(x\) and \(y\), subsequently shifting the orientation and position of the line in the Cartesian plane.
This way, any chosen \(t\) will give a specific line through the same general formulation, creating a spectrum of lines with shared geometric properties.

Distance Formula in Geometry

The distance from a point to a line is a crucial concept in geometry. The formula provides a way to calculate how far a point is from a given line.
The distance from a point \((x_0, y_0)\) to a line given by the equation \(ax + by + c = 0\) is determined using:

• Numerator: the absolute value of the expression \(ax_0 + by_0 + c\).
• Denominator: the square root of the sum of squares of the coefficients \(a\) and \(b\), i.e., \(\sqrt{a^2 + b^2}\).

Let’s say we have a line and want to know how far it is from a point (2,3). Plug in the values into the formula as follows:\[\text{Distance} = \frac{| (2 + 3t) \cdot 2 + (1 - 2t) \cdot 3 + 4 |}{\sqrt{(2 + 3t)^2 + (1 - 2t)^2}}\]This enables precise distance calculations, essential in applications such as determining if two objects will intersect.

Maximizing Distance

Maximizing the distance between a point and a line involves mathematical insights to adjust the function’s parameters optimally.
In our problem, maximizing the distance involves minimizing the denominator of the distance formula:\[\sqrt{13t^2 + 8t + 5}\]Choosing values for \(t\) that keep this expression small will maximize the separation distance towards the line. This step entails derivative calculations:

• Differentiate the denominator expression.
• Set the derivative to zero to find potential maximums or minimums.
• Solve for \(t\) to identify optimal parameter values.

The strategy typically yields specialized choices making it effective in practical optimizations.

Parameter in Line Equation

Parameters in line equations essentially determine the particular line from a family. By addressing each parameter value's impact on equation elements, critical insights into line behavior can be obtained.
In our family of lines \[(2+3t)x + (1-2t)y + 4 = 0\]every value of \(t\) lets the coefficients \(2 + 3t\) and \(1 - 2t\) to adjust incrementally, thus modifying line slope and position.
Choosing specific \(t\) optimizes the scenarios being evaluated, such as maximizing distances in our task. Through careful analysis and trial, you determine the exact parameter value, influencing geometry and outcomes directly.
Thus, parameters are more than just placeholders—they are dynamic factors in the line's geometry, alterable to meet desired criteria.