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

Question

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 equation is (C) 14x - 21y = 0.

Step 1: Identify Distance Formula

The distance of a point \((x_1, y_1)\) from the line \(Ax + By + C = 0\) is given by: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \]

Step 2: Substitute Line Equation Coefficients

Substitute the coefficients from the given family of lines \((2 + 3t)x + (1 - 2t)y + 4 = 0\) into the distance formula, treating the coefficients as functions of \(t\), \(A = 2+3t\), \(B = 1-2t\), and \(C = 4\).

Step 3: Calculate Distance From Point to Line

Substitute the point \((2,3)\) into the distance formula: \[ d = \frac{|(2 + 3t)2 + (1 - 2t)3 + 4|}{\sqrt{(2 + 3t)^2 + (1 - 2t)^2}}. \] Simplifying the numerator, it becomes: \( 4 + 6t + 3 - 6t + 4 = 11 \). So, \[ d = \frac{|11|}{\sqrt{(2 + 3t)^2 + (1 - 2t)^2}}. \]

Step 4: Maximize the Denominator

To maximize the distance, minimize the denominator \(\sqrt{(2 + 3t)^2 + (1 - 2t)^2}\). This simplifies to: \[ (2 + 3t)^2 + (1 - 2t)^2 = 4 + 12t + 9t^2 + 1 - 4t + 4t^2 = 13t^2 + 8t + 5 \]. To find the extrema, calculate the derivative \(\frac{d}{dt}(13t^2 + 8t + 5) = 26t + 8 \), set it to zero \(26t + 8 = 0\), solve for \(t\), \(t = -\frac{8}{26} = -\frac{4}{13}\).

Step 5: Substitute to Find Specific Line

Substitute \(t = -\frac{4}{13}\) back into the line equation: \((2 + 3(-\frac{4}{13}))x + (1 - 2(-\frac{4}{13}))y + 4 = 0\). Simplify to get the line as \(\frac{14}{13}x + \frac{21}{13}y = 0\) which can be written as \(14x + 21y = 0\).

Step 6: Match with Options

Compare the resulting line equation to the provided options. The matching one is \((C) 14x - 21y = 0\).

Key Concepts

Distance Formula

To solve the given problem involving a family of lines, it's essential to understand how to calculate the distance from a point to a line. The distance formula for a point \(x_1, y_1\) from a line with the equation \(Ax + By + C = 0\) is given by: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \]

• This formula lets us find the shortest length between the point and the line.
• The numerator represents the value of the line equation at the given point, which is then treated as an absolute value to ensure distance is non-negative.
• The denominator accounts for the magnitude of the direction vector \[A, B\].

For the problem at hand, using this formula enables us to compute the distance from the point \(2,3\) to any line in our family.

Parameterization

The family of lines we are dealing with is described through parameterization. This is achieved by using a parameter \(t\) to represent the different lines in the family equation: \[(2+3t)x + (1-2t)y + 4 = 0.\]

• Parameterization allows us to express each line in this family by substituting different values for \(t\).
• As \(t\) varies, the line coefficients change, portraying an infinite set of lines.
• This concept is crucial because it provides control over the line's orientation and position through a single variable.

By specifically focusing on how \(t\) alters the line equation, we can analyze how the family intersects or approaches features like the maximum or minimum distance to a point.

Maximizing Distance

To solve the problem, maximizing the distance from a point to a member of the family of lines requires focusing on minimizing a particular expression derived from the distance formula. We have to consider the denominator in the distance formula: \[ \sqrt{(2 + 3t)^2 + (1 - 2t)^2}. \]

• To maximize the whole expression, the denominator needs to be minimized since distance is inversely proportional to it.
• Minimizing this expression gives us the maximum possible value for the distance.
• The squared terms here are particularly useful because they contribute monotonically, making differentiation straightforward.

Therefore, we proceed to use calculus on the expression \(13t^2 + 8t + 5\) to find its minimum.

Differentiation

Differentiation helps us identify the minimum or maximum values of a function. In this context, we determine the minimum of \(13t^2 + 8t + 5\) to maximize the distance. The derivative of this function, \[ \frac{d}{dt}(13t^2 + 8t + 5) = 26t + 8, \] gives the rate of change of the function. To find critical points that would yield minimums, we set it to zero: \[ 26t + 8 = 0. \]

• Solving gives \(t = -\frac{4}{13}\), which indicates where the expression attains its minimum.
• This critical \(t\) clarifies which specific line in our parameterized family is the furthest from the point.

Differentiation makes it possible to use calculus for real-world problems requiring maximal distances.

Line Equation

Once the optimal parameter \(t = -\frac{4}{13}\) is identified, finding the specific line equation is straightforward. By substituting this value back into the parameterized line equation, we form a concrete line: \[(2 + 3\left(-\frac{4}{13}\right))x + (1 - 2\left(-\frac{4}{13}\right))y + 4 = 0.\]

• Simplifying further yields \((\frac{14}{13})x + (\frac{21}{13})y = 0\) or equivalently \(14x + 21y = 0\), which is among the options.
• This result demonstrates how algebra and analysis jointly solve the geometric query of maximum distance from a family line to a point.

Once reduced to a standard form, this line can be easily matched against potential answers to confirm the solution, in this case, choice \(C\).