Problem 73 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation of the straight line passing through the point \((4,3)\) and making intercepts on the co-ordinate axes whose sum is \(-1\) is (A) \(\frac{x}{2}+\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\) (B) \(\frac{x}{2}-\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\) (C) \(\frac{x}{2}+\frac{y}{3}=1\) and \(\frac{x}{2}+\frac{y}{1}=1\) (D) \(\frac{x}{2}-\frac{y}{3}=1\) and \(\frac{x}{-2}+\frac{y}{1}=1\)

Step-by-Step Solution

Answer

The correct equations are found in options (A): \(\frac{x}{-2} + \frac{y}{1} = -1\) and (B): \(\frac{x}{-2} + \frac{y}{1} = -1\).

Step 1: Understanding the Equation of a Line with Intercepts

The equation of a line making intercepts on the x-axis and y-axis is of the form \(\frac{x}{a} + \frac{y}{b} = 1\), where \(a\) and \(b\) are the x-intercept and y-intercept, respectively. The sum of intercepts is given as \(a + b = -1\).

Step 2: Setting Up Equations from Intercepts Condition

Since \(a + b = -1\), we set up the equations using the intercept form: \(\frac{x}{a} + \frac{y}{(-1-a)} = 1\). Because the line passes through the point (4,3), substituting these values into the equation gives \(\frac{4}{a} + \frac{3}{-1-a} = 1\).

Step 3: Solving Substitution Equation

From \(\frac{4}{a} + \frac{3}{-1-a} = 1\), clear the fractions by multiplying through by \(a(-1-a)\): \(4(-1-a) + 3a = a(-1-a)\). Solve internally: \(-4 - 4a + 3a = a^2 + a\) simplifies to \(a^2 + 5a + 4 = 0\).

Step 4: Solving Quadratic for Intercepts

The quadratic \(a^2 + 5a + 4 = 0\) can be solved using the quadratic formula: \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting \(a = 1, b = 5, c = 4\), gives roots \(a = -1\) and \(a = -4\).

Step 5: Finding Corresponding Intercepts

If \(a = -1\), then \(b = 0\) (because \(a + b = -1\)); if \(a = -4\), then \(b = 3\). Thus the intercepts are (\(-1, 0\)) and (\(-4, 3\)).

Step 6: Writing Equations for Each Pair of Intercepts

For intercepts \((-1, 0)\), the line's equation is \(\frac{x}{-1} + \frac{y}{0} = 1\), simplifying to \(x + 0 = -1\) (which looks odd). For intercepts \((-4, 3)\), the line equation follows \(\frac{x}{-4} + \frac{y}{3} = 1\), simplifying becomes \(\frac{x}{-2} + \frac{y}{1} = -1\).

Step 7: Comparing with Provided Options

From option choices, the valid equations that match our results are \(\frac{x}{-2} + \frac{y}{1} = -1\), which corresponds to options (A): \(\frac{x}{-2} + \frac{y}{1} = -1\) and (B): \(\frac{x}{-2} + \frac{y}{1} = -1\).

Key Concepts

Coordinate Geometry

Coordinate geometry is all about connecting geometry and algebra. This branch focuses on utilizing coordinates to represent geometric figures like points, lines, and shapes within a plane. A key component in coordinate geometry is the Cartesian coordinate system. Here, every point in a plane is represented by coordinates, usually labeled \(x\) and \(y\), which correspond to their positions along the horizontal and vertical axes, respectively.

Imagine, you have a flat piece of paper. By plotting points using \(x\) and \(y\) values, we can visualize shapes and lines easily. For instance, in plotting straight lines, different forms of equations might be used. A common form is the slope-intercept form \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. However, in some problems, like the one in our exercise, intercept form is often more convenient. Knowing how to interpret and use these forms allows us to solve various geometric problems using algebraic methods.

Intercepts in Coordinate Geometry

Intercepts represent the points where a line crosses the coordinate axes. In coordinate geometry, understanding intercepts provides insight into the behavior of the line. The x-intercept is where the line crosses the x-axis, and here the \(y\) value is zero. Similarly, the y-intercept is where the line crosses the y-axis, and at this point, the \(x\) value is zero.

Mathematically, a line with known intercepts is expressed using the formula: \( \frac{x}{a} + \frac{y}{b} = 1 \), where \(a\) and \(b\) are the intercepts. When we know the sum of intercepts, such as in the problem statement \(a + b = -1\), this equation allows us to determine each intercept explicitly.

• For an x-intercept \(a\), substitute \(y = 0\) in the equation.
• For a y-intercept \(b\), substitute \(x = 0\).

Calculating intercepts provides a method to understand the position and orientation of a line, thus making it easier to write its equation and solve real-world problems effectively.

Quadratic Formula

When working with second-degree equations, also known as quadratic equations, the quadratic formula is indispensable. These equations generally take the form \(ax^2 + bx + c = 0\). The quadratic formula provides a straightforward approach to find the roots of such equations. It is expressed as:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In the context of solving for intercepts from a derived equation, this formula helps in determining potential values for \(a\) and \(b\). For example, if upon rearranging and simplifying the intercept-related equation, you end up with \(a^2 + 5a + 4 = 0\), applying the quadratic formula will yield solutions for \(a\). These roots can provide the intercepts needed for further analysis.

Understanding the quadratic formula and its use is fundamental in solving equations derived from geometrical contexts safely, as it simplifies what could otherwise be a complex algebraic task into manageable calculations. This remains crucial to solving a variety of geometry problems effectively.