Problem 51
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
Verified Answer
The correct option is (b).
1Step 1: Identify the Equation Form
To solve the problem, we need to use the intercept form of a line, which is \( \frac{x}{a} + \frac{y}{b} = 1 \), where \(a\) and \(b\) are the x and y intercepts respectively. The exercise states that their sum is \(-1\), so \(a + b = -1\).
2Step 2: Express Intercepts in Terms of a Variable
We can express one intercept in terms of the other using the equation \(a + b = -1\). Let's assume \(a = t\), then \(b = -1 - t\). Our line equation in intercept form becomes \( \frac{x}{t} + \frac{y}{-1-t} = 1 \).
3Step 3: Plug the Point into the Equation
Using the point \((4, 3)\), plug in \(x = 4\) and \(y = 3\) into the line equation: \[ \frac{4}{t} + \frac{3}{-1-t} = 1 \].
4Step 4: Solve for t
To solve the equation \( \frac{4}{t} + \frac{3}{-1-t} = 1 \), clear the fractions by multiplying through by \(t(-1-t)\): \[ 4(-1-t) + 3t = t(-1-t) \].Simplifying gives: \[-4 - 4t + 3t = -t - t^2 \].Which simplifies to:\[ -4 - t = -t - t^2 \].
5Step 5: Rearrange the Equation
Cancel \(t\) from both sides, simplify to find:\[ -4 = -t^2 \].This implies that \(t^2 = 4\), so \(t = \pm2\).
6Step 6: Find the Solution Equations
Plug \(t = 2\) back into intercept form: \(a = 2\), \(b = -3\), equation is \( \frac{x}{2} + \frac{y}{-3} = 1 \). For \(t = -2\), \(a = -2\), \(b = 1\), equation is \( \frac{x}{-2} + \frac{y}{1} = 1 \). The correct option (b) is \( \frac{x}{2} - \frac{y}{3} = -1 \) and \( \frac{x}{-2} + \frac{y}{1} = -1\), equivalent to our found equations.
Key Concepts
Intercept Form of a LineEquation of a LineProblem Solving in Geometry
Intercept Form of a Line
In coordinate geometry, the intercept form of a line is a useful representation for lines cutting through the x- and y-axes. This form is expressed as \( \frac{x}{a} + \frac{y}{b} = 1 \), where \(a\) is the x-intercept and \(b\) is the y-intercept of the line.
The x-intercept \(a\) is the point where the line crosses the x-axis, and similarly, the y-intercept \(b\) is where the line crosses the y-axis. These intercepts define the location of the line in a coordinate plane.
Understanding these intercepts helps us visualize the slope and direction of the line. The intercept form is particularly useful when given information about intercepts or when required to find them. This form ties directly to the problem-solving strategies in the intercept-based equations, often transforming complex geometric problems into straightforward linear equations.
When dealing with intercepts, always ensure the sum of the intercepts equates correctly if given, as in cases where constraints are defined, like the sum of intercepts equaling a specific value.
The x-intercept \(a\) is the point where the line crosses the x-axis, and similarly, the y-intercept \(b\) is where the line crosses the y-axis. These intercepts define the location of the line in a coordinate plane.
Understanding these intercepts helps us visualize the slope and direction of the line. The intercept form is particularly useful when given information about intercepts or when required to find them. This form ties directly to the problem-solving strategies in the intercept-based equations, often transforming complex geometric problems into straightforward linear equations.
When dealing with intercepts, always ensure the sum of the intercepts equates correctly if given, as in cases where constraints are defined, like the sum of intercepts equaling a specific value.
Equation of a Line
The equation of a line is core to understanding its behavior and properties in geometry. Various forms exist, each serving specific purposes and applicable in different scenarios. Besides intercept form, common ones include slope-intercept form \(y = mx + c\) and point-slope form \(y - y_1 = m(x - x_1)\).
Each form brings out different aspects of the line:
Developing line equations requires identifying the most suitable form based on known data, leading to the efficient solving of geometric problems.
Each form brings out different aspects of the line:
- Slope-intercept form: Directly shows the slope \(m\) and the y-intercept \(c\).
- Point-slope form: Utilizes a known point \((x_1, y_1)\) and slope \(m\).
Developing line equations requires identifying the most suitable form based on known data, leading to the efficient solving of geometric problems.
Problem Solving in Geometry
Problem solving in geometry often involves employing different strategies and forms of equations to find solutions. The intercept form of a line serves as a crucial tool under specific conditions where intercepts are known or calculable.
In this exercise, the problem was to find the line intercepts, given a point and the sum of intercepts. By defining the intercepts as variables and solving through substitution and algebraic manipulation, the solution was obtained.
When solving such problems:
In this exercise, the problem was to find the line intercepts, given a point and the sum of intercepts. By defining the intercepts as variables and solving through substitution and algebraic manipulation, the solution was obtained.
When solving such problems:
- Identify the given information and the form it fits best.
- Use substitution methods to link knowns and unknowns.
- Manipulate equations to isolate variables for solutions.
Other exercises in this chapter
Problem 49
A straight line through the point \(\mathrm{A}(3,4)\) is such that its intercept between the axes is bisected at \(\mathrm{A}\). Its equation is (a) \(x+y=7\) (
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Let \(A(2,-3)\) and \(B(-2,3)\) be vertices of a triangle \(A B C\). If the centroid of this triangle moves on the line \(2 x+3 y=1\), then the locus of the ver
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Locus of mid point of the portion between the axes of \(x \cos \alpha+y \sin \alpha=p\) whre \(p\) is constant is [2002] (a) \(x^{2}+y^{2}=\frac{4}{p^{2}}\) (b)
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