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Chapter 4: Problem 57

A line with slope \(m(m<0)\) passes through the point \((a, b)\) in the first quadrant. Express the area of the triangle bounded by this line and the axes in terms of \(m\).

Short Answer

Expert verified
The area of the triangle is \( A = \frac{-ma^2 + 2mab - b^2}{2m} \).

Step by step solution

01

Equation of the Line

To find the equation of the line, use the point-slope form: \( y - b = m(x - a) \). Simplify this equation to the slope-intercept form: \( y = mx - ma + b \).
02

X-Intercept

The x-intercept occurs when \( y = 0 \). So, set \( 0 = mx - ma + b \) and solve for \( x \): \[ x = \frac{ma - b}{m} \].
03

Y-Intercept

The y-intercept is found by plugging \( x = 0 \) into the line equation: \( y = m(0) - ma + b = b - ma \).
04

Area of the Triangle

The triangle formed by the line and the axes has vertices at \((0,0)\), \((\frac{ma - b}{m}, 0)\), and \((0, b-ma)\). The base of the triangle on the x-axis is \( \frac{ma-b}{m} \) and the height is \( b-ma \). The area \( A \) is: \[ A = \frac{1}{2} \left(\frac{ma-b}{m}\right)(b-ma) \].
05

Simplification

Simplify the area formula:\[ A = \frac{1}{2m} (ma - b)(b - ma) \].Perform the multiplication inside the parentheses:\[ A = \frac{1}{2m} (mab - ma^2 - b^2 + mab) \].Combine like terms:\[ A = \frac{1}{2m} (-ma^2 + 2mab - b^2) \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation of a Line
In analytic geometry, the equation of a line is crucial as it describes a straight line on a coordinate plane. One common form is the point-slope form, which we can use when we know a point on the line and the slope. Given a point \((a, b)\) and a negative slope \(m\), the equation is:
  • Point-slope form: \(y - b = m(x - a)\)
  • Solving for \(y\) gives slope-intercept form: \(y = mx - ma + b\)
Breaking down, \(mx\) is the slope term, \(-ma + b\) is the intercept c. This form helps us understand how changes in \(x\) affect \(y\). Finding the equation aids in defining other essential points, like intercepts, useful for solving geometric problems like area of triangles.
Triangle Area
Finding the area of triangles using coordinate geometry is a practical skill. In this scenario, the line intersects x and y axes forming a triangle with these axes. The base and height are easily derived from intercepts:
  • Base: \(\frac{ma - b}{m}\) from x-intercept
  • Height: \(b - ma\) from y-intercept
Finally, the area \(A\) is calculated using:\[A = \frac{1}{2} \times \text{base} \times \text{height}\]Substituting the known values, \[A = \frac{1}{2m}(-ma^2 + 2mab - b^2)\] This representation is versatile and combines parameters like slope and specific coordinates, showcasing the connection between algebra and geometry.
X-Intercept and Y-Intercept
Intercepts are fundamental in understanding the behavior of linear equations.
The x-intercept of a line is where the line crosses the x-axis, which happens when \(y=0\). For our line:
  • x-intercept: \(x = \frac{ma-b}{m}\)
The y-intercept crosses the y-axis where \(x=0\):
  • y-intercept: \(y = b-ma\)
These points are critical as they mark the triangle's limits with the axes in coordinate geometry. Calculating these intercepts provides the triangle's base and height needed for area calculations.
Coordinate Geometry
Coordinate geometry, or analytic geometry, connects algebra and geometry by using a coordinate plane to describe geometric shapes with equations.
Understanding problems, like finding the triangle's area bounded by axes and a line, requires this blending of algebraic equations with geometric interpretations.
This approach helps simplify seeing relationships between points and forms: how a line's equation can tell about triangle dimensions.
  • It uses intercepts to define dimensions
  • Derives areas with algebraic forms
  • Illustrates real-world scenarios through geometric visualizations
By interpreting coordinates in this dual fashion, complex geometric information becomes manageable, showcasing coordinate geometry's power to demystify spatial algorithms.

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Most popular questions from this chapter

This exercise shows that if we have a table generated by a linear function and the \(x\) -values are equally spaced, then the first differences of the \(y\) -values are constant. (a) In the following data table, the three \(x\) -entries are equally spaced. Compute the three entries in the \(f(x)\) row assuming that \(f\) is the linear function given by \(f(x)=m x+b\) \(f(x)=m x+b .\) (Don't worry about the fact that your answers contain all four of the letters \(m, b, a, \text { and } h .)\) \begin{tabular}{llll} \hline\(x\) & \(a\) & \(a+h\) & \(a+2 h\) \\ \(f(x)\) & & & \\ \hline \end{tabular} (b) Compute the first differences for the three quantities that you listed in the \(f(x)\) row in part (a). (The two first differences that you obtain should turn out to be equal, as required.)

Find quadratic functions satisfying the given conditions. The vertex is \((3,-1),\) and one \(x\) -intercept is 1.

(a) Determine the \(x\) - and \(y\) -intercepts and the excluded regions for the graph of the given function. Specify your results using a sketch similar to Figure \(16(a) .\) In Exercises \(31-34\) you will first need to factor the polynomial. (b) Graph each function. $$y=x^{3}+3 x^{2}-4 x-12$$

A power station is on one side of a river that is \(1 / 2\) mi. wide, and a factory is \(6 \mathrm{mi}\). downstream on the other side. It costs \(\$ 6\) per foot to run power lines overland and \(\$ 8\) per foot to run them underwater. Find the most economical path for the transmission line from the power station to the factory.

First use the graph to estimate the \(x\) -intercepts. Then use algebra to determine each \(x\) -intercept. If an intercept involves a radical, give that answer as well as a calculator approximation rounded to three decimal places. Be sure to check that your results are consistent with the initial graphical estimates. $$y=x^{3}-3 x^{2}-5 x$$ (GRAPH CAN'T COPY)
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