Chapter 7

What is the graph of \(x=0\) ? What is the graph of \(y=0\) ? Explain your answers.

If the ratio of rise to run is to be \(\frac{3}{5}\) for some steps and the rise is 19 centimeters, find the run to the nearest centimeter.

Is a graph symmetric with respect to the origin if it is symmetric with respect to both axes? Defend your answer.

If the ratio of rise to run is to be \(\frac{2}{3}\) for some steps, and the run is 28 centimeters, find the rise to the nearest centimeter.

Is a graph symmetric with respect to both axes if it is symmetric with respect to the origin? Defend your answer.

Suppose that a county ordinance requires a \(2 \frac{1}{4} \%\) "fall" for a sewage pipe from the house to the main pipe at the street. How much vertical drop must there be for a horizontal distance of 45 feet? Express the answer to the nearest tenth of a foot.

Graph the equation \(y=\frac{1}{x}\) (Example 4) using the following boundaries. (a) \(-15 \leq x \leq 15\) and \(-10 \leq y \leq 10\) (b) \(-10 \leq x \leq 10\) and \(-10 \leq y \leq 10\) (c) \(-5 \leq x \leq 5\) and \(-5 \leq y \leq 5\)

How would you explain the concept of slope to someone who was absent from class the day it was discussed?

Graph the equation \(y=\frac{-2}{x^{2}}\) (Example 5), using the following boundaries. (a) \(-15 \leq x \leq 15\) and \(-10 \leq y \leq 10\) (b) \(-5 \leq x \leq 5\) and \(-10 \leq y \leq 10\) (c) \(-5 \leq x \leq 5\) and \(-10 \leq y \leq 1\)

If one line has a slope of \(\frac{2}{5}\), and another line has a slope of \(\frac{3}{7}\), which line is steeper? Explain your answer.

Graph the two equations \(y=\pm \sqrt{x}\) (Example 3 ) on the same set of axes, using the following boundaries. (Let \(Y_{1}=\sqrt{x}\) and \(Y_{2}=-\sqrt{x}\) ) (a) \(-15 \leq x \leq 15\) and \(-10 \leq y \leq 10\) (b) \(-1 \leq x \leq 15\) and \(-10 \leq y \leq 10\) (c) \(-1 \leq x \leq 15\) and \(-5 \leq y \leq 5\)

A company uses 7 pounds of fertilizer for a lawn that measures 5000 square feet and 12 pounds for a lawn that measures 10,000 square feet. Let \(y\) represent the pounds of fertilizer and \(x\) the square footage of the lawn.

Suppose that a line has a slope of \(\frac{2}{3}\) and contains the point \((4,7)\). Are the points \((7,9)\) and \((1,3)\) also on the line? Explain your answer.

Graph \(y=\frac{1}{x}, y=\frac{5}{x}, y=\frac{10}{x}\), and \(y=\frac{20}{x}\) on the same set of axes. (Choose your own boundaries.) What effect does increasing the constant seem to have on the graph?

Sometimes it is necessary to find the coordinate of a point on a number line that is located somewhere between two given points. For example, suppose that we want to find the coordinate \((x)\) of the point located twothirds of the distance from 2 to 8 . Because the total distance from 2 to 8 is \(8-2=6\) units, we can start at 2 and move \(\frac{2}{3}(6)=4\) units toward 8 . Thus \(x=2+\frac{2}{3}(6)=\) \(2+4=6\). For each of the following, find the coordinate of the indicated point on a number line. (a) Two-thirds of the distance from 1 to 10 (b) Three-fourths of the distance from \(-2\) to 14 (c) One-third of the distance from \(-3\) to 7 (d) Two-fifths of the distance from \(-5\) to 6 (e) Three-fifths of the distance from \(-1\) to \(-11\) (f) Five-sixths of the distance from 3 to \(-7\)

Graph \(y=\frac{10}{x}\) and \(y=\frac{-10}{x}\) on the same set of axes. What relationship exists between the two graphs?

Now suppose that we want to find the coordinates of point \(P\), which is located two-thirds of the distance from \(A(1,2)\) to \(B(7,5)\) in a coordinate plane. We have plotted the given points \(A\) and \(B\) in Figure \(7.42\) to help with the analysis of this problem. Point \(D\) is twothirds of the distance from \(A\) to \(C\) because parallel lines cut off proportional segments on every transversal that intersects the lines. Thus \(\overline{A C}\) can be treated as a segment of a number line, as shown in Figure 7.43. Therefore, $$ x=1+\frac{2}{3}(7-1)=1+\frac{2}{3}(6)=5 $$ Similarly, \(\overline{C B}\) can be treated as a segment of a number line, as shown in Figure 7.44. Therefore, For each of the following, find the coordinates of the indicated point in the \(x y\) plane. (a) One-third of the distance from \((2,3)\) to \((5,9)\) (b) Two-thirds of the distance from \((1,4)\) to \((7,13)\) (c) Two-fifths of the distance from \((-2,1)\) to \((8,11)\) (d) Three-fifths of the distance from \((2,-3)\) to \((-3,8)\) (e) Five-eighths of the distance from \((-1,-2)\) to \((4,-10)\) (f) Seven-eighths of the distance from \((-2,3)\) to \((-1,-9)\)

Graph \(y=\frac{10}{x^{2}}\) and \(y=\frac{-10}{x^{2}}\) on the same set of axes. What relationship exists between the two graphs?

An accountant has a schedule of depreciation for some business equipment. The schedule shows that after 12 months the equipment is worth \(\$ 7600\) and that after 20 months it is worth \(\$ 6000\). Let \(y\) represent the worth and \(x\) represent the time in months.

Suppose we want to find the coordinates of the midpoint of a line segment. Let \(P(x, y)\) represent the midpoint of the line segment from \(A\left(x_{1}, y_{1}\right)\) to \(B\left(x_{2}, y_{2}\right)\). Using the method in Problem 68, the formula for the \(x\) coordinate of the midpoint is \(x=x_{1}+\frac{1}{2}\left(x_{2}-x_{1}\right)\). This formula can be simplified algebraically to produce a simpler formula. $$ \begin{aligned} &x=x_{1}+\frac{1}{2}\left(x_{2}-x_{1}\right) \\ &x=x_{1}+\frac{1}{2} x_{2}-\frac{1}{2} x_{1} \\ &x=\frac{1}{2} x_{1}+\frac{1}{2} x_{2} \\ &x=\frac{x_{1}+x_{2}}{2} \end{aligned} $$ Hence the \(x\) coordinate of the midpoint can be interpreted as the average of the \(x\) coordinates of the endpoints of the line segment. A similar argument for the \(y\) coordinate of the midpoint gives the following formula. $$ y=\frac{y_{1}+y_{2}}{2} $$ For each of the pairs of points, use the formula to find the midpoint of the line segment between the points. (a) \((3,1)\) and \((7,5)\) (b) \((-2,8)\) and \((6,4)\) (c) \((-3,2)\) and \((5,8)\) (d) \((4,10)\) and \((9,25)\) (e) \((-4,-1)\) and \((-10,5)\) (f) \((5,8)\) and \((-1,7)\)

What does it mean to say that two points determine a line?

Remember that we did some work with parallel lines back in the graphing calculator activities in Problem Set 7.1. Now let's do some work with perpendicular lines. Be sure to set your boundaries so that the distance between tick marks is the same on both axes. (a) Graph \(y=4 x\) and \(y=-\frac{1}{4} x\) on the same set of axes. Do they appear to be perpendicular lines? (b) Graph \(y=3 x\) and \(y=\frac{1}{3} x\) on the same set of axes. Do they appear to be perpendicular lines? (c) Graph \(y=\frac{2}{5} x-1\) and \(y=-\frac{5}{2} x+2\) on the same set of axes. Do they appear to be perpendicular lines? (d) Graph \(y=\frac{3}{4} x-3, y=\frac{4}{3} x+2\), and \(y=-\frac{4}{3} x+2\) on the same set of axes. Does there appear to be a pair of perpendicular lines? (e) On the basis of your results in parts (a) through (d), make a statement about how we can recognize perpendicular lines from their equations.

How would you help a friend determine the equation of the line that is perpendicular to \(x-5 y=7\) and contains the point \((5,4)\) ?

For each of the following pairs of equations, (1) predict whether they represent parallel lines, perpendicular lines, or lines that intersect but are not perpendicular, and (2) graph each pair of lines to check your prediction. (a) \(5.2 x+3.3 y=9.4\) and \(5.2 x+3.3 y=12.6\) (b) \(1.3 x-4.7 y=3.4\) and \(1.3 x-4.7 y=11.6\) (c) \(2.7 x+3.9 y=1.4\) and \(2.7 x-3.9 y=8.2\) (d) \(5 x-7 y=17\) and \(7 x+5 y=19\) (e) \(9 x+2 y=14\) and \(2 x+9 y=17\) (f) \(2.1 x+3.4 y=11.7\) and \(3.4 x-2.1 y=17.3\)

\text { Explain how you would find the slope of the line } y=4 \text {. }

The equation of a line that contains the two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is \(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\). We often refer to this as the two-point form of the equation of a straight line. Use the two- point form and write the equation of the line that contains each of the indicated pairs of points. Express final equations in standard form. (a) \((1,1)\) and \((5,2)\) (b) \((2,4)\) and \((-2,-1)\) (c) \((-3,5)\) and \((3,1)\) (d) \((-5,1)\) and \((2,-7)\)

Let \(A x+B y=C\) and \(A^{\prime} x+B^{\prime} y=C^{\prime}\) represent two lines. Change both of these equations to slopeintercept form, and then verify each of the following properties. (a) If \(\frac{A}{A^{\prime}}=\frac{B}{B^{\prime}} \neq \frac{C}{C^{\prime}}\), then the lines are parallel. (b) If \(A A^{\prime}=-B B^{\prime}\), then the lines are perpendicular.

The properties in Problem 75 provide us with another way to write the equation of a line parallel or perpendicular to a given line that contains a given point not on the line. For example, suppose that we want the equation of the line perpendicular to \(3 x+4 y=6\) that contains the point \((1,2)\). The form \(4 x-3 y=k\), where \(k\) is a constant, represents a family of lines perpendicular to \(3 x+4 y=6\) because we have satisfied the condition \(A A^{\prime}=-B B^{\prime}\). Therefore, to find what specific line of the family contains \((1,2)\), we substitute 1 for \(x\) and 2 for \(y\) to determine \(k\). $$ \begin{array}{r} 4 x-3 y=k \\ 4(1)-3(2)=k \\ -2=k \end{array} $$ Thus the equation of the desired line is \(4 x-3 y=-2\). Use the properties from Problem 75 to help write the equation of each of the following lines. (a) Contains \((1,8)\) and is parallel to \(2 x+3 y=6\) (b) Contains \((-1,4)\) and is parallel to \(x-2 y=4\) (c) Contains \((2,-7)\) and is perpendicular to \(3 x-\) \(5 y=10\) (d) Contains \((-1,-4)\) and is perpendicular to \(2 x+\) \(5 y=12\)

The problem of finding the perpendicular bisector of a line segment presents itself often in the study of analytic geometry. As with any problem of writing the equation of a line, you must determine the slope of the line and a point that the line passes through. A perpendicular bisector passes through the midpoint of the line segment and has a slope that is the negative reciprocal of the slope of the line segment. The problem can be solved as follows: Find the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\). The midpoint of the line segment is \(\left(\frac{1+7}{2}, \frac{-2+8}{2}\right)\) \(=(4,3)\). \(=(4,3)\). The slope of the line segment is \(m=\frac{8-(-2)}{7-1}\) \(=\frac{10}{6}=\frac{5}{3}\) Hence the perpendicular bisector will pass through the point \((4,3)\) and have a slope of \(m=-\frac{3}{5}\). $$ \begin{aligned} y-3 &=-\frac{3}{5}(x-4) \\ 5(y-3) &=-3(x-4) \\ 5 y-15 &=-3 x+12 \\ 3 x+5 y &=27 \end{aligned} $$ Thus the equation of the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\) is \(3 x+5 y=27 .\) Find the perpendicular bisector of the line segment between the points for the following. Write the equation in standard form. (a) \((-1,2)\) and \((3,0)\) (b) \((6,-10)\) and \((-4,2)\) (c) \((-7,-3)\) and \((5,9)\) (d) \((0,4)\) and \((12,-4)\)

Predict whether each of the following pairs of equations represents parallel lines, perpendicular lines, or lines that intersect but are not perpendicular. Then graph each pair of lines to check your predictions. (The properties presented in Problem 75 should be very helpful.) (a) \(5.2 x+3.3 y=9.4\) and \(5.2 x+3.3 y=12.6\) (b) \(1.3 x-4.7 y=3.4\) and \(1.3 x-4.7 y=11.6\) (c) \(2.7 x+3.9 y=1.4\) and \(2.7 x-3.9 y=8.2\) (d) \(5 x-7 y=17\) and \(7 x+5 y=19\) (e) \(9 x+2 y=14\) and \(2 x+9 y=17\) (f) \(2.1 x+3.4 y=11.7\) and \(3.4 x-2.1 y=17.3\) (g) \(7.1 x-2.3 y=6.2\) and \(2.3 x+7.1 y=9.9\) (h) \(-3 x+9 y=12\) and \(9 x-3 y=14\) (i) \(2.6 x-5.3 y=3.4\) and \(5.2 x-10.6 y=19.2\) (j) \(4.8 x-5.6 y=3.4\) and \(6.1 x+7.6 y=12.3\)