QuanticaAlgebra IOpen in the courseWhere a line is and how steep it is are two different questions, and 7.2 answered only the first. Steepness is a comparison of two changes, how much \(y\) changes from one point of a line to another and how much \(x\) changes between the same two points. Both are subtractions of coordinates, so any two points are enough to make the comparison.
Problem
A line passes through \((2,5)\) and \((6,17)\). Going from the first point to the second, find how much \(y\) changes and how much \(x\) changes, then enter the value of the change in \(y\) divided by the change in \(x\).
Show a hint
- Each change is a subtraction, the second point's coordinate minus the first point's. For \(y\) that is \(17-5\).
- The change in \(y\) is \(12\) and the change in \(x\) is \(6-2\). Divide the first by the second.
Show the full solution
The change in \(y\) is \(17-5=12\) and the change in \(x\) is \(6-2=4\), so the quotient is \(\frac{12}{4}=\boxed{3}\). A quotient of \(3\) means \(y\) changes by \(3\) for every \(1\) that \(x\) changes.
Problem
Three more points are claimed to lie on that same line, \((-1,-4)\), \((5,14)\) and \((9,25)\). From \((2,5)\) to each of them, compute the change in \(y\) divided by the change in \(x\). Two of the three quotients come out \(3\), the value from the previous problem, and one does not. Enter the \(x\)-coordinate of the point whose quotient is not \(3\).
Show a hint
- Run the same two subtractions three times, always starting from \((2,5)\), and compare each quotient with \(3\).
- From \((2,5)\) to \((-1,-4)\) both changes are negative, \(-9\) and \(-3\), and a negative divided by a negative is positive. Check the other two the same way.
Show the full solution
From \((2,5)\) the quotient to \((-1,-4)\) is \(\frac{-9}{-3}=3\), the quotient to \((5,14)\) is \(\frac{9}{3}=3\), and the quotient to \((9,25)\) is \(\frac{20}{7}\). Only the third differs from \(3\), so the \(x\)-coordinate asked for is \(\boxed{9}\). Two different second points on the line both give \(3\), so the quotient does not depend on which pair is used. The one point that gives something else is off the line.
Problem
A line passes through \((-4,9)\) and \((2,-9)\). Enter its slope.
Show a hint
- Take \((-4,9)\) first in both subtractions. The change in \(y\) is then \(-9-9\).
- The change in \(y\) is \(-18\) and the change in \(x\) is \(2-(-4)=6\).
Show the full solution
Taking \((-4,9)\) first, \(m=\frac{-9-9}{2-(-4)}=\frac{-18}{6}=\boxed{-3}\). Between these points \(y\) decreases while \(x\) increases, so the top of the fraction is negative, the bottom is positive, and the slope is negative.
Problem
A line passes through \((3,-4)\) and \((9,8)\). A student writes \(\frac{-4-8}{9-3}\) and reports \(-2\), subtracting the \(y\)-values in one order and the \(x\)-values in the other order. Enter the correct slope.
Show a hint
- Both differences have to be taken in the same order, so pick one point to come first and use that point first on the top and on the bottom.
- With \((9,8)\) first in both, the top is \(8-(-4)\) and the bottom is \(9-3\).
Show the full solution
With \((9,8)\) first in both differences, \(m=\frac{8-(-4)}{9-3}=\frac{12}{6}=\boxed{2}\). The student's \(-2\) has the right size and the wrong sign, because only the top was reversed, and a reversed top is the negative of the correct top.
Writing the left-hand point first makes the change in \(x\) positive, since the other point has the larger \(x\). The sign of the quotient then depends only on the change in \(y\), positive when the right-hand point is higher and negative when it is lower. So the sign of a slope follows from where two points are, with no coordinates needed.
Problem
Points \(P\) and \(Q\) lie on the same line, with \(Q\) to the right of \(P\) and lower than \(P\). No coordinates are given, so no number can be computed for the slope of that line, but its sign can be named. Enter the word positive or negative.
Show a hint
- Take \(P\) first in both subtractions. \(Q\) is to the right of \(P\), so the change in \(x\) is positive.
- \(Q\) is lower than \(P\), so the change in \(y\) is a negative number while the change in \(x\) stays positive. Think about the sign of a fraction whose top and bottom have opposite signs.
Show the full solution
Taking \(P\) first, the change in \(x\) is positive because \(Q\) is to the right, and the change in \(y\) is negative because \(Q\) is lower, so the slope is \(\boxed{\text{negative}}\). Taking \(Q\) first flips the sign of both changes, so the quotient is negative either way.
Problem
A line passes through \((-6,-5)\) and \((10,7)\). Enter its slope as a fraction in lowest terms.
Show a hint
- Take \((-6,-5)\) as the first point and subtract in that same order on the top and the bottom. Both differences involve subtracting a negative number.
- The top and the bottom share a common factor, so the fraction you get first is not in lowest terms. Divide both by that factor.
Show the full solution
Taking \((-6,-5)\) first, \(m=\frac{7-(-5)}{10-(-6)}=\frac{12}{16}=\boxed{\frac{3}{4}}\). Both differences are positive because the second point is above and to the right of the first. Stopping at \(\frac{12}{16}\) is the common miss, since the answer was asked for in lowest terms.
Two kinds of line from 7.2 have not been put through the definition, the horizontal and the vertical. Every point of a horizontal line has the same \(y\) and every point of a vertical line has the same \(x\), so one of the two changes is \(0\) in each case. A \(0\) on top and a \(0\) underneath are not the same, as the next two problems show.
Problem
From 7.2 the graph of \(y=6\) is a horizontal line, and \((-8,6)\) and \((15,6)\) are two of its points. Enter its slope, or the word undefined if it has none.
Show a hint
- The two points have the same second coordinate, so start with the change in \(y\).
- The change in \(y\) is \(6-6\) and the change in \(x\) is \(15-(-8)\). Work both out, then divide the change in \(y\) by the change in \(x\).
Show the full solution
The change in \(y\) is \(6-6=0\) and the change in \(x\) is \(15-(-8)=23\), so \(m=\frac{0}{23}=\boxed{0}\). Zero divided by a nonzero number is \(0\), so every horizontal line has slope \(0\), however far apart its two points are.
Problem
From 7.2 the graph of \(x=-3\) is a vertical line, and \((-3,4)\) and \((-3,-9)\) are two of its points. Enter its slope, or the word undefined if it has none.
Show a hint
- The two points have the same first coordinate, so look at the bottom of the fraction before the top.
- The change in \(y\) is \(-9-4=-13\) and the change in \(x\) is \(-3-(-3)=0\). Ask what number multiplied by \(0\) gives \(-13\).
Show the full solution
The change in \(y\) is \(-9-4=-13\) and the change in \(x\) is \(-3-(-3)=0\), so the slope would be \(\frac{-13}{0}\), and division by zero has no value. The slope is \(\boxed{\text{undefined}}\). Answering \(0\) is the standard slip. A slope of \(0\) is a number and belongs to a horizontal line, while a vertical line has no slope at all.
A slope is one change divided by another, which is what a constant rate was in chapter 6. When \(x\) is a time and \(y\) is a measured amount, the slope is how much \(y\) changes per one unit of \(x\), and its units are the units of \(y\) over the units of \(x\). The next two problems ask for a rate from two readings, then the steeper of two lines.
Problem
A tank fills at a steady rate. At \(6\) minutes it holds \(23\) liters and at \(20\) minutes it holds \(58\) liters. Liters plotted against minutes give a straight line. Enter the slope of that line, in liters per minute.
Show a hint
- The change in \(y\) is a number of liters and the change in \(x\) is a number of minutes.
- The tank gains \(58-23\) liters over \(20-6\) minutes. Reduce the quotient.
Show the full solution
The change is \(58-23=35\) liters over \(20-6=14\) minutes, so \(m=\frac{35}{14}=\boxed{\frac{5}{2}}\). The units of a slope are the units of \(y\) over the units of \(x\), so this line's slope is \(2.5\) liters per minute.
Problem
Line \(g\) passes through \((1,2)\) and \((5,20)\). Line \(h\) passes through \((-3,11)\) and \((2,-14)\). One of the two lines is steeper than the other. Enter the slope of the steeper line.
Show a hint
- Compute both slopes first. Which line is steeper depends on the size of each slope and not on its sign.
- Line \(g\) has slope \(\frac{9}{2}\), which is \(4.5\). Compare that with the size of the slope of line \(h\).
Show the full solution
Line \(g\) has slope \(\frac{20-2}{5-1}=\frac{18}{4}=\frac{9}{2}\) and line \(h\) has slope \(\frac{-14-11}{2-(-3)}=\frac{-25}{5}=-5\). Steepness is measured by \(|m|\), and \(|-5|=5\) is greater than \(\left|\frac{9}{2}\right|=4.5\), so the steeper line has slope \(\boxed{-5}\). The steeper line here is the one with the smaller slope, since a negative slope can still be the larger in size.
The last two problems use the definition backwards. In each one the slope is fixed by other information and a missing coordinate is what the question asks for. Writing the slope condition down leaves one equation in one letter, which is chapter 4's work.
Problem
The points \((-2,3)\), \((4,k)\) and \((10,23)\) are collinear, so one straight line passes through all three. Only the middle point has an unknown coordinate. Enter \(k\).
Show a hint
- Two of the three points are fully known, so the slope of the line comes from those two alone.
- That slope is \(\frac{20}{12}=\frac{5}{3}\), and \((4,k)\) is \(6\) to the right of \((-2,3)\).
Show the full solution
The outer points give \(m=\frac{23-3}{10-(-2)}=\frac{20}{12}=\frac{5}{3}\). From \((-2,3)\) to \((4,k)\) the change in \(x\) is \(6\), so the change in \(y\) is \(6\cdot\frac{5}{3}=10\) and \(k=3+10=\boxed{13}\). Checking the other pair, \((4,13)\) to \((10,23)\) gives \(\frac{10}{6}=\frac{5}{3}\), the same slope.
Problem
Line \(u\) passes through \((-6,4)\) and \((2,-8)\). Line \(v\) passes through \((3,-7)\) and \((n,5)\). Line \(v\) is exactly as steep as line \(u\), and \(v\) rises from left to right while \(u\) falls. Enter \(n\).
Show a hint
- Find \(u\)'s slope first. Equal steepness means the two slopes have the same size, and a line that rises has a positive slope.
- Line \(u\) has slope \(-\frac{3}{2}\), so \(v\) has slope \(\frac{3}{2}\), and \(v\)'s change in \(y\) is \(5-(-7)\).
Show the full solution
Line \(u\) has slope \(\frac{-8-4}{2-(-6)}=\frac{-12}{8}=-\frac{3}{2}\), so both slopes have size \(\frac{3}{2}\), and \(v\) rises, so its slope is \(+\frac{3}{2}\). Then \(\frac{5-(-7)}{n-3}=\frac{12}{n-3}=\frac{3}{2}\) gives \(3(n-3)=24\), so \(n-3=8\) and \(n=\boxed{11}\). Reusing \(-\frac{3}{2}\) for \(v\) gives \(n=-5\), a line falling just like \(u\).
Every slope here came from two points of a line already drawn. A slope can also be known first. Exactly one line passes through a given point with a given slope. Lesson 7.4, Finding the Equation of a Line, turns a point and a slope into that line's equation.
Practice these ideas
Practice
A line passes through \((-1,-2)\) and \((3,18)\). Enter its slope.
Show the solution
\(m=\frac{18-(-2)}{3-(-1)}=\frac{20}{4}=\boxed{5}\). Subtracting a negative is the same as adding, so the numerator is \(20\) and not \(16\).
Practice
A line passes through \((-5,13)\) and \((1,-11)\). Enter its slope.
Show the solution
\(m=\frac{-11-13}{1-(-5)}=\frac{-24}{6}=\boxed{-4}\). A negative slope means the line falls from left to right.
Practice
A line passes through \((-5,-4)\) and \((7,4)\). Enter its slope as a fraction in lowest terms.
Show the solution
\(m=\frac{4-(-4)}{7-(-5)}=\frac{8}{12}=\boxed{\frac{2}{3}}\). A slope is reported in lowest terms, so \(\frac{8}{12}\) is not the final form.
Practice
A line passes through \((-2,5)\) and \((6,-11)\). A student computes \(\frac{5-(-11)}{6-(-2)}\) and reports \(2\). Enter the correct slope.
Show the solution
Putting the second point first in both, \(m=\frac{-11-5}{6-(-2)}=\frac{-16}{8}=\boxed{-2}\). Subtracting in opposite orders on the top and the bottom flips the sign, so the student's \(2\) has the right size and the wrong sign.
Practice
\(U\) and \(V\) are two points on a line, and \(U\) is to the right of \(V\) and higher than \(V\). Enter the word positive or negative for the sign of the slope of that line.
Show the solution
Taking \(V\) first, both changes are positive, since \(U\) is to the right of \(V\) and above it, so the slope is \(\boxed{\text{positive}}\). With \(U\) first both changes are negative and the quotient is the same positive number.
Practice
Four lines are given by two points each, \((5,-2)\) and \((5,6)\), then \((-7,3)\) and \((9,3)\), then \((2,-1)\) and \((6,5)\), then \((-8,1)\) and \((-8,-6)\). Enter how many of the four have slope \(0\).
Show the solution
Only \((-7,3)\) and \((9,3)\) share a \(y\)-coordinate, giving \(\frac{0}{16}=0\), so the count is \(\boxed{1}\). The first and fourth pairs share an \(x\)-coordinate, so those lines are vertical and have no slope rather than slope \(0\), and the third pair gives \(\frac{6}{4}=\frac{3}{2}\).
Practice
The line through \((-6,w)\) and \((11,-8)\) has slope \(0\). Enter \(w\).
Show the solution
The slope is \(\frac{-8-w}{11-(-6)}=\frac{-8-w}{17}\), and a fraction equals \(0\) only when its numerator is \(0\), so \(-8-w=0\) and \(w=\boxed{-8}\). The line is horizontal, and every point on a horizontal line has the same \(y\)-coordinate.
Practice
The line through \((c,4)\) and \((-7,19)\) has no slope. Enter \(c\).
Show the solution
The slope is undefined only when the two \(x\)-coordinates are equal, so \(c=\boxed{-7}\). Both points then lie on the vertical line \(x=-7\). The change in \(x\) is \(0\), and division by \(0\) is undefined.
Practice
A candle burns down at a steady rate. After \(9\) minutes it is \(23\) cm tall and after \(24\) minutes it is \(17\) cm tall. Height plotted against time gives a straight line. Enter its slope, in centimeters per minute, as a fraction in lowest terms.
Show the solution
\(m=\frac{17-23}{24-9}=\frac{-6}{15}=\boxed{-\frac{2}{5}}\). The candle gets \(0.4\) cm shorter each minute, and the slope is negative because the height decreases as time passes.
Practice
A shipping firm charges by weight. An \(8\) kg parcel costs \(37\) dollars and a \(20\) kg parcel costs \(64\) dollars. Cost plotted against weight gives a straight line. Enter the slope of that line, in dollars per kilogram, as a fraction in lowest terms.
Show the solution
\(m=\frac{64-37}{20-8}=\frac{27}{12}=\boxed{\frac{9}{4}}\). Each extra kilogram adds \(2.25\) dollars.
Practice
Four lines have slopes \(-\frac{5}{2}\), \(\frac{12}{5}\), \(-\frac{7}{4}\) and \(\frac{8}{3}\). Enter the slope of the least steep one.
Show the solution
The sizes are \(\frac{5}{2}=2.5\), \(\frac{12}{5}=2.4\), \(\frac{7}{4}=1.75\) and \(\frac{8}{3}\approx 2.67\), and the smallest is \(\frac{7}{4}\), so the least steep line has slope \(\boxed{-\frac{7}{4}}\). The minus sign stays, since the question asks for the slope and not for its size.
Practice
Line \(g\) passes through \((0,0)\) and \((7,-24)\). Line \(h\) passes through \((1,-4)\) and \((6,13)\). Enter the slope of the steeper line.
Show the solution
Line \(g\) has slope \(\frac{-24-0}{7-0}=-\frac{24}{7}\) and line \(h\) has slope \(\frac{13-(-4)}{6-1}=\frac{17}{5}\). Steepness is measured by \(|m|\), and \(\frac{24}{7}=\frac{120}{35}\) is greater than \(\frac{17}{5}=\frac{119}{35}\), so the steeper line is \(g\), with slope \(\boxed{-\frac{24}{7}}\). Both sizes round to \(3.4\) at one decimal place, so a rounded check is not enough here.
Practice
A line has slope \(7\) and passes through \((2,-5)\). Another of its points has \(x\)-coordinate \(8\). Enter that point's \(y\)-coordinate.
Show the solution
The change in \(x\) is \(8-2=6\), and \(\frac{y-(-5)}{6}=7\) gives \(y+5=42\), so \(y=\boxed{37}\). Checking, \((2,-5)\) to \((8,37)\) gives \(\frac{42}{6}=7\).
Practice
A line passes through \((4,-11)\) and \((4,6)\). Enter its slope, or the word undefined if it has none.
Show the solution
Both points have \(x\)-coordinate \(4\), so the change in \(x\) is \(4-4=0\) and the slope would be \(\frac{6-(-11)}{0}\), a division by zero, so the slope is \(\boxed{\text{undefined}}\). A vertical line has no slope at all, which is not the same as having slope \(0\), and that is the horizontal case.
Practice
The points \((-7,9)\), \((1,-7)\) and \((r,-16)\) lie on one straight line. Enter \(r\).
Show the solution
The first two points give \(m=\frac{-7-9}{1-(-7)}=\frac{-16}{8}=-2\). From \((1,-7)\) to \((r,-16)\) the change in \(y\) is \(-9\), so \(\frac{-9}{r-1}=-2\) gives \(r-1=\frac{9}{2}\) and \(r=\boxed{\frac{11}{2}}\). The unknown here is an \(x\)-coordinate, so it is in the denominator of the slope, which makes the last step a division rather than a multiplication.
Practice
The line through \((-4,3)\) and \((8,12)\) is exactly as steep as the line through \((2,6)\) and \((n,-3)\), and the second line falls from left to right. Enter \(n\).
Show the solution
The first line has slope \(\frac{12-3}{8-(-4)}=\frac{9}{12}=\frac{3}{4}\), so the second slope has size \(\frac{3}{4}\), and since that line falls the second slope is \(-\frac{3}{4}\). Then \(\frac{-3-6}{n-2}=\frac{-9}{n-2}=-\frac{3}{4}\) gives \(3(n-2)=36\), so \(n=\boxed{14}\). Taking the second slope as \(+\frac{3}{4}\) instead gives \(n=-10\), and that line rises from left to right.
Practice
The points \((-9,2)\), \((-9,7)\) and \((c,15)\) lie on one straight line. Enter \(c\).
Show the solution
The first two points both have \(x\)-coordinate \(-9\), so the line through them is the vertical line \(x=-9\), and every point of that line has first coordinate \(-9\), giving \(c=\boxed{-9}\). The change in \(x\) from \((-9,2)\) to \((-9,7)\) is \(0\), so that slope is undefined and the equal-slopes check does not apply.