QuanticaAlgebra IOpen in the courseA slope is a number computed from a line that is already drawn. It can also be known before any line is drawn. One point and a slope, or two points, are enough to write an equation whose graph is that line, and once that equation is written, arithmetic gives every other point of the line.
Problem
A line has slope \(3\) and passes through \((2,5)\). The point \((8,y)\) is on that same line, and the slope between any two points of a line is the same. Enter the value of \(y\).
Show a hint
- The slope between \((2,5)\) and \((8,y)\) is \(\frac{y-5}{8-2}\), and that has to come out to \(3\).
- The denominator is \(6\), so the equation is \(\frac{y-5}{6}=3\). Multiply both sides by \(6\).
Show the full solution
The slope between the two points is \(\frac{y-5}{8-2}=\frac{y-5}{6}\), and that equals \(3\), so \(y-5=18\) and \(y=\boxed{23}\). The slope formula from 7.3 is used the other way round here, with the slope known and one coordinate unknown.
Problem
A line has slope \(4\) and passes through \((5,3)\), and \((x,y)\) is any other point on it. The slope between those two points is \(4\), so \(\frac{y-3}{x-5}=4\). Multiply both sides by \(x-5\), expand, and solve for \(y\). The result reads \(y=4x+c\) for one number \(c\). Enter \(c\).
Show a hint
- Multiplying both sides by \(x-5\) leaves \(y-3=4(x-5)\), and the \(4\) multiplies both terms inside the parentheses.
- Expanding gives \(y-3=4x-20\). Now add \(3\) to both sides.
Show the full solution
Multiplying by \(x-5\) gives \(y-3=4(x-5)=4x-20\), and adding \(3\) to both sides gives \(y=4x-17\), so \(c=\boxed{-17}\). Check at the given point, \(4(5)-17=3\). The constant is not the \(3\) from the point, because the \(-20\) from expanding is added to it.
Problem
A line has slope \(-4\) and passes through \((-3,8)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(5\).
Show a hint
- Write point-slope with \((x_1,y_1)=(-3,8)\). Because \(x_1\) is negative, the factor \(x-x_1\) is \(x+3\) and not \(x-3\).
- The line is \(y-8=-4(x+3)\). At \(x=5\) the bracket is \(8\), so multiply that by \(-4\) and then solve for \(y\).
Show the full solution
Point-slope with \((x_1,y_1)=(-3,8)\) and \(m=-4\) gives \(y-8=-4(x+3)\), and at \(x=5\) the right side is \(-4(8)=-32\), so \(y=8-32=\boxed{-24}\). With \(x_1=-3\) the factor \(x-x_1\) is \(x+3\), and using \(x-3\) instead gives \(-4(2)=-8\) and the wrong value \(0\).
Problem
A line has slope \(\frac{2}{7}\) and passes through \((3,5)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(8\). It is not a whole number, so enter it as a fraction in lowest terms.
Show a hint
- The run from \(x=3\) to \(x=8\) is \(5\), and the rise is the slope times the run.
- The rise is \(\frac{2}{7}(5)=\frac{10}{7}\). Write \(5\) as \(\frac{35}{7}\) before adding.
Show the full solution
The run is \(8-3=5\) and the rise is \(\frac{2}{7}(5)=\frac{10}{7}\), so \(y=5+\frac{10}{7}=\frac{35}{7}+\frac{10}{7}=\boxed{\frac{45}{7}}\). A fractional slope is not a special case. It only means most runs give a \(y\) that is not a whole number.
Point-slope form is built from a point and a slope, and two points of a line give both. Either of the two is the point, and \(m\) is the slope computed from the pair. So two points are enough to write the equation of the line through them, and the next two problems do exactly that.
Problem
A line passes through \((1,10)\) and \((6,-5)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(8\).
Show a hint
- The slope is not stated here. Compute it from \((1,10)\) and \((6,-5)\) first, then use point-slope form with either of those points.
- The slope is \(\frac{-5-10}{6-1}=-3\). Now work from \((6,-5)\) with a run of \(2\).
Show the full solution
The slope is \(\frac{-5-10}{6-1}=\frac{-15}{5}=-3\). From \((6,-5)\), the run out to \(x=8\) is \(2\) and the rise is \(-3(2)=-6\), so \(y=-5-6=\boxed{-11}\). Reversing only one of the two differences gives \(+3\) for the slope, and the answer then comes out \(1\) instead.
Problem
A line passes through \((-4,3)\) and \((8,11)\). Exactly one point of that line has \(y\)-coordinate \(15\). Enter that point's \(x\)-coordinate.
Show a hint
- The number given here is a \(y\)-coordinate, so the unknown is \(x\). Start with the slope of the two given points.
- The slope is \(\frac{11-3}{8-(-4)}=\frac{2}{3}\), so \(15-11=\frac{2}{3}(x-8)\).
Show the full solution
The slope is \(\frac{11-3}{8-(-4)}=\frac{8}{12}=\frac{2}{3}\). Using \((8,11)\), \(15-11=\frac{2}{3}(x-8)\), so \(4=\frac{2}{3}(x-8)\), then \(x-8=6\) and \(x=\boxed{14}\). Checking with \((-4,3)\), \(\frac{15-3}{14-(-4)}=\frac{12}{18}=\frac{2}{3}\), the same slope. A slope of \(\frac{2}{3}\) does not force a fractional answer, since the run \(\frac{3}{2}\cdot 4\) is whole.
There are two choices for \((x_1,y_1)\), one for each of the two points, and every problem so far used only one of them. Choosing the other point puts different numbers into point-slope form, so the two equations do not look alike until one of them is expanded. The next problem rebuilds an earlier line from its other point.
Problem
The line through \((1,10)\) and \((6,-5)\) has slope \(-3\). Substitute the point \((6,-5)\) and that slope into point-slope form, expand, and solve for \(y\). The result reads \(y=-3x+c\). Enter \(c\).
Show a hint
- With \((x_1,y_1)=(6,-5)\), the left side is \(y-(-5)\), which is \(y+5\).
- The equation is \(y+5=-3(x-6)\). Expand the right side before moving the \(5\).
Show the full solution
Substituting gives \(y-(-5)=-3(x-6)\), which is \(y+5=-3x+18\), so \(y=-3x+13\) and \(c=\boxed{13}\). Starting from \((1,10)\) instead gives \(y-10=-3(x-1)\), and that expands to \(y=-3x+13\) as well. The two point-slope equations look different and describe the same line.
Problem
A line passes through \((-6,-2)\) and \((4,3)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(15\), as a fraction in lowest terms.
Show a hint
- Take \((-6,-2)\) first in both differences of the slope. Both involve subtracting a negative number, so write them out before simplifying.
- The slope is \(\frac{3-(-2)}{4-(-6)}=\frac{1}{2}\), so point-slope from \((4,3)\) gives \(y-3=\frac{1}{2}(x-4)\). Put \(x=15\) into that.
Show the full solution
The slope is \(\frac{3-(-2)}{4-(-6)}=\frac{5}{10}=\frac{1}{2}\), so point-slope from \((4,3)\) gives \(y-3=\frac{1}{2}(x-4)\). At \(x=15\) that reads \(y-3=\frac{1}{2}(11)=\frac{11}{2}\), so \(y=\boxed{\frac{17}{2}}\). The \(x\) asked for sits to the right of both given points, and the work is no different from finding \(y\) at an \(x\) between them.
Every pair of points so far had two different \(x\)-coordinates. When the two \(x\)-coordinates are equal, \(x_2-x_1\) is \(0\), the quotient is a division by zero, and the line has no slope at all. Point-slope form contains \(m\), so with no slope there is nothing to substitute into it. The next problem is that case.
Problem
A line passes through \((-5,11)\) and \((-5,-6)\). One point on that line has \(y\)-coordinate \(60\). Enter that point's \(x\)-coordinate.
Show a hint
- Start with the slope through the two given points, and look hard at the denominator \(x_2-x_1\) before dividing.
- There is no slope, so point-slope form cannot be used here at all. Compare the two first coordinates, then recall the vertical lines from 7.2.
Show the full solution
Both given points have first coordinate \(-5\), so this is the vertical line \(x=-5\), and the point with \(y\)-coordinate \(60\) is \((-5,60)\), whose \(x\)-coordinate is \(\boxed{-5}\). The slope \(\frac{-6-11}{-5-(-5)}\) has denominator \(0\), so this line has no slope and point-slope form does not apply. The second coordinate is unrestricted here, so the \(60\) is not needed.
Problem
The points \((-4,9)\), \((2,-3)\) and \((k,-13)\) all lie on one straight line. Enter \(k\).
Show a hint
- Two of the three points are fully known, and two points are enough to pin the line. Start there.
- Find the slope from the two known points, then put that slope and \((2,-3)\) into point-slope form and substitute \(x=k\) and \(y=-13\).
Show the full solution
The first two points give slope \(\frac{-3-9}{2-(-4)}=\frac{-12}{6}=-2\), and from \((2,-3)\), \(-13-(-3)=-2(k-2)\), so \(-10=-2(k-2)\), then \(k-2=5\) and \(k=\boxed{7}\). The line is \(y=-2x+1\), and \(-2(7)+1=-13\) checks it. Three points lie on one line exactly when the third satisfies the equation of the line through the other two.
A quantity that changes at a constant rate has a straight line for its graph, and the rate is the slope of that line. Two readings of the quantity are two points of that line, so two readings are enough to determine the whole relationship, including its value at a moment nobody measured.
Problem
A machine winds cable onto a drum at a constant rate. Three minutes after the machine starts, the drum holds \(52\) metres of cable, and ten minutes after it starts, the drum holds \(115\) metres. The drum was not empty when the machine started. Enter how many metres of cable the drum held at the moment the machine started.
Show a hint
- Time and length are the two coordinates. The readings are the points \((3,52)\) and \((10,115)\), and the constant rate is the slope.
- The rate is \(\frac{115-52}{10-3}=9\) metres per minute, so step back \(3\) minutes from the first reading.
Show the full solution
The readings are the points \((3,52)\) and \((10,115)\), so the rate is \(\frac{115-52}{10-3}=\frac{63}{7}=9\) metres per minute. Stepping back \(3\) minutes from \((3,52)\) gives \(52-9(3)=\boxed{25}\). Dividing \(115\) by \(10\) gives \(11.5\), which is the rate only for a drum that starts empty.
Problem
A line passes through \((-2,11)\) and \((6,-5)\). Exactly one point of that line has its \(y\)-coordinate equal to three times its \(x\)-coordinate. Enter that point's \(x\)-coordinate, as a fraction in lowest terms.
Show a hint
- Find the equation of the line first. The stated condition describes one point of it and says nothing extra about the line.
- The line is \(y=-2x+7\). A point on it whose \(y\)-coordinate is three times its \(x\)-coordinate satisfies \(3x=-2x+7\).
Show the full solution
The slope is \(\frac{-5-11}{6-(-2)}=\frac{-16}{8}=-2\), and from \((-2,11)\), \(y-11=-2(x+2)\), so \(y=-2x+7\). The condition is \(y=3x\), so \(3x=-2x+7\), then \(5x=7\) and \(x=\boxed{\frac{7}{5}}\). The point is \(\left(\frac{7}{5},\frac{21}{5}\right)\), and \(-2\left(\frac{7}{5}\right)+7=\frac{21}{5}\) confirms it.
Nearly every line here came from a point and a slope, and each of those equations was left in point-slope form or solved for \(y\). The exception was the vertical line, which has no slope, so it is written \(x=k\) instead. Lesson 7.5, Intercepts and Standard Forms, gives the two standard ways of writing a line and reads the slope straight off each one.
Practice these ideas
Practice
A line has slope \(5\) and passes through \((1,2)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(3\).
Show the solution
The run is \(3-1=2\) and the rise is \(5(2)=10\), so \(y=2+10=\boxed{12}\).
Practice
A line has slope \(3\) and passes through \((-1,4)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(5\).
Show the solution
Point-slope gives \(y-4=3(x-(-1))\), and at \(x=5\) the right side is \(3(6)=18\), so \(y=\boxed{22}\). Checking, the slope from \((-1,4)\) to \((5,22)\) is \(\frac{18}{6}=3\).
Practice
A line has slope \(-4\) and passes through \((6,2)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(9\).
Show the solution
Point-slope with \(m=-4\) and \((6,2)\) gives \(y-2=-4(x-6)\). At \(x=9\) that reads \(y-2=-4(3)=-12\), so \(y=\boxed{-10}\). The slope is negative, so the point three to the right of \((6,2)\) lies below it rather than above.
Practice
A line has slope \(5\) and passes through \((4,-3)\). Solved for \(y\), its equation reads \(y=5x+c\). Enter \(c\).
Show the solution
Point-slope form gives \(y+3=5(x-4)=5x-20\), so \(y=5x-23\) and \(c=\boxed{-23}\). Reading \(c\) off \(y+3=5x-20\) gives \(-20\), which skips the last subtraction. Checking, \(5(4)-23=-3\).
Practice
A line has slope \(\frac{3}{5}\) and passes through \((5,1)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(9\), as a fraction in lowest terms.
Show the solution
The run is \(4\) and the rise is \(\frac{3}{5}(4)=\frac{12}{5}\), so \(y=1+\frac{12}{5}=\frac{5}{5}+\frac{12}{5}=\boxed{\frac{17}{5}}\).
Practice
A line has slope \(4\) and passes through \((3,-1)\). Enter the \(x\)-coordinate of the point on that line whose \(y\)-coordinate is \(19\).
Show the solution
From \((3,-1)\) with slope \(4\), \(19-(-1)=4(x-3)\), so \(20=4(x-3)\), then \(x-3=5\) and \(x=\boxed{8}\).
Practice
A line passes through \((2,1)\) and \((5,10)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(8\).
Show the solution
The slope is \(\frac{10-1}{5-2}=3\), so the line through \((5,10)\) is \(y-10=3(x-5)\). At \(x=8\), \(y-10=3\cdot 3=9\), so \(y=\boxed{19}\). Using \((2,1)\) instead gives \(y-1=3(x-2)\) and the same \(19\) at \(x=8\).
Practice
A line passes through \((-3,8)\) and \((1,0)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(4\).
Show the solution
The slope is \(\frac{0-8}{1-(-3)}=\frac{-8}{4}=-2\). From \((1,0)\) the run out to \(x=4\) is \(3\), so the rise is \(3\cdot(-2)=-6\) and \(y=0-6=\boxed{-6}\). Measuring the run from \(x=0\) instead of from \(x=1\) gives a run of \(4\) and the wrong value \(-8\).
Practice
A line passes through \((-1,-2)\) and \((3,10)\). Enter the \(x\)-coordinate of the point on that line whose \(y\)-coordinate is \(28\).
Show the solution
The slope is \(\frac{10-(-2)}{3-(-1)}=\frac{12}{4}=3\). From \((3,10)\), \(28-10=3(x-3)\), so \(18=3(x-3)\), then \(x-3=6\) and \(x=\boxed{9}\).
Practice
A line passes through \((-4,3)\) and \((-4,11)\). Enter the \(x\)-coordinate of the point on that line whose \(y\)-coordinate is \(-60\).
Show the solution
Both given points have first coordinate \(-4\), so the change in \(x\) is \(0\) and there is no slope. The line is \(x=-4\), so every point on it has \(x=\boxed{-4}\). There is no slope to substitute into point-slope form, so it does not apply here, and the \(-60\) is not needed.
Practice
A line passes through \((-7,5)\) and \((4,5)\). Solved for \(y\), its equation reads \(y=ax+b\) for two numbers \(a\) and \(b\). Enter \(a\).
Show the solution
The change in \(y\) is \(5-5=0\) and the change in \(x\) is \(4-(-7)=11\), so the slope is \(\frac{0}{11}=0\) and the coefficient of \(x\) is \(\boxed{0}\). The line is horizontal, and every point on it has \(y=5\) whatever \(x\) is.
Practice
The line through \((1,5)\) and \((4,17)\) also passes through \((9,k)\). Enter \(k\).
Show the solution
The slope is \(\frac{17-5}{4-1}=4\), so point-slope from \((1,5)\) gives \(y-5=4(x-1)\). At \(x=9\), \(k=5+4(8)=\boxed{37}\).
Practice
A line passes through \((-3,2)\) and \((7,10)\). Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(4\), as a fraction in lowest terms.
Show the solution
The slope is \(\frac{10-2}{7-(-3)}=\frac{8}{10}=\frac{4}{5}\), so the line is \(y-2=\frac{4}{5}(x+3)\). At \(x=4\) that reads \(y-2=\frac{4}{5}(7)=\frac{28}{5}\), so \(y=\boxed{\frac{38}{5}}\).
Practice
A line passes through \((-6,14)\) and \((2,-2)\). Solve for \(y\) to write its equation as \(y=mx+c\) for two numbers \(m\) and \(c\). Enter \(c\).
Show the solution
The slope is \(\frac{-2-14}{2-(-6)}=\frac{-16}{8}=-2\). From \((2,-2)\), \(y+2=-2(x-2)=-2x+4\), so \(y=-2x+2\) and \(c=\boxed{2}\). From \((-6,14)\), \(y-14=-2(x+6)=-2x-12\) gives \(y=-2x+2\) as well.
Practice
A candle burns down at a constant rate. It stands \(21\) cm tall \(12\) minutes after it is lit and \(16\) cm tall \(32\) minutes after it is lit. Enter its height in centimeters at the moment it was lit.
Show the solution
The rate is \(\frac{16-21}{32-12}=\frac{-5}{20}=-\frac{1}{4}\) cm per minute. Over the \(12\) minutes before the first reading the candle lost \(3\) cm, so its height at the moment it was lit is \(\boxed{24}\). Checking, \(24-\frac{1}{4}\cdot 32=16\), which matches the second reading.
Practice
A tank drains at a constant rate, and time is measured in minutes from the moment it starts draining. At \(3\) minutes it holds \(92\) liters and at \(11\) minutes it holds \(68\) liters. Enter the number of minutes at which it holds \(50\) liters.
Show the solution
The rate is \(\frac{68-92}{11-3}=\frac{-24}{8}=-3\) liters per minute. From \((3,92)\), \(50-92=-3(t-3)\), so \(-42=-3(t-3)\), then \(t-3=14\) and \(t=\boxed{17}\).
Practice
A line passes through \((-6,13)\) and \((-6,-2)\), and the point \((4k,40)\) is on that same line. Enter \(k\).
Show the solution
Both given points have first coordinate \(-6\), so the line has no slope and its equation is \(x=-6\). A point is on it exactly when its first coordinate is \(-6\), so \(4k=-6\) and \(k=\boxed{-\frac{3}{2}}\). The \(40\) is not needed, since \(x=-6\) is true for every value of \(y\).