QuanticaAlgebra IOpen in the course7.4 built a line from a point and a slope, and expanding the result left \(y\) alone on one side, equal to a multiple of \(x\) plus a constant. Only two numbers differ from one such equation to the next, the coefficient of \(x\) and the constant term. The next two problems locate both of them on the graph.
Problem
The graph of \(y=4x-9\) is a straight line, and this equation is already solved for \(y\). The line meets the \(y\)-axis at exactly one point. Enter the \(y\)-coordinate of that point.
Show a hint
- From 7.2, every point of the \(y\)-axis has \(x=0\). That point is on the line too, so its coordinates satisfy \(y=4x-9\).
- Replace \(x\) with \(0\) in \(y=4x-9\). The product \(4(0)\) is \(0\), so the right side becomes a single subtraction from \(0\).
Show the full solution
Substituting \(x=0\) gives \(y=4(0)-9=0-9=\boxed{-9}\). Reading the \(9\) without its minus sign gives \(9\) and puts the crossing above the origin instead of below it.
Problem
The line \(y=\frac{2}{5}x+6\) passes through the point with \(x=0\) and the point with \(x=10\). Work out the \(y\)-coordinate of each, then use 7.3's definition to enter the slope of the line.
Show a hint
- Substitute each \(x\) into the equation to get the two points, then divide the change in \(y\) by the change in \(x\).
- At \(x=0\) the point is \((0,6)\), and at \(x=10\) the term \(\frac{2}{5}(10)\) is \(4\).
Show the full solution
At \(x=0\) the point is \((0,6)\) and at \(x=10\) it is \((10,10)\), so \(m=\frac{10-6}{10-0}=\frac{4}{10}=\boxed{\frac{2}{5}}\). The slope equals the coefficient of \(x\) already written in the equation, and the constant term \(6\) is again the \(y\)-coordinate at \(x=0\).
Problem
The equation \(y=-8x+13\) is already solved for \(y\), so the slope of its graph can be read straight off the equation without computing anything. Enter that slope.
Show a hint
- The slope is the whole coefficient of \(x\), and the sign written in front of it is part of that coefficient.
- Check with two points. Substituting \(x=0\) gives \(y=13\), and substituting \(x=1\) gives \(y=5\).
Show the full solution
The coefficient of \(x\) is \(-8\), so the slope is \(\boxed{-8}\). The points \((0,13)\) and \((1,5)\) give \(\frac{5-13}{1-0}=-8\) as well. A slope of \(8\) would mean \(y\) rises as \(x\) rises, and this line falls.
Problem
A line has slope \(-\frac{3}{4}\) and \(y\)-intercept \(14\), so its equation in slope-intercept form can be written down with no rearranging. Enter the \(y\)-coordinate of the point on that line whose \(x\)-coordinate is \(-12\).
Show a hint
- Putting \(m=-\frac{3}{4}\) and \(b=14\) into \(y=mx+b\) gives the equation, and one substitution finishes it.
- The equation is \(y=-\frac{3}{4}x+14\), and at \(x=-12\) the product \(-\frac{3}{4}(-12)\) is positive.
Show the full solution
The line is \(y=-\frac{3}{4}x+14\), and substituting \(x=-12\) gives \(y=-\frac{3}{4}(-12)+14=9+14=\boxed{23}\). Two negatives multiply to a positive, so the term \(-\frac{3}{4}x\) is \(9\) here and the result is larger than \(14\), not smaller.
Not every line is written with \(y\) by itself. 7.2 wrote linear equations as \(ax+by=c\), both letters on the left, and that arrangement has its own name. Substituting \(0\) for either variable still leaves a one-letter equation, so the intercepts stay cheap, and one rearrangement is enough for the slope. The next two problems do one of each.
Problem
Exactly one point of the graph of \(7x-4y=42\) lies on the \(x\)-axis. Enter the \(x\)-coordinate of that point, not the point itself.
Show a hint
- That point is the \(x\)-intercept, so start by asking what the second coordinate equals at every point of the \(x\)-axis.
- Replace \(y\) with \(0\) and simplify. The \(42\) is divided by the coefficient of \(x\), not by \(-4\).
Show the full solution
Substituting \(y=0\) gives \(7x-4(0)=42\), so \(7x=42\) and \(x=\boxed{6}\). Dividing \(42\) by \(-4\) instead gives \(-\frac{21}{2}\), which is where the line meets the \(y\)-axis, not the \(x\)-axis.
Problem
To find the slope of \(4x+9y=63\), solve the equation for \(y\) and read the coefficient of \(x\). A student skips that and divides the coefficient of \(x\) by the coefficient of \(y\) as they stand, reporting \(\frac{4}{9}\). Enter the correct slope.
Show a hint
- Reading the coefficients where they stand does not give the slope. Get \(y\) by itself on one side first, and then the coefficient of \(x\) is the slope.
- Subtracting \(4x\) from both sides gives \(9y=-4x+63\). Now divide every term by \(9\).
Show the full solution
Subtracting \(4x\) gives \(9y=-4x+63\), and dividing every term by \(9\) gives \(y=-\frac{4}{9}x+7\), so the slope is \(\boxed{-\frac{4}{9}}\). The student's \(\frac{4}{9}\) is the negative of the correct slope, because subtracting \(4x\) makes that term negative before the division by \(9\).
Any line that has a slope can be written either way, and each direction is one short rearrangement. Standard form to slope-intercept form is solving for \(y\), which the last problem already did. The other direction means collecting both letters on one side and clearing any fraction, since \(A\), \(B\) and \(C\) are wanted as integers.
Problem
Write \(y=\frac{5}{6}x-4\) in the form \(Ax+By=C\), where \(A\), \(B\) and \(C\) are integers with no common factor greater than \(1\) and \(A>0\). Those conditions fix the three values completely. Enter \(C\).
Show a hint
- The three constants have to be integers, so clear the fraction first by multiplying every term by \(6\).
- After multiplying, the equation is \(6y=5x-24\). Move the \(x\) term to the left, then make the coefficient of \(x\) positive.
Show the full solution
Multiplying every term by \(6\) gives \(6y=5x-24\), and subtracting \(5x\) gives \(-5x+6y=-24\). Multiplying through by \(-1\) makes \(A>0\), leaving \(5x-6y=24\), so \(C=\boxed{24}\). The coefficients \(5\), \(-6\) and \(24\) have no common factor above \(1\), so with \(A>0\) this is the only such equation for the line.
Problem
Solve \(14x+4y=9\) for \(y\) to put it in slope-intercept form. Enter the slope of the line, as a fraction in lowest terms.
Show a hint
- Slope-intercept form needs \(y\) alone with coefficient \(1\), so subtract the \(x\) term first and then divide every term by the coefficient of \(y\).
- The equation becomes \(4y=-14x+9\). Dividing by \(4\) gives a coefficient of \(x\) that is not yet in lowest terms.
Show the full solution
Subtracting \(14x\) gives \(4y=-14x+9\), and dividing every term by \(4\) gives \(y=-\frac{14}{4}x+\frac{9}{4}\), so in lowest terms the slope is \(\boxed{-\frac{7}{2}}\). Reading \(-\frac{A}{B}\) straight off gives \(-\frac{14}{4}\), the same number, and it still has to be reduced.
Substituting \(y=0\) in \(Ax+By=C\) leaves \(Ax=C\), so the \(x\)-intercept is \(\frac{C}{A}\), and \(x=0\) leaves \(By=C\), so the \(y\)-intercept is \(\frac{C}{B}\). That is 7.2's rule here, and each is one division. Exactly one line passes through two distinct points, so those two are the quickest pair to plot for a line in standard form.
Problem
The equation \(6x=-15\) is in standard form \(Ax+By=C\) with \(A=6\), \(B=0\) and \(C=-15\). Enter the slope of its graph, or the word undefined if it has none.
Show a hint
- The rule \(-\frac{A}{B}\) came from solving \(Ax+By=C\) for \(y\), and that step meant dividing by \(B\). Check whether that division can be done here.
- There is no \(y\) term, so the equation says \(x=-\frac{5}{2}\) and nothing about \(y\). Two points of the graph are \(\left(-\frac{5}{2},1\right)\) and \(\left(-\frac{5}{2},4\right)\), so work out the change in \(x\) between them.
Show the full solution
With no \(y\) term the equation is \(x=-\frac{5}{2}\), a vertical line, and every point of it has the same \(x\)-coordinate, so the change in \(x\) in the slope fraction is \(0\) and the slope is \(\boxed{\text{undefined}}\). The rule \(-\frac{A}{B}\) would give \(-\frac{6}{0}\), which is not a number, since that rule came from dividing by \(B\) and is valid only when \(B \neq 0\).
Problem
The line \(2x+9y=-36\) is in standard form, and it crosses the \(x\)-axis at \((p,0)\) and the \(y\)-axis at \((0,q)\). Enter the value of \(p+q\).
Show a hint
- The two crossings come from two separate substitutions, \(y=0\) for the first and \(x=0\) for the second.
- Setting \(y=0\) leaves \(2x=-36\), and setting \(x=0\) leaves \(9y=-36\).
Show the full solution
Setting \(y=0\) gives \(2x=-36\), so \(p=-18\), and setting \(x=0\) gives \(9y=-36\), so \(q=-4\), and their sum is \(-18+(-4)=\boxed{-22}\). Answering \(-18\) reports only the \(x\)-axis crossing, and dropping the minus on \(q\) gives \(-14\) instead.
Which form to use depends on the question. From standard form, each intercept is one division. From slope-intercept form, the slope and the value of \(y\) at \(x=0\) are both readable with no work, and when \(y\) is an amount changing at a constant rate as \(x\) grows, from 7.3 those two numbers are the rate and the amount at the start.
Problem
Water runs into a tank at a steady rate. After \(x\) minutes the tank holds \(y=\frac{11}{2}x+38\) liters. Enter how many liters the tank gains each minute.
Show a hint
- The equation is already in the form \(y=mx+b\). The value of \(y\) at \(x=0\) is the amount at the start, and the gain each minute is the change in \(y\) when \(x\) goes up by \(1\).
- Work out what the tank holds at \(x=0\) and what it holds at \(x=1\). The difference between those two amounts is the gain in one minute.
Show the full solution
The equation is already in the form \(y=mx+b\), and the change in \(y\) for each extra \(1\) of \(x\) is \(m\), so the number of liters gained each minute is \(\boxed{\frac{11}{2}}\). At \(x=0\) the tank holds \(38\) liters and at \(x=1\) it holds \(\frac{11}{2}+38\), a difference of \(\frac{11}{2}\). The \(38\) is the water in the tank before any ran in, so it is no part of the rate.
Problem
A line has slope \(-\frac{7}{4}\) and its \(x\)-intercept is \(12\). Written as \(Ax+By=C\) with \(A\), \(B\) and \(C\) integers having no common factor greater than \(1\) and \(A>0\), enter \(C\).
Show a hint
- An \(x\)-intercept of \(12\) is the point \((12,0)\), so a point and a slope are both known and 7.4's point-slope form applies.
- Point-slope gives \(y=-\frac{7}{4}(x-12)\). Multiply both sides by \(4\), expand the right side, then move the \(x\) term across.
Show the full solution
From \((12,0)\) with slope \(-\frac{7}{4}\), point-slope gives \(y-0=-\frac{7}{4}(x-12)\), so \(4y=-7x+84\) and \(7x+4y=84\), giving \(C=\boxed{84}\). Here \(A=7\), \(B=4\) and \(C=84\) have no common factor greater than \(1\) and \(A>0\), so no rescaling is needed.
Each line here was taken on its own, and either form is enough to get its slope. With two lines at once there are further questions, whether they ever meet, whether they stay the same distance apart, and whether they cross at a right angle. Lesson 7.6, Parallel, Perpendicular, and Comparing Lines, answers all three by comparing the two slopes.
Practice these ideas
Practice
Enter the \(y\)-intercept of the line \(y=3x+17\), meaning the \(y\)-coordinate of the point where the line crosses the \(y\)-axis.
Show the solution
Substituting \(x=0\) gives \(y=3(0)+17=\boxed{17}\). In slope-intercept form \(y=mx+b\) the constant term \(b\) is the \(y\)-intercept, so it can be read off without substituting.
Practice
Enter the slope of the line \(y=-\frac{5}{8}x+2\).
Show the solution
The coefficient of \(x\) is \(-\frac{5}{8}\), so the slope is \(\boxed{-\frac{5}{8}}\). The \(2\) is the \(y\)-intercept, not the slope.
Practice
Enter the slope of the line \(y=21-\frac{1}{2}x\).
Show the solution
Reordering gives \(y=-\frac{1}{2}x+21\), so the slope is \(\boxed{-\frac{1}{2}}\). The coefficient of \(x\) is the slope whichever order the two terms are written in, and \(21\) is the constant term, so it is the \(y\)-intercept.
Practice
The line \(y=\frac{2}{3}x-5\) contains a point whose \(x\)-coordinate is \(18\). Enter that point's \(y\)-coordinate.
Show the solution
Substituting \(x=18\) gives \(y=\frac{2}{3}(18)-5=12-5=\boxed{7}\). The denominator \(3\) divides \(18\) exactly, which is why an \(x\)-value that is a multiple of the denominator keeps the arithmetic in whole numbers.
Practice
The equation \(5y=35x-80\) is not in slope-intercept form, since \(y\) has a coefficient of \(5\). Rewrite it as \(y=mx+b\) and enter the value of \(b\).
Show the solution
Dividing every term by \(5\) gives \(y=7x-16\), so \(b=\boxed{-16}\). Reading \(-80\) off the original equation is the common slip, since that constant still has to be divided by \(5\).
Practice
Enter the \(x\)-coordinate of the point where the line \(4x+9y=52\) crosses the \(x\)-axis.
Show the solution
Substituting \(y=0\) gives \(4x=52\), so \(x=\boxed{13}\). In standard form \(Ax+By=C\) that quotient is \(\frac{C}{A}\), and the coefficient \(9\) of \(y\) takes no part in it.
Practice
Enter the \(y\)-coordinate of the point where the line \(4x-7y=77\) crosses the \(y\)-axis.
Show the solution
Substituting \(x=0\) gives \(-7y=77\), so \(y=\boxed{-11}\). In \(Ax+By=C\) the \(y\)-intercept is \(\frac{C}{B}\). Here \(B=-7\) and \(C=77\), so a positive \(C\) still gives a negative intercept.
Practice
A student says the slope of \(9x+8y=19\) is \(\frac{9}{8}\), dividing the coefficient of \(x\) by the coefficient of \(y\). Enter the correct slope.
Show the solution
Solving for \(y\) gives \(8y=-9x+19\), then \(y=-\frac{9}{8}x+\frac{19}{8}\), so the slope is \(\boxed{-\frac{9}{8}}\). For \(Ax+By=C\) the slope is \(-\frac{A}{B}\) and not \(\frac{A}{B}\), so the student's value has the wrong sign.
Practice
Enter the slope of the line \(18x-4y=25\) as a fraction in lowest terms.
Show the solution
Subtracting \(18x\) leaves \(-4y=-18x+25\), and dividing every term by \(-4\) gives \(y=\frac{9}{2}x-\frac{25}{4}\), so the slope is \(\boxed{\frac{9}{2}}\). The rule \(-\frac{A}{B}\) gives the same value, since \(A=18\) and \(B=-4\) make \(-\frac{18}{-4}=\frac{9}{2}\).
Practice
Write \(y=\frac{7}{4}x+\frac{1}{2}\) in the form \(Ax+By=C\), where \(A\), \(B\) and \(C\) are integers with no common factor greater than \(1\) and \(A>0\). Enter \(C\).
Show the solution
Multiplying every term by \(4\) gives \(4y=7x+2\), and subtracting \(7x\) gives \(-7x+4y=2\). Multiplying through by \(-1\) makes \(A>0\), leaving \(7x-4y=-2\), so \(C=\boxed{-2}\). Multiplying by \(4\) rather than by \(8\) keeps the coefficients free of a common factor.
Practice
The graph of \(9x=-51\) is a vertical line. Enter the \(x\)-coordinate of the point where it crosses the \(x\)-axis, as a fraction in lowest terms.
Show the solution
Dividing both sides by \(9\) gives \(x=-\frac{51}{9}\), and \(51\) and \(9\) share the factor \(3\), so \(x=\boxed{-\frac{17}{3}}\). Written as \(Ax+By=C\) this line has \(B=0\), so it has no slope and never meets the \(y\)-axis.
Practice
The graph of \(11y=-88\) is a horizontal line. Enter its slope, or the word undefined if it has none.
Show the solution
Dividing by \(11\) gives \(y=-8\), a horizontal line, and any two of its points have the same \(y\), so the change in \(y\) is \(0\) and the slope is \(\boxed{0}\). The rule agrees here, since \(-\frac{A}{B}=-\frac{0}{11}=0\), and it is allowed because \(B \neq 0\).
Practice
A cartridge holds \(V\) milliliters of ink after printing \(p\) pages, where \(3p+2V=480\). Enter how many milliliters of ink each page uses.
Show the solution
Solving for \(V\) gives \(V=-\frac{3}{2}p+240\), and the size of the coefficient of \(p\) is the milliliters lost per page, so each page uses \(\boxed{\frac{3}{2}}\). The \(240\) is the value of \(V\) at \(p=0\), the ink in a full cartridge, and it is not the rate.
Practice
Enter the slope of the line \(8y-3x=5y+24\).
Show the solution
Subtracting \(5y\) gives \(3y-3x=24\), so \(3y=3x+24\) and \(y=x+8\), which has slope \(\boxed{1}\). Reading \(-\frac{A}{B}\) off the original equation gives \(\frac{3}{8}\), which is wrong because the two \(y\) terms had not been collected yet.
Practice
The line \(7x+2y=42\) crosses the two axes at two points. Those two points and the origin are the vertices of a right triangle. Enter its area.
Show the solution
Setting \(y=0\) gives \(x=6\) and setting \(x=0\) gives \(y=21\), so the legs have lengths \(6\) and \(21\), and the area is \(\frac{1}{2}(6)(21)=\boxed{63}\). The right angle sits at the origin, where the two axes meet and the two intercept distances are the legs.
Practice
The graph of \(kx+6y=23\) has slope \(\frac{5}{3}\), where \(k\) is a constant. Enter \(k\).
Show the solution
The slope is \(-\frac{k}{6}\), so \(-\frac{k}{6}=\frac{5}{3}\), and multiplying both sides by \(-6\) gives \(k=\boxed{-10}\). Checking, \(-10x+6y=23\) becomes \(y=\frac{5}{3}x+\frac{23}{6}\), whose slope is \(\frac{5}{3}\).
Practice
A line passes through \((-8,3)\) and \((4,-6)\). Written as \(Ax+By=C\) with \(A\), \(B\) and \(C\) integers having no common factor greater than \(1\) and \(A>0\), enter \(C\).
Show the solution
The slope is \(\frac{-6-3}{4-(-8)}=\frac{-9}{12}=-\frac{3}{4}\), so from \((4,-6)\), \(y+6=-\frac{3}{4}(x-4)\), and multiplying by \(4\) gives \(4y+24=-3x+12\), so \(3x+4y=\boxed{-12}\). Moving the \(x\) term the other way instead gives \(-3x-4y=12\), where \(A\) is negative, so \(12\) is the wrong value to report.