QuanticaAlgebra IOpen in the courseSlope-intercept form and standard form each describe a single line. With two lines on one plane there are three further questions, whether they ever meet, whether they stay the same distance apart, and whether they cross at a right angle. All three are settled by comparing the two slopes, and the first one starts here.
Problem
The lines \(y=4x-1\) and \(y=4x+7\) are two different lines. A point lying on both has a single \(y\)-coordinate, so its \(x\) satisfies \(4x-1=4x+7\). Enter how many points lie on both lines.
Show a hint
- Both lines have slope \(4\), so both sides of \(4x-1=4x+7\) contain the same \(4x\) term. Subtract \(4x\) from both sides and look at what remains.
- The subtraction leaves \(-1=7\), an equation containing no \(x\). Decide whether that statement is true or false, then count the shared points that follow.
Show the full solution
Subtracting \(4x\) from both sides of \(4x-1=4x+7\) leaves \(-1=7\), false for every \(x\), so no point lies on both lines and the count is \(\boxed{0}\). The two slopes are equal, which is why the subtraction removes every \(x\) term and leaves only the constants.
Problem
The lines \(y=3x-5\) and \(y=-2x+15\) have different slopes. At a point on both, the two right sides are equal, so \(3x-5=-2x+15\). Enter the \(y\)-coordinate of the point that lies on both lines.
Show a hint
- Solve \(3x-5=-2x+15\) for \(x\) first. The question asks for a \(y\)-coordinate, so that value of \(x\) is not the answer.
- Adding \(2x\) to both sides gives \(5x-5=15\). Once \(x\) is known, put it into either equation to get \(y\), and both equations should give the same \(y\).
Show the full solution
Adding \(2x\) to both sides of \(3x-5=-2x+15\) gives \(5x-5=15\), so \(5x=20\) and \(x=4\). Then \(y=3(4)-5=\boxed{7}\). Putting \(x=4\) into the other equation gives \(-2(4)+15=7\), the same \(y\). The slopes differ, so the \(x\) terms do not cancel and one value of \(x\) is left, with one \(y\) to match.
Problem
Solving \(5x-2y=8\) for \(y\) gives \(y=\frac{5}{2}x-4\), so those two equations describe one line and every point of that line satisfies both. Of the five points \((0,-4)\), \((2,1)\), \((4,6)\), \((6,11)\) and \((-2,-8)\), enter how many satisfy both equations.
Show a hint
- The two equations describe one line, so each listed point satisfies both of them or neither of them. That is one test per point, not two.
- Put a point's \(x\) into \(y=\frac{5}{2}x-4\) and compare the result with that point's own \(y\). The point counts only when the two agree.
Show the full solution
The points \((0,-4)\), \((2,1)\), \((4,6)\) and \((6,11)\) all satisfy \(y=\frac{5}{2}x-4\), while \(\frac{5}{2}(-2)-4=-9\), not \(-8\), so the count is \(\boxed{4}\). One line written twice has infinitely many points on it, and a listed point still has to be one of them.
Problem
The line \(9x+ky=40\) is parallel to the line \(y=\frac{3}{4}x-2\). From 7.5 the first line has slope \(-\frac{9}{k}\). Enter \(k\).
Show a hint
- Parallel lines have equal slopes, so set the slope of \(9x+ky=40\) equal to the slope of the other line and solve for \(k\).
- The equation to solve is \(-\frac{9}{k}=\frac{3}{4}\). Cross-multiply, keeping the minus sign attached to the \(9\), then divide both sides by \(3\).
Show the full solution
Parallel lines have equal slopes, so \(-\frac{9}{k}=\frac{3}{4}\), which gives \(3k=-36\) and \(k=\boxed{-12}\). Then \(9x-12y=40\) reads \(y=\frac{3}{4}x-\frac{10}{3}\), a different \(y\)-intercept from \(-2\), so the two lines are parallel rather than one line.
Chapter 5 counted the solutions of a system of two linear equations and found three answers, exactly one, none at all, and infinitely many. A solution is a pair \((x,y)\) that satisfies both equations, and such a pair is a point lying on both lines. The three panels above are those three counts drawn.
Problem
In the system \(10x+4y=26\) and \(15x+7y=39\), the \(x\)-coefficient and the constant of the second equation are both \(\frac{3}{2}\) times those of the first. Enter the number of solutions the system has.
Show a hint
- The \(x\)-coefficients and the constants being in proportion is not enough on its own. Compare the two slopes instead, using \(-\frac{A}{B}\) on each equation.
- The slopes are \(-\frac{10}{4}\) and \(-\frac{15}{7}\). Reduce both and compare them, rather than judging by how alike the two equations look.
Show the full solution
The slopes are \(-\frac{10}{4}=-\frac{5}{2}\) and \(-\frac{15}{7}\), which differ, so the lines cross once and the number of solutions is \(\boxed{1}\). The \(\frac{3}{2}\) scaling of the \(x\)-coefficients and the constants means nothing by itself, since the \(y\)-coefficients are in the ratio \(\frac{7}{4}\), and all three ratios must match for the two equations to be the same line.
Problem
The system \(kx+8y=5\) and \(10x+4y=7\) has no solution for exactly one value of \(k\). Deciding it needs the slopes of both lines, which 7.5 reads off standard form. Enter \(k\).
Show a hint
- No solution means the two lines are parallel, so their slopes are equal and their \(y\)-intercepts are not.
- Set \(-\frac{k}{8}\) equal to the slope of \(10x+4y=7\), then solve for \(k\).
Show the full solution
No solution means the lines are parallel, so the slopes match. The slope of \(10x+4y=7\) is \(-\frac{10}{4}=-\frac{5}{2}\), so \(-\frac{k}{8}=-\frac{5}{2}\) and \(k=\boxed{20}\). Doubling \(10x+4y=7\) gives \(20x+8y=14\), and the first equation has \(5\) on the right, so the lines are parallel rather than the same line.
Two lines that cross form four angles, and when all four are right angles the two lines are perpendicular. A quarter turn about the crossing point sends each line onto the other, so it sends a slope triangle under one line to a slope triangle under the other. That is enough to get the second slope from the first.
Problem
A segment from the origin to \((5,2)\) has run \(5\) and rise \(2\), so it lies along a line of slope \(\frac{2}{5}\). A quarter turn counterclockwise about the origin leaves the origin in place and sends \((5,2)\) to \((-2,5)\), so the turned segment runs from the origin to \((-2,5)\) along a line perpendicular to the first. Enter the slope of the turned line, as a fraction in lowest terms.
Show a hint
- Slope is rise over run, and here both are measured from the origin out to the far end of the turned segment. Going out to that end means going left first, so one of the two is negative.
- Divide the rise by the run and reduce, keeping the minus sign in front of the fraction. As a check, that slope times \(\frac{2}{5}\) has to come out to \(-1\).
Show the full solution
The run is \(-2\) and the rise is \(5\), so the slope is \(\frac{5}{-2}=\boxed{-\frac{5}{2}}\). That is \(\frac{2}{5}\) flipped over with its sign changed, and the product \(\frac{2}{5}\cdot\left(-\frac{5}{2}\right)\) is \(-1\), as it must be for perpendicular lines.
Problem
In the previous problem the perpendicular slope came out as the first slope flipped over with its sign changed. The lines \(5x+6y=13\) and \(6x+cy=5\) are perpendicular. Enter \(c\).
Show a hint
- Read the slope of \(5x+6y=13\) off standard form, then flip it over and change its sign.
- The perpendicular slope is \(\frac{6}{5}\), and the second line has slope \(-\frac{6}{c}\).
Show the full solution
The line \(5x+6y=13\) has slope \(-\frac{5}{6}\), so a line perpendicular to it has slope \(\frac{6}{5}\). The second line has slope \(-\frac{6}{c}\), and \(-\frac{6}{c}=\frac{6}{5}\) gives \(6c=-30\), so \(c=\boxed{-5}\). Flipping without changing the sign would give \(c=5\), which makes the two lines parallel instead.
Deciding how two lines sit takes two comparisons and no graph. Compute the slope of each line, whatever form it is written in. Equal slopes means parallel or the same line, and comparing \(y\)-intercepts settles which. Different slopes means one crossing point, a right angle exactly when the two slopes multiply to \(-1\).
Problem
One line is \(6x+15y=20\) and the other is \(y-4=\frac{5}{2}(x+3)\). Enter one word, parallel, perpendicular, same or neither, for the relationship between the two lines.
Show a hint
- Comparing two lines means comparing their slopes. One equation is in standard form and the other is in point-slope form, and neither has to be solved for \(y\) first.
- The slopes are \(-\frac{6}{15}\) and \(\frac{5}{2}\). Put the first in lowest terms, then multiply the two.
Show the full solution
The line \(6x+15y=20\) has slope \(-\frac{6}{15}=-\frac{2}{5}\), and \(y-4=\frac{5}{2}(x+3)\) has slope \(\frac{5}{2}\). The slopes differ and their product is \(-1\), so the lines are \(\boxed{\text{perpendicular}}\). Reduce before comparing, since \(-\frac{6}{15}\) and \(-\frac{2}{5}\) are the same number but only the second one reads as the negative reciprocal of \(\frac{5}{2}\).
Problem
Four pairs of lines are listed. Pair 1 is \(y=3x-2\) and \(y=-\frac{1}{3}x+5\). Pair 2 is \(y=7\) and \(x=-4\). Pair 3 is \(y=\frac{4}{9}x\) and \(y=\frac{9}{4}x-1\). Pair 4 is \(y=-5x+2\) and \(y=-\frac{1}{5}x\). Enter how many of the four pairs are perpendicular.
Show a hint
- Three of the pairs can be settled by multiplying the two slopes. In the one that is left a line has no slope, so there is no product to take, and that pair has to be decided from the two lines themselves.
- From 7.2 the graph of \(y=7\) is a horizontal line and the graph of \(x=-4\) is a vertical one. Picture those two and read the angle where they cross.
Show the full solution
Pair 1 gives \(3\cdot\left(-\frac{1}{3}\right)=-1\), and pair 2 is a horizontal line crossing a vertical one, so both of those are perpendicular. Pair 3 gives \(\frac{4}{9}\cdot\frac{9}{4}=1\) and pair 4 gives \(-5\cdot\left(-\frac{1}{5}\right)=1\), so the count is \(\boxed{2}\). The product test needs two slopes, and pair 2 supplies only one.
Chapter 7 began with one point on the plane and ends with two lines compared. Pinning down a single line takes a point and a slope. Comparing two lines takes their slopes, plus the \(y\)-intercepts when those slopes are equal. The last two problems run that both ways, choosing a constant to force a relationship and then finding a crossing point.
Problem
The line \(kx+6y=11\) is parallel to \(y=\frac{2}{3}x+5\) for one value of \(k\), and perpendicular to it for a different value. Enter the value of \(k\) that makes the two lines perpendicular.
Show a hint
- The slope of \(kx+6y=11\) is \(-\frac{k}{6}\), and the other line has slope \(\frac{2}{3}\). Perpendicular is the case where the two slopes multiply to \(-1\).
- The first line needs slope \(-\frac{3}{2}\), so solve \(-\frac{k}{6}=-\frac{3}{2}\). The parallel value of \(k\) is a different number and is not what is asked for.
Show the full solution
The line \(kx+6y=11\) has slope \(-\frac{k}{6}\), and perpendicular to a slope of \(\frac{2}{3}\) means a slope of \(-\frac{3}{2}\). Then \(-\frac{k}{6}=-\frac{3}{2}\) gives \(2k=18\), so \(k=\boxed{9}\). Parallel would need \(-\frac{k}{6}=\frac{2}{3}\) instead, giving \(k=-4\), so the two values have opposite signs.
Problem
Line \(g\) passes through \((6,1)\) and is perpendicular to \(4x+3y=18\). Line \(h\) passes through \((-2,5)\) and is parallel to \(4x+3y=18\). Enter the \(x\)-coordinate of the point where \(g\) and \(h\) cross, as a fraction in lowest terms.
Show a hint
- Get the slope of \(4x+3y=18\) first. Line \(h\) has that same slope and line \(g\) has its negative reciprocal, and each line also comes with a point of its own.
- The two equations are \(y=1+\frac{3}{4}(x-6)\) and \(y=5-\frac{4}{3}(x+2)\). Setting them equal and multiplying through by \(12\) clears both fractions at once.
Show the full solution
The line \(4x+3y=18\) has slope \(-\frac{4}{3}\), so \(g\) has slope \(\frac{3}{4}\) and \(h\) has slope \(-\frac{4}{3}\). Setting \(1+\frac{3}{4}(x-6)=5-\frac{4}{3}(x+2)\) and multiplying by \(12\) gives \(9x-42=28-16x\), so \(25x=70\) and \(x=\boxed{\frac{14}{5}}\). Both lines give \(y=-\frac{7}{5}\) there. Since \(g\) is perpendicular to \(4x+3y=18\) and \(h\) is parallel to it, \(g\) and \(h\) are themselves perpendicular.
Every equation in this chapter has been a statement that two quantities are equal, and the graph of each one has been a line. Chapter 8 replaces that equal sign with \(<\) or \(>\). Lesson 8.1, The Basics of Inequality, starts with what those signs mean and what chapter 4's moves do to them.
Practice these ideas
Practice
Enter the slope of a line parallel to \(y=-7x+4\).
Show the solution
Parallel lines have equal slopes, and \(y=-7x+4\) has slope \(\boxed{-7}\). The \(4\) is the \(y\)-intercept, not the slope, and two distinct parallel lines differ only in that constant.
Practice
A line is perpendicular to the line \(y=\frac{1}{3}x-8\). Enter the slope of the perpendicular line.
Show the solution
The given line has slope \(\frac{1}{3}\), and perpendicular slopes multiply to \(-1\), so the other slope is \(-1\div\frac{1}{3}=\boxed{-3}\). Flipping \(\frac{1}{3}\) without changing the sign gives \(3\), and \(\frac{1}{3}\cdot 3=1\), not \(-1\).
Practice
The lines \(y=5x-2\) and \(y=5x+11\) are parallel, perpendicular, the same line, or none of those. Enter one word, parallel, perpendicular, same or neither.
Show the solution
Both lines have slope \(5\), and the \(y\)-intercepts \(-2\) and \(11\) are different, so the lines never meet and they are \(\boxed{\text{parallel}}\). If the \(y\)-intercepts were equal as well, the two equations would be the same line.
Practice
Enter one word, parallel, perpendicular, same or neither, for the lines \(y=\frac{3}{5}x-2\) and \(3x-5y=10\).
Show the solution
Solving \(3x-5y=10\) for \(y\) gives \(-5y=-3x+10\), then \(y=\frac{3}{5}x-2\), which matches the other equation term for term, so the answer is \(\boxed{\text{same}}\). Same slope and same \(y\)-intercept means one line, since every point on one is on the other.
Practice
Are the lines \(y=2x+6\) and \(y=-2x+6\) parallel, perpendicular, the same line, or none of those? Enter one word, parallel, perpendicular, same or neither.
Show the solution
The slopes are \(2\) and \(-2\). Since they are not equal, the lines are not parallel and not the same line, and \(2\cdot(-2)=-4\) rather than \(-1\), so they are not perpendicular and the answer is \(\boxed{\text{neither}}\). Opposite signs alone do not make a right angle. Only a product of exactly \(-1\) does.
Practice
A line is perpendicular to \(2x+7y=15\). Enter the slope of the perpendicular line, as a fraction in lowest terms.
Show the solution
The line \(2x+7y=15\) has slope \(-\frac{2}{7}\), so the perpendicular line has slope \(\boxed{\frac{7}{2}}\). The product of the two slopes is \(-\frac{2}{7}\cdot\frac{7}{2}=-1\), which is the test for perpendicular lines.
Practice
The lines \(y=4x-9\) and \(y=x+30\) cross at one point. Enter the \(x\)-coordinate of that point, not the point itself.
Show the solution
Setting \(4x-9=x+30\) gives \(3x=39\), so \(x=\boxed{13}\). Substituting back gives \(4(13)-9=43\) and \(13+30=43\), so the crossing point is \((13,43)\), and answering \(43\) reports the \(y\)-coordinate instead of the \(x\)-coordinate.
Practice
The line \(y=kx+2\) is parallel to \(3x-4y=8\). Enter \(k\) as a fraction in lowest terms.
Show the solution
Solving \(3x-4y=8\) for \(y\) gives \(y=\frac{3}{4}x-2\), and parallel lines have equal slopes, so \(k=\boxed{\frac{3}{4}}\). Dividing \(-3x\) by \(-4\) makes the sign positive, so reporting \(-\frac{3}{4}\) is the common slip.
Practice
Consider the two equations \(3x-y=6\) and \(y=3x-6\). Of the six points \((0,-6)\), \((4,6)\), \((-1,-9)\), \((2,1)\), \((5,10)\) and \((-3,-14)\), enter how many satisfy both equations.
Show the solution
Rearranging \(3x-y=6\) gives \(y=3x-6\), the second equation exactly, so the two describe one line and a point satisfies both or neither. Testing each, \((0,-6)\), \((4,6)\) and \((-1,-9)\) fit while the other three do not, so the count is \(\boxed{3}\). One line written twice still only contains the points that satisfy it.
Practice
The line \(cx+10y=7\) is perpendicular to \(y=\frac{5}{4}x-1\). Enter \(c\).
Show the solution
A line perpendicular to a line of slope \(\frac{5}{4}\) has slope \(-\frac{4}{5}\), and \(cx+10y=7\) has slope \(-\frac{c}{10}\), so multiplying \(-\frac{c}{10}=-\frac{4}{5}\) by \(-10\) gives \(c=\boxed{8}\). Checking, \(8x+10y=7\) gives \(y=-\frac{4}{5}x+\frac{7}{10}\), and \(\frac{5}{4}\cdot\left(-\frac{4}{5}\right)=-1\).
Practice
The vertical line \(x=-4\) and the horizontal line \(y=9\) cross at a right angle, even though the product of their slopes cannot be formed, since a vertical line has no slope. A third line is perpendicular to \(y=9\). Enter the slope of that third line, or the word undefined if it has none.
Show the solution
A line perpendicular to a horizontal line is vertical, and any two points of a vertical line have the same \(x\)-coordinate, so the change in \(x\) is \(0\) and the slope is \(\boxed{\text{undefined}}\). This is the pair the product rule cannot reach, since one of the two slopes does not exist to be multiplied.
Practice
For exactly one value of \(p\) the system \(5x-2y=9\) and \(15x+py=4\) has no solution. Enter \(p\).
Show the solution
No solution means parallel, so \(-\frac{15}{p}=\frac{5}{2}\), giving \(5p=-30\) and \(p=\boxed{-6}\). Dividing \(15x-6y=4\) by \(3\) gives \(5x-2y=\frac{4}{3}\), which differs from \(9\), so the lines are parallel and not the same.
Practice
Line \(w\) is parallel to \(7x+4y=9\) and passes through \((-4,5)\). Enter the \(y\)-coordinate of the point where \(w\) crosses the \(y\)-axis.
Show the solution
Line \(w\) has slope \(-\frac{7}{4}\), so going from \((-4,5)\) right by \(4\) to \(x=0\) drops \(y\) by \(7\), leaving \(\boxed{-2}\).
Practice
Line \(n\) is perpendicular to \(y=\frac{5}{3}x+1\) and passes through \((10,4)\). Enter the \(y\)-coordinate of the point where \(n\) crosses the \(y\)-axis.
Show the solution
Line \(n\) has slope \(-\frac{3}{5}\), so \(y=-\frac{3}{5}x+b\), and putting in \(x=10\) and \(y=4\) gives \(4=-6+b\), which makes the \(y\)-intercept \(\boxed{10}\). Using \(\frac{5}{3}\) instead of \(-\frac{3}{5}\) is the common slip, and the check is that perpendicular slopes multiply to \(-1\).
Practice
Line \(a\) is perpendicular to the line \(3x+12y=20\) and passes through \((4,7)\). Enter the \(y\)-coordinate of the point on line \(a\) whose \(x\)-coordinate is \(12\).
Show the solution
Solving \(3x+12y=20\) for \(y\) gives slope \(-\frac{1}{4}\), so line \(a\) has slope \(4\), since \(-\frac{1}{4}\cdot 4=-1\). From \((4,7)\) a run of \(8\) gives \(y=7+4(8)=\boxed{39}\).
Practice
For exactly one value of \(t\) the system \(10x-15y=t\) and \(4x-6y=14\) has infinitely many solutions. Enter \(t\).
Show the solution
Multiplying \(4x-6y=14\) by \(\frac{5}{2}\) gives \(10x-15y=35\), so \(t=\boxed{35}\). Infinitely many solutions means both equations describe the same line, which needs all three numbers proportional and not just the \(x\) and \(y\) coefficients. With any other \(t\) the two lines have the same slope but different \(y\)-intercepts, so they are parallel and there is no solution.
Practice
For each of \(k=-8\), \(k=-2\), \(k=0\), \(k=3\), \(k=8\) and \(k=12\), the equations \(kx+6y=21\) and \(4x-3y=9\) form a system. Enter how many of those six systems have exactly one solution.
Show the solution
Only the \(k=-8\) system does not have exactly one solution, so the count is \(\boxed{5}\). In slope-intercept form the slopes are \(-\frac{k}{6}\) and \(\frac{4}{3}\), equal only when \(k=-8\), and different slopes mean exactly one crossing point. Scaling \(4x-3y=9\) by \(-2\) gives \(-8x+6y=-18\), and \(-18 \neq 21\), so at \(k=-8\) the two lines are parallel and that system has no solution.