QuanticaAlgebra IOpen in the courseChapter 7 is about pictures of equations, and drawing one takes a way to name a single point by numbers. Two number lines set at right angles are enough. One runs left and right, the other up and down, so every point has two counts, how far along and how far up. The first problem reads those counts off a walk, the second turns on their order.
Problem
The pair naming a point lists the count along first and the count up second, so a place 6 units along and 4 units up from where the two number lines cross is written \((6,4)\). A beetle starts at that crossing point, walks 11 units along, turns around and walks 3 units back the way it came, then walks 5 units up. Enter the first number of the pair naming where the beetle stops.
Show a hint
- The first number of the pair is how far along the beetle ends up, not how far it walked in total. Work that count out from the moves along and leave the 5 units up out of it.
- Walking 3 units back the way it came undoes part of the 11 units out, so the ending count along is smaller than 11.
Show the full solution
The beetle ends up \(11-3=8\) units along and \(5\) units up, so its pair is \((8,5)\) and the first number is \(\boxed{8}\). Adding the two moves along instead of subtracting gives \(14\), which is the total distance walked along and not the place the beetle stops.
Problem
Three points are \((6,21)\), \((21,6)\) and \((13,27)\). One of them sits furthest to the right. Enter the second number of that point's pair.
Show a hint
- How far right a point sits comes from the first number of its pair, not the second. Compare the three first numbers before anything else.
- The first numbers are \(6\), \(21\) and \(13\), so one point is clearly furthest right. Read off that point's second number.
Show the full solution
The first number of a pair is how far along, and those are \(6\), \(21\) and \(13\), so \((21,6)\) sits furthest to the right and its second number is \(\boxed{6}\). Swapping the two roles would make \((13,27)\) look furthest right and give \(27\), which is why the order inside a pair is fixed.
Problem
A point sits \(6\) units to the left of the \(y\)-axis and \(3\) units below the \(x\)-axis. A second point sits \(10\) units further left, at the same height. Enter the \(x\)-coordinate of the second point.
Show a hint
- Fix both coordinates of the first point before moving anything, and remember that left of the \(y\)-axis puts its \(x\)-coordinate below zero.
- Moving further left subtracts, so the second \(x\)-coordinate is \(10\) less than the first. The height is the same for both points and is not needed here.
Show the full solution
The first point is \(6\) to the left, so \(x=-6\), and \(3\) below, so \(y=-3\). Moving \(10\) further left subtracts \(10\), giving \(-6-10=\boxed{-16}\). Further left always means a smaller \(x\)-coordinate. Adding the \(10\) instead gives \(4\), which sits on the other side of the \(y\)-axis entirely.
Problem
One of the points \((0,-19)\) and \((-19,0)\) lies on the \(x\)-axis, and the other lies on the \(y\)-axis. Enter the \(y\)-coordinate of the one that lies on the \(y\)-axis.
Show a hint
- The origin \((0,0)\) lies on the \(y\)-axis, and so does every point directly above or below it. Work out what those points all share in their first coordinate.
- Both pairs contain a \(0\), so the position of that \(0\) is what decides. A first coordinate of \(0\) means \(x=0\), which is the \(y\)-axis, and a second coordinate of \(0\) means \(y=0\), which is the \(x\)-axis.
Show the full solution
Every point on the \(y\)-axis has \(x\)-coordinate \(0\), so \((0,-19)\) is the one on the \(y\)-axis, and its \(y\)-coordinate is \(\boxed{-19}\). The common mistake is the reversal, since a coordinate of \(0\) puts a point on the axis named by the other letter. So \(x=0\) gives the \(y\)-axis, \(y=0\) gives the \(x\)-axis, and picking \((-19,0)\) would have given \(0\).
Those four regions are the quadrants, numbered counterclockwise from the upper right, as in the figure. Which side of the \(y\)-axis a point is on depends only on the sign of its \(x\)-coordinate, and which side of the \(x\)-axis only on the sign of its \(y\)-coordinate. The next two problems need nothing beyond those two signs.
Problem
A point \((m,n)\) has \(m<0\) and \(n>0\), and nothing else about the two values is known. Enter the number of the quadrant the point lies in.
Show a hint
- A negative \(x\)-coordinate means the point is left of the \(y\)-axis, and a positive \(y\)-coordinate means it is above the \(x\)-axis. Name that corner of the plane first.
- Left and above is the upper left region. Count counterclockwise from the upper right to reach its number.
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Negative \(x\) means left of the \(y\)-axis and positive \(y\) means above the \(x\)-axis, so the point is in the upper left region, quadrant \(\boxed{2}\). Neither value is needed, only the two signs. Every point with that sign pattern is in the same quadrant, however large or small \(m\) and \(n\) are.
Problem
The point \((a,b)\) lies in quadrant two, and nothing else about \(a\) and \(b\) is known. Enter the number of the quadrant that \((b,a)\) lies in.
Show a hint
- Get the sign of \(a\) and the sign of \(b\) from quadrant two first, which is left of the \(y\)-axis and above the \(x\)-axis.
- So \(a\) is negative and \(b\) is positive. In \((b,a)\) the positive number is the \(x\)-coordinate and the negative one is the \(y\)-coordinate.
Show the full solution
Quadrant two is left of the \(y\)-axis and above the \(x\)-axis, so \(a<0\) and \(b>0\). Then \((b,a)\) has a positive \(x\)-coordinate and a negative \(y\)-coordinate, which is right of the \(y\)-axis and below the \(x\)-axis, quadrant \(\boxed{4}\). The quadrant depends on the signs alone, so no values for \(a\) and \(b\) are needed. In \((a,b)\) the negative number is first and in \((b,a)\) it is second, so the two points lie in different quadrants.
Any two points have a distance between them, and when they share a coordinate that distance is already a length on a single number line. Two points at the same height differ only in how far along they sit, so the gap between them is the difference of their \(x\)-coordinates. A distance is never negative, so subtract the smaller from the larger.
Problem
The points \((-12,5)\) and \((21,5)\) have the same \(y\)-coordinate, so the segment joining them is horizontal. Enter its length.
Show a hint
- Only the \(x\)-values differ, so the length is the distance from \(-12\) to \(21\) on the number line. Counting across \(0\) is a safe way to do it.
- The length is \(21-(-12)\). Subtracting a negative adds, so the result is larger than \(21\), not smaller.
Show the full solution
Both endpoints have \(y=5\), so the length is the difference of the \(x\)-coordinates, \(21-(-12)=21+12=\boxed{33}\). Computing \(21-12=9\) instead is what comes of dropping the sign on \(-12\). One endpoint is \(12\) units left of the \(y\)-axis and the other \(21\) units right, so those two distances add.
Problem
The points \(A=(2,-1)\) and \(B=(14,4)\) have no coordinate in common, so a single subtraction will not give the distance. The point \(C=(14,-1)\) lies directly to the right of \(A\) and directly below \(B\), so triangle \(ABC\) has a right angle at \(C\), one leg horizontal and one leg vertical. Find the two leg lengths, then use the Pythagorean theorem. Enter the distance from \(A\) to \(B\).
Show a hint
- A horizontal segment has length equal to the difference of the two \(x\)-coordinates, and a vertical segment the difference of the two \(y\)-coordinates. Take each difference in the order that makes it positive, and watch the subtraction of \(-1\).
- The legs come out \(12\) and \(5\). The theorem gives the square of \(AB\) as the sum of the squares of those two.
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The horizontal leg is \(14-2=12\) and the vertical leg is \(4-(-1)=5\), so \(AB^2=12^2+5^2=144+25=169\) and \(AB=\sqrt{169}=\boxed{13}\). One leg is horizontal and the other vertical, so the angle between them is right and the theorem applies. Adding the legs gives \(17\), the length of the path through \(C\) rather than the direct distance.
The point halfway along the segment joining two points is their midpoint. On a single number line, halfway between two numbers is their average. How far along a point sits and how high it sits are each measured on a number line of their own, so averaging the two coordinates separately is worth a try, and the next two problems check it.
Problem
A segment has endpoints \((-6,-5)\) and \((10,19)\). Enter the \(y\)-coordinate of the midpoint of that segment.
Show a hint
- The midpoint has two coordinates and only one of them is wanted here. The \(-6\) and the \(10\) give the \(x\)-coordinate of the midpoint, and the \(y\)-coordinate comes from the other two numbers.
- The \(y\)-coordinate of the midpoint is the average of \(-5\) and \(19\). Add the two heights first, keeping the sign on \(-5\), then halve the total.
Show the full solution
The \(y\)-coordinate of the midpoint is the average of the two endpoint heights, \(\frac{-5+19}{2}=\frac{14}{2}=\boxed{7}\). Counting from the lower end gives the same value, since \(19-(-5)=24\) and half of that added to \(-5\) is \(-5+12=7\).
Problem
The midpoint of the segment joining \(A=(-3,9)\) and \(B\) is \(M=(5,-4)\). Enter the \(y\)-coordinate of \(B\).
Show a hint
- The midpoint's \(y\)-coordinate is the average of the two endpoints' \(y\)-coordinates. Here that average is known and one of the two numbers being averaged is not.
- Call the \(y\)-coordinate of \(B\) simply \(y\) and write \(\frac{9+y}{2}=-4\), then multiply both sides by \(2\) to clear the fraction and solve.
Show the full solution
Write \(B=(x,y)\). The midpoint's \(y\)-coordinate is the average of the two endpoints', so \(\frac{9+y}{2}=-4\), giving \(9+y=-8\) and \(y=\boxed{-17}\). As a check, the drop from \(9\) to \(-4\) is \(13\), and another \(13\) below \(-4\) is \(-17\). Averaging the two given heights gives \(\frac{9+(-4)}{2}=\frac{5}{2}\), the height halfway from \(A\) to \(M\) rather than the height of \(B\).
Everything in this lesson comes from the two coordinates of a point. Naming a quadrant takes only their signs, a distance between two points takes the two differences, and a midpoint takes the two averages. The last two problems use more than one of the three at once.
Problem
A segment runs from \(A=(-7,-2)\) to \(B=(5,14)\), and \(M\) is the midpoint of that segment. Enter the distance from \(A\) to \(M\).
Show a hint
- The midpoint is halfway along, so its distance from an endpoint is half the whole length. Finding the whole length first avoids computing the midpoint at all.
- The two differences from \(A\) to \(B\) are \(12\) and \(16\). Put those through the distance formula, then halve.
Show the full solution
The differences are \(5-(-7)=12\) and \(14-(-2)=16\), so the whole length is \(\sqrt{12^2+16^2}=\sqrt{400}=20\), and half of it is \(\boxed{10}\). The other route agrees, since the midpoint is \((-1,6)\) and \(\sqrt{6^2+8^2}=\sqrt{100}=10\). Each difference from \(A\) to \(M\) is half the matching difference from \(A\) to \(B\), so halving the whole length gives the same value.
Problem
The midpoint of the segment joining \(A=(-3,7)\) and \(B\) lies on the \(x\)-axis, and the \(x\)-coordinate of \(B\) is \(45\). Enter the distance from \(A\) to \(B\).
Show a hint
- Only the second coordinate of \(B\) is unknown, so write \(B=(45,k)\). A point of the \(x\)-axis has \(y\)-coordinate \(0\), so the midpoint's second coordinate is known even though the midpoint is not. That is one equation for \(k\).
- The two second coordinates average to \(0\), so \(\frac{7+k}{2}=0\). Once \(B\) is known, the two differences are \(48\) and \(-14\).
Show the full solution
Write \(B=(45,k)\). A point of the \(x\)-axis has \(y\)-coordinate \(0\), so \(\frac{7+k}{2}=0\), giving \(k=-7\) and \(B=(45,-7)\). The differences are \(45-(-3)=48\) and \(-7-7=-14\), so the distance is \(\sqrt{48^2+(-14)^2}=\sqrt{2304+196}=\sqrt{2500}=\boxed{50}\). Setting \(x=0\) instead puts the zero on the wrong coordinate, and the \(x\)-coordinate of \(B\) is already given.
The plane is built. Every ordered pair names one point, every point has exactly one pair, and both the distance formula and the midpoint formula come out of that naming alone. Lesson 7.2, Graphing Linear Equations, is about the pairs that make an equation true.
Practice these ideas
Practice
A point lies \(24\) units to the left of the \(y\)-axis and \(6\) units above the \(x\)-axis. Enter its \(x\)-coordinate.
Show the solution
The horizontal position is the \(x\)-coordinate, and \(24\) units left of the \(y\)-axis is \(24\) units in the negative direction, so the \(x\)-coordinate is \(\boxed{-24}\). The \(6\) units above the \(x\)-axis is the vertical position, so the point is \((-24,6)\).
Practice
The points \((20,-11)\) and \((-11,20)\) are built from the same two numbers in opposite orders. Exactly one of them sits above the \(x\)-axis. Enter the \(x\)-coordinate of that point.
Show the solution
The second numbers are \(-11\) and \(20\), so \((-11,20)\) is the point above the \(x\)-axis, and its \(x\)-coordinate is \(\boxed{-11}\). Answering \(20\) means reading the pair in the wrong order, since \(20\) is that point's \(y\)-coordinate.
Practice
A point lies on the \(y\)-axis, \(26\) units above the origin. Enter its \(x\)-coordinate.
Show the solution
A point on the \(y\)-axis sits directly above or below the origin with no horizontal step, so its \(x\)-coordinate is \(\boxed{0}\). The \(26\) is the vertical distance, so the point is \((0,26)\). The common slip is to read \(y\)-axis as \(y=0\), but the coordinate that is \(0\) there is \(x\).
Practice
Seven points are \((-9,2)\), \((0,7)\), \((3,-4)\), \((-1,-1)\), \((8,0)\), \((-5,-6)\) and \((2,9)\). Enter how many of them lie inside a quadrant rather than on an axis.
Show the solution
Of the seven, \((0,7)\) and \((8,0)\) have a zero coordinate and so are on an axis, which leaves \(\boxed{5}\). A point is on an axis whenever either of its coordinates is zero, and no point on an axis is inside a quadrant.
Practice
Enter the number of the quadrant containing the point \((-2.5,-40)\).
Show the solution
The first coordinate is negative, so the point is left of the \(y\)-axis, and the second is negative, so the point is below the \(x\)-axis. That is the lower left region, quadrant \(\boxed{3}\). The sizes \(2.5\) and \(40\) make no difference, since a quadrant depends only on the two signs.
Practice
The product of a point's two coordinates is positive, and its \(x\)-coordinate is positive. Enter the number of the quadrant the point lies in.
Show the solution
A positive product means the two coordinates share a sign, and the \(x\)-coordinate is positive, so the \(y\)-coordinate is positive as well. Right of the \(y\)-axis and above the \(x\)-axis is quadrant \(\boxed{1}\). Without the fact about \(x\), both coordinates could be negative instead, which is quadrant three.
Practice
The point \((c+5,\ c+12)\) lies on the \(x\)-axis. Enter \(c\).
Show the solution
A point on the \(x\)-axis has \(y\)-coordinate \(0\), so \(c+12=0\) and \(c=\boxed{-12}\). The point is then \((-7,0)\). Setting the first entry to zero instead gives \(c=-5\), which puts the point on the \(y\)-axis rather than the \(x\)-axis.
Practice
Enter the distance between \((-8,-6)\) and \((15,-6)\).
Show the solution
Both points have \(y=-6\), so the segment is horizontal and the distance is \(15-(-8)=15+8=\boxed{23}\). When two points share a coordinate, only the other pair of coordinates matters, and the distance is the larger minus the smaller.
Practice
Enter the distance between \((7,-21)\) and \((7,9)\).
Show the solution
Both points have \(x=7\), so the distance is \(9-(-21)=9+21=\boxed{30}\). A shared \(x\)-coordinate means a vertical segment, and a shared \(y\)-coordinate means a horizontal one.
Practice
Enter the distance between \((2,-3)\) and \((26,4)\).
Show the solution
The differences are \(26-2=24\) and \(4-(-3)=7\), so the distance is \(\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=\boxed{25}\). Reading \(4-(-3)\) as \(1\) is the usual slip, and it gives \(\sqrt{577}\) instead.
Practice
Enter the distance from the origin to the point \((-9,-40)\).
Show the solution
Taking \((x_1,y_1)=(0,0)\), the differences are \(-9\) and \(-40\), so the distance is \(\sqrt{(-9)^2+(-40)^2}=\sqrt{81+1600}=\sqrt{1681}=\boxed{41}\). With the origin as the first point, \(x_1\) and \(y_1\) are both zero, so the legs of the right triangle have lengths \(9\) and \(40\), the sizes of the point's own coordinates.
Practice
The distance between \((-5,3)\) and \((4,21)\) simplifies to \(a\sqrt{5}\), where \(a\) is a whole number. Enter \(a\).
Show the solution
The differences are \(4-(-5)=9\) and \(21-3=18\), so the distance is \(\sqrt{9^2+18^2}=\sqrt{81+324}=\sqrt{405}\). Since \(405=81\cdot 5\), that is \(\sqrt{81}\sqrt{5}=9\sqrt{5}\), so \(a=\boxed{9}\). Splitting \(405\) as \(9\cdot 45\) gives \(3\sqrt{45}\), the same number in a form that is not simplified, so take the largest perfect square factor, \(81\).
Practice
Enter the \(x\)-coordinate of the midpoint of the segment joining \((-13,-20)\) and \((49,6)\).
Show the solution
The \(x\)-coordinates average to \(\frac{-13+49}{2}=\frac{36}{2}=\boxed{18}\). Each coordinate of the midpoint is averaged on its own, so the \(-20\) and the \(6\) are used only for the midpoint's \(y\)-coordinate, which is \(-7\).
Practice
Enter the \(y\)-coordinate of the midpoint of the segment joining \((6,-3)\) and \((20,10)\).
Show the solution
The \(y\)-coordinates average to \(\frac{-3+10}{2}=\boxed{\frac{7}{2}}\). With whole-number coordinates, the average is a whole number only when the two are both even or both odd, so a fraction here is correct rather than a sign of a slip.
Practice
The segment with endpoints \(P=(-7,3)\) and \(Q\) has midpoint \((4,-6)\). Enter the \(x\)-coordinate of \(Q\).
Show the solution
Writing \(Q=(q,r)\), the \(x\)-coordinates average to \(4\), so \(\frac{-7+q}{2}=4\), giving \(-7+q=8\) and \(q=\boxed{15}\). As a check, \(-7\) to \(4\) is a step of \(11\), and \(4+11=15\).
Practice
The midpoint of the segment joining \((-31,4)\) and \((t,10)\) lies on the \(y\)-axis. Enter \(t\).
Show the solution
A point on the \(y\)-axis has \(x\)-coordinate \(0\), so \(\frac{-31+t}{2}=0\), giving \(-31+t=0\) and \(t=\boxed{31}\). The \(y\)-axis is the line \(x=0\), not \(y=0\), so only the first coordinates matter here. Averaging \(4\) and \(10\) instead gives \(7\) for every \(t\), so that reading has no solution.
Practice
A segment has one endpoint at \((-6,1)\) and its midpoint at \((14,22)\). Enter the length of the whole segment.
Show the solution
The differences from \((-6,1)\) to \((14,22)\) are \(20\) and \(21\), so half the segment is \(\sqrt{20^2+21^2}=\sqrt{400+441}=\sqrt{841}=29\) and the whole segment is \(\boxed{58}\). Working the long way gives the same number. The other endpoint is \((34,43)\), and \(\sqrt{40^2+42^2}=\sqrt{3364}=58\).