NCERT Solutions · Class 9 Mathematics Orienting Yourself: The Use of Coordinates

Use what each axis is a set of.The x-axis is the set of all points whose distance from the x-axis is zero, that is, all points of the form \(\displaystyle (x, 0)\); every point on it has y-coordinate 0.The y-axis is the set of all points of the form \(\displaystyle (0, y)\); every point on it has x-coordinate 0.The point where the two axes cross lies on both lines at once. So it must satisfy both conditions at the same time: its y-coordinate is $\displaystyle 0$ (because it is on the x-axis) and its x-coordinate is $\displaystyle 0$ (because it is on the y-axis). Only one point can do that.\[\text{point of intersection} = (0,\ 0). \]This point is given a special name: the origin, O. It is the point from which all distances along both axes are measured, which is why both of its coordinates have to be zero - it is zero units from each axis.Answer: the x-coordinate is $\displaystyle 0$ and the y-coordinate is $\displaystyle 0$; the point of intersection of the two axes is the origin O$\displaystyle (0, 0)$.

A line parallel to the y-axis is a line of constant x.W has x-coordinate \(\displaystyle -5\), so \(\displaystyle \mathrm{W} = (-5,\ w)\) for some number \(\displaystyle w\) that we are not told.The y-axis is vertical, so a line parallel to the y-axis is vertical too. Travelling along a vertical line changes only how high you are; it never changes how far left or right of the y-axis you are. Since the x-coordinate of a point is its distance from the y-axis (with a sign for the side), every point on that line has the same x-coordinate as W, namely \(\displaystyle -5\).\[\mathrm{H} = (-5,\ k) \quad \text{for some number } k. \]So we can predict one coordinate exactly and not the other. The x-coordinate must be \(\displaystyle -5\). The y-coordinate \(\displaystyle k\) cannot be predicted, because the question tells us which line H is on but not where along it H sits - H could be anywhere on that vertical line.Which quadrants can H lie in? Its x-coordinate is \(\displaystyle -5\), which is negative, so H is on the left of the y-axis. Of the four quadrants, only two lie on the left: H can never be in Quadrant I or Quadrant IV, since those need a positive x-coordinate.Answer: \(\displaystyle \mathrm{H} = (-5, k)\) - its x-coordinate must be \(\displaystyle -5\), but its y-coordinate cannot be predicted. H lies in Quadrant II if its y-coordinate is positive and in Quadrant III if it is negative; if the y-coordinate is $\displaystyle 0$ then H is the point \(\displaystyle (-5,0)\) on the x-axis, which is in no quadrant.

Predict from the coordinates first, then plot to confirm.The four points are \(\displaystyle \mathrm{R}(3,0)\), \(\displaystyle \mathrm{A}(0,-2)\), \(\displaystyle \mathrm{M}(-5,-2)\), \(\displaystyle \mathrm{P}(-5,2)\), joined in that order, so the four sides are RA, AM, MP and PR.(i) Two sides perpendicular to each other.Look for equal coordinates. A\(\displaystyle (0,-2)\) and M\(\displaystyle (-5,-2)\) share the same y-coordinate*, \(\displaystyle -2\). Points with the same y-coordinate lie on a line parallel to the x-axis, so AM is horizontal, with \(\displaystyle \mathrm{AM} = |0-(-5)| = 5\). M\(\displaystyle (-5,-2)\) and P\(\displaystyle (-5,2)\) share the same x-coordinate*, \(\displaystyle -5\), so MP is vertical, with \(\displaystyle \mathrm{MP} = 2-(-2) = 4\).A horizontal line and a vertical line always meet at a right angle, and AM and MP meet at M. So \(\displaystyle \mathrm{AM} \perp \mathrm{MP}\), and the right angle is at M.The other two sides are slanting, and they are not perpendicular to anything here. You can test any corner with the Baudhāyana-Pythagoras theorem, since a right angle at a corner would force (leg)\(\displaystyle ^2\) + (leg)\(\displaystyle ^2\) = (opposite side)\(\displaystyle ^2\). For instance at R, \(\displaystyle \mathrm{RA}^2 = 3^2+2^2 = 13\) and \(\displaystyle \mathrm{RP}^2 = 8^2+2^2 = 68\), while \(\displaystyle \mathrm{AP}^2 = 5^2+4^2 = 41\); and \(\displaystyle 13 + 68 = 81 \neq 41\), so the angle at R is not a right angle. So M is the only right angle.(ii) One side parallel to one of the axes.From the same observation: AM is parallel to the x-axis (both A and M have \(\displaystyle y=-2\), so the whole side lies on the line \(\displaystyle y=-2\)). Equally, MP is parallel to the y-axis (both have \(\displaystyle x=-5\)), so either side may be quoted.(iii) Two points that are mirror images of each other in one axis.Reflecting in the x-axis sends \(\displaystyle (x,y)\) to \(\displaystyle (x,-y)\): the point flips to the other side, keeping the same x. Reflecting in the y-axis sends \(\displaystyle (x,y)\) to \(\displaystyle (-x,y)\). So look for two of the four points that agree in one coordinate and are opposite in the other.\[\mathrm{M}(-5,\,-2) \quad \text{and} \quad \mathrm{P}(-5,\,2): \ \text{same } x = -5,\ \text{opposite } y. \]So M and P are mirror images, and the mirror is the x-axis. (Check: M is $\displaystyle 2$ units below the x-axis and P is $\displaystyle 2$ units above it, both on the same vertical line \(\displaystyle x=-5\).)Now plot and verify. Mark R\(\displaystyle (3,0)\) on the positive x-axis, A\(\displaystyle (0,-2)\) on the negative y-axis, M\(\displaystyle (-5,-2)\) in Quadrant III and P\(\displaystyle (-5,2)\) in Quadrant II. Joining R \(\displaystyle \to\) A \(\displaystyle \to\) M \(\displaystyle \to\) P \(\displaystyle \to\) R you can see all three predictions at once: AM runs flat along \(\displaystyle y=-2\), MP runs straight up the line \(\displaystyle x=-5\), they meet in a square corner at M, and M and P sit at equal distances below and above the x-axis, so folding the paper along the x-axis lands one exactly on the other.Answer: (i) AM \(\displaystyle \perp\) MP, the right angle being at M. (ii) AM is parallel to the x-axis (and MP is parallel to the y-axis). (iii) M\(\displaystyle (-5,-2)\) and P\(\displaystyle (-5,2)\) are mirror images of each other in the x-axis.

Build the right angle out of the directions of the axes.(The question says answers may differ from person to person - this is one correct choice; yours may be different and still be right.)There is a way to be certain of a right angle without measuring anything: make one side of the triangle parallel to the x-axis and the other side parallel to the y-axis. Those two directions are perpendicular, because the axes are drawn at right angles to each other. Two points have a horizontal join exactly when their y-coordinates are equal, and a vertical join exactly when their x-coordinates are equal.Plot \(\displaystyle \mathrm{Z}(5,-6)\). Its x-coordinate is positive and its y-coordinate is negative, so Z lies in Quadrant IV: $\displaystyle 5$ units right of the y-axis, $\displaystyle 6$ units below the x-axis.Now put the right angle at Z: Since ZI is vertical and ZN is horizontal, \(\displaystyle \angle \mathrm{IZN} = 90^\circ\), so IZN is right-angled at Z.The three side lengths.Both ZI and ZN join points that share a coordinate, so each is just a difference:\[\mathrm{ZI} = |-2 - (-6)| = 4\ \text{units}, \qquad \mathrm{ZN} = |8 - 5| = 3\ \text{units}. \]For the hypotenuse IN, use the distance formula:\[\mathrm{IN} = \sqrt{(8-5)^2 + \big(-6-(-2)\big)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5\ \text{units}. \]Check with the Baudhāyana-Pythagoras theorem: \(\displaystyle 3^2 + 4^2 = 9 + 16 = 25 = 5^2\). It fits, as it must.Other answers are equally correct. For example, taking \(\displaystyle \mathrm{I}(5,0)\) and \(\displaystyle \mathrm{N}(11,-6)\) gives \(\displaystyle \mathrm{ZI}=6\), \(\displaystyle \mathrm{ZN}=6\) and \(\displaystyle \mathrm{IN}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\); the triangle is still right-angled at Z. The lengths will differ from person to person - what must be true for everybody is that the squares of the two shorter sides add up to the square of the longest side.Answer: with \(\displaystyle \mathrm{Z}(5,-6)\), \(\displaystyle \mathrm{I}(5,-2)\) and \(\displaystyle \mathrm{N}(8,-6)\), the triangle is right-angled at Z with \(\displaystyle \mathrm{ZI}=4\), \(\displaystyle \mathrm{ZN}=3\) and \(\displaystyle \mathrm{IN}=5\) units (and \(\displaystyle 3^2+4^2=5^2\)). Any other choice of I and N giving a right angle - most easily by keeping one side vertical and the other horizontal - is equally acceptable.

What the minus sign is actually doing.In the system we have built, a coordinate carries two pieces of information at once: how far from an axis, and which side. The size of the number gives the distance; the sign gives the side - right or left of the y-axis for x, above or below the x-axis for y.Now suppose negative numbers did not exist. Then every coordinate would have to be $\displaystyle 0$ or positive, and a pair \(\displaystyle (x,y)\) with \(\displaystyle x \ge 0\) and \(\displaystyle y \ge 0\) names exactly those points that are on or to the right of the y-axis and on or above the x-axis. Those points make up\[\text{Quadrant I} \ +\ \text{the positive halves of the two axes} \ +\ \text{the origin}, \]which is only one quarter of the plane. It would be a single-quadrant system: a corner, not a whole plane.So points such as \(\displaystyle \mathrm{Q}(-5,3)\) in Quadrant II, or a point $\displaystyle 5$ units left and $\displaystyle 3$ units below O, would simply have no name. Two different points - one $\displaystyle 3$ units right of O and one $\displaystyle 3$ units left of O, both on the x-axis - would both have to be called "$\displaystyle 3$", and we could not tell them apart.No, such a system could not locate all the points of a $\displaystyle 2$-D plane.There are two ways round it, and they are worth noticing.1. Move the origin. Put O at one corner of the region you care about, so that everything you want to describe already lies up and to the right. This is exactly how a seating chart, a chess board, or the street grid of the Sindhu-Sarasvati cities works when measured from one corner. It works, but only for a bounded piece of the plane, and you must know in advance where the corner is. 2. Attach a direction word to every number, for example "$\displaystyle 5$ left, $\displaystyle 3$ down". This does work for the whole plane - but writing a direction beside each number is clumsy, and you cannot then calculate with the coordinates.A negative sign is just the second idea written compactly, in a form you can also do arithmetic with. That is why Brahmagupta's treatment of negative numbers as genuine algebraic quantities had to come before the four-quadrant plane could exist at all.Answer: without negative numbers we would have only the first quadrant - a quarter-plane of points on or above the x-axis and on or to the right of the y-axis. It would let us locate points in that corner only, so no, it would not allow us to locate all the points of a $\displaystyle 2$-D plane; the negative sign is what records which side of each axis a point lies on.

Collinearity by distances: the longest must equal the sum of the other two.The method (no plotting needed). Take any three points. If they are not on one line they form a triangle, and in a triangle every side is strictly shorter than the sum of the other two - a detour through a third corner is always longer than going straight. If they are on one line, one of them lies between the other two, and the journey from the first to the last passes through the middle one, so the longest distance is exactly the sum of the other two. So:\[\text{three points are collinear} \iff \text{(largest distance)} = \text{(sum of the other two distances)}. \]All three distances come from the Baudhāyana-Pythagoras distance formula, so this is pure arithmetic.Applying it to M\(\displaystyle (-3,-4)\), A\(\displaystyle (0,0)\), G\(\displaystyle (6,8)\).\[\mathrm{MA} = \sqrt{\big(0-(-3)\big)^2 + \big(0-(-4)\big)^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5. \]\[\mathrm{AG} = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36+64} = \sqrt{100} = 10. \]\[\mathrm{MG} = \sqrt{\big(6-(-3)\big)^2 + \big(8-(-4)\big)^2} = \sqrt{9^2 + 12^2} = \sqrt{81+144} = \sqrt{225} = 15. \]The largest is MG. And\[\mathrm{MA} + \mathrm{AG} = 5 + 10 = 15 = \mathrm{MG}. \]They add up exactly, so the three points are collinear, with A lying between M and G.A second method, often quicker: compare the steps. Instead of distances, look at how you travel from one point to the next. The second step is the first step doubled: same direction, just twice as far. Since you never turn at A, all three points lie on one straight line. (If the two steps had not been multiples of each other - say \(\displaystyle (3,4)\) then \(\displaystyle (6,7)\) - the direction would have changed at A and the points would not be collinear.) This method uses only whole-number arithmetic, with no square roots at all.Answer: yes, M, A and G lie on the same straight line, with A between M and G. Method: use the distance formula to get MA \(\displaystyle =5\), AG \(\displaystyle =10\), MG \(\displaystyle =15\) and check that the largest equals the sum of the other two, \(\displaystyle 5+10=15\); or, more simply, check that the step from M to A, \(\displaystyle (3,4)\), and the step from A to G, \(\displaystyle (6,8)\), are multiples of each other, so the direction of travel never changes.

The same two tests - and this time the plot cannot decide it.The points are \(\displaystyle \mathrm{R}(-5,-1)\), \(\displaystyle \mathrm{B}(-2,-5)\), \(\displaystyle \mathrm{C}(4,-12)\).Distance test.\[\mathrm{RB} = \sqrt{\big(-2-(-5)\big)^2 + \big(-5-(-1)\big)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = \sqrt{25} = 5. \]\[\mathrm{BC} = \sqrt{\big(4-(-2)\big)^2 + \big(-12-(-5)\big)^2} = \sqrt{6^2+(-7)^2} = \sqrt{36+49} = \sqrt{85}. \]\[\mathrm{RC} = \sqrt{\big(4-(-5)\big)^2 + \big(-12-(-1)\big)^2} = \sqrt{9^2+(-11)^2} = \sqrt{81+121} = \sqrt{202}. \]Numerically \(\displaystyle \sqrt{85} \approx 9.2195\) and \(\displaystyle \sqrt{202} \approx 14.2127\), so\[\mathrm{RB} + \mathrm{BC} \approx 5 + 9.2195 = 14.2195, \qquad \mathrm{RC} \approx 14.2127. \]These are not equal, so the points are not collinear - but they differ by only about \(\displaystyle 0.007\) of a unit, far less than the thickness of a pencil line.We can settle it exactly, without decimals. Suppose \(\displaystyle 5 + \sqrt{85} = \sqrt{202}\) were true. Squaring both sides, \[25 + 10\sqrt{85} + 85 = 202 \ \Longrightarrow\ 10\sqrt{85} = 92 \ \Longrightarrow\ \sqrt{85} = 9.2 \ \Longrightarrow\ 85 = 84.64, \] which is false. So \(\displaystyle \mathrm{RB} + \mathrm{BC} \neq \mathrm{RC}\), for certain.Step test (cleaner, and it needs no square roots at all). If the direction had not changed at B, then a step twice as long sideways (\(\displaystyle 6 = 2\times 3\)) would have to be twice as long vertically too, that is \(\displaystyle (6,-8)\). It is \(\displaystyle (6,-7)\), one unit short. So the direction does change at B, and R, B, C are not collinear.Now plot both sets and check. R\(\displaystyle (-5,-1)\) and B\(\displaystyle (-2,-5)\) are in Quadrant III and C\(\displaystyle (4,-12)\) is in Quadrant IV. On the graph the three look as good as straight - the bend at B is a single unit spread over a journey of more than $\displaystyle 14$ units, so the eye cannot see it. Plotting M\(\displaystyle (-3,-4)\), A\(\displaystyle (0,0)\), G\(\displaystyle (6,8)\) from Problem $\displaystyle 6$, by contrast, gives a line that is genuinely straight, matching \(\displaystyle 5+10=15\) exactly.That contrast is the real lesson of this pair of questions: a picture can suggest collinearity, but only the calculation can prove it or disprove it.Answer: no - R, B and C are not on the same straight line. RB \(\displaystyle =5\), BC \(\displaystyle =\sqrt{85}\approx 9.22\), RC \(\displaystyle =\sqrt{202}\approx 14.21\), and \(\displaystyle \mathrm{RB}+\mathrm{BC}\approx 14.22 \neq \mathrm{RC}\); equivalently, the step \(\displaystyle (3,-4)\) from R to B would have to be followed by \(\displaystyle (6,-8)\), but the actual step from B to C is \(\displaystyle (6,-7)\). They form a very flat triangle, which is exactly why plotting alone cannot settle it.

Let the axes do the work.(Both parts have many correct answers; these are worked examples, and a different choice that meets the conditions is equally right.)(i) A right-angled isosceles triangle with a vertex at the origin.Such a triangle needs a right angle and two equal sides. The easiest way to guarantee the right angle is to lay the two equal sides along the axes, since the axes are perpendicular to each other by construction. Take\[\mathrm{O}(0,0), \qquad \mathrm{A}(4,0), \qquad \mathrm{B}(0,4). \] Check: \(\displaystyle 4^2 + 4^2 = 32 = (4\sqrt2)^2\). So OAB is right-angled at O and isosceles.Another valid answer, with the right angle not at the origin: \(\displaystyle \mathrm{O}(0,0)\), \(\displaystyle (3,3)\), \(\displaystyle (6,0)\). Here the two sides from \(\displaystyle (3,3)\) both have length \(\displaystyle \sqrt{3^2+3^2}=\sqrt{18}\), the third side is $\displaystyle 6$, and \(\displaystyle 18+18=36=6^2\), so the right angle is at \(\displaystyle (3,3)\).(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.The origin is the third vertex. Quadrant III needs both coordinates negative; Quadrant IV needs x positive and y negative. A neat way to force the two sides from O to be equal is to choose the two vertices as mirror images in the y-axis: reflecting in the y-axis does not change a point's distance from O, so the two distances come out equal automatically. Take\[\mathrm{O}(0,0), \qquad \mathrm{P}(-3,-4)\ \text{in Quadrant III}, \qquad \mathrm{Q}(3,-4)\ \text{in Quadrant IV}. \]\[\mathrm{OP} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9+16} = 5, \qquad \mathrm{OQ} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5. \]\[\mathrm{PQ} = \sqrt{\big(3-(-3)\big)^2 + \big(-4-(-4)\big)^2} = \sqrt{36+0} = 6. \]Since \(\displaystyle \mathrm{OP} = \mathrm{OQ} = 5\), the triangle is isosceles. It is a genuine triangle (not three points in a line), because P and Q lie on the line \(\displaystyle y = -4\) while O is $\displaystyle 4$ units above it.Answer: (i) for example \(\displaystyle \mathrm{O}(0,0)\), \(\displaystyle (4,0)\) and \(\displaystyle (0,4)\) - legs of $\displaystyle 4$ units along the two axes, right angle at O, hypotenuse \(\displaystyle 4\sqrt2\). (ii) for example \(\displaystyle \mathrm{O}(0,0)\), \(\displaystyle (-3,-4)\) in Quadrant III and \(\displaystyle (3,-4)\) in Quadrant IV, giving \(\displaystyle \mathrm{OP}=\mathrm{OQ}=5\) and \(\displaystyle \mathrm{PQ}=6\). Many other answers are correct; in (ii) any two points that are mirror images in the y-axis, one below-left and one below-right, will do.

The midpoint is the average of the two endpoints.The test to use. M is the midpoint of ST only if two things hold: M lies on the segment ST, and \(\displaystyle \mathrm{SM} = \mathrm{MT}\). Checking \(\displaystyle \mathrm{SM}=\mathrm{MT}\) alone is not enough - a point can be equally far from S and from T while sitting well off the line.The safest single check is arithmetic: work out the average of the endpoints, \[\left(\frac{x_\mathrm{S}+x_\mathrm{T}}{2},\ \frac{y_\mathrm{S}+y_\mathrm{T}}{2}\right), \] and see whether it is the given M. If it is, M is the midpoint; if it is not, M is not.Row by row.Row $\displaystyle 1$: S\(\displaystyle (-3,0)\), T\(\displaystyle (3,0)\). Average \(\displaystyle =\left(\frac{-3+3}{2}, \frac{0+0}{2}\right) = (0,0)\), which is M. All three points are on the x-axis, and \(\displaystyle \mathrm{SM}=3=\mathrm{MT}\). Yes.Row $\displaystyle 2$: S\(\displaystyle (2,3)\), T\(\displaystyle (4,5)\). Average \(\displaystyle =\left(\frac{2+4}{2}, \frac{3+5}{2}\right) = (3,4)\), which is M. Notice also that the step S \(\displaystyle \to\) M is \(\displaystyle (1,1)\) and the step M \(\displaystyle \to\) T is \(\displaystyle (1,1)\) - identical, so M is on ST and exactly halfway. Yes.Row $\displaystyle 3$: S\(\displaystyle (0,0)\), T\(\displaystyle (0,-10)\). Average \(\displaystyle =\left(0, \frac{0+(-10)}{2}\right) = (0,-5)\), but M is \(\displaystyle (0,5)\). No. All three are on the y-axis, but M is $\displaystyle 5$ units above S while T is $\displaystyle 10$ units below it, so M is not even between S and T; here \(\displaystyle \mathrm{SM}=5\) and \(\displaystyle \mathrm{MT}=15\).Row $\displaystyle 4$: S\(\displaystyle (-8,7)\), T\(\displaystyle (6,-3)\). Average \(\displaystyle =\left(\frac{-8+6}{2}, \frac{7+(-3)}{2}\right) = (-1,2)\), but M is \(\displaystyle (0,-2)\). No. (The distances are nowhere near equal either: \(\displaystyle \mathrm{SM}=\sqrt{8^2+9^2}=\sqrt{145}\approx 12.04\) while \(\displaystyle \mathrm{MT}=\sqrt{6^2+1^2}=\sqrt{37}\approx 6.08\).)

S	M	T	Is M the midpoint of ST?	Reason
\(\displaystyle (-3,0)\)	\(\displaystyle (0,0)\)	\(\displaystyle (3,0)\)	Yes	\(\displaystyle \left(\frac{-3+3}{2},\frac{0+0}{2}\right)=(0,0)=\mathrm{M}\); \(\displaystyle \mathrm{SM}=\mathrm{MT}=3\) on the x-axis
\(\displaystyle (2,3)\)	\(\displaystyle (3,4)\)	\(\displaystyle (4,5)\)	Yes	\(\displaystyle \left(\frac{2+4}{2},\frac{3+5}{2}\right)=(3,4)=\mathrm{M}\); equal steps \(\displaystyle (1,1)\) then \(\displaystyle (1,1)\)
\(\displaystyle (0,0)\)	\(\displaystyle (0,5)\)	\(\displaystyle (0,-10)\)	No	the midpoint is \(\displaystyle (0,-5)\), not \(\displaystyle (0,5)\); M is not even between S and T (\(\displaystyle \mathrm{SM}=5\), \(\displaystyle \mathrm{MT}=15\))
\(\displaystyle (-8,7)\)	\(\displaystyle (0,-2)\)	\(\displaystyle (6,-3)\)	No	the midpoint is \(\displaystyle (-1,2)\), not \(\displaystyle (0,-2)\); also \(\displaystyle \mathrm{SM}=\sqrt{145}\), \(\displaystyle \mathrm{MT}=\sqrt{37}\)

The connection between the coordinates. In the two "Yes" rows, each coordinate of M is the average of the corresponding coordinates of S and T:\[x_\mathrm{M} = \frac{x_\mathrm{S}+x_\mathrm{T}}{2}, \qquad y_\mathrm{M} = \frac{y_\mathrm{S}+y_\mathrm{T}}{2}. \]Why this must be so: travelling from S to T, the total sideways change is \(\displaystyle x_\mathrm{T}-x_\mathrm{S}\). Halfway along, you have made half of it, so \[x_\mathrm{M} = x_\mathrm{S} + \tfrac{1}{2}\left(x_\mathrm{T}-x_\mathrm{S}\right) = \tfrac{1}{2}\left(x_\mathrm{S}+x_\mathrm{T}\right), \] and the same argument on the vertical change gives the formula for \(\displaystyle y_\mathrm{M}\).Answer: Row $\displaystyle 1$ - Yes. Row $\displaystyle 2$ - Yes. Row $\displaystyle 3$ - No (the midpoint is \(\displaystyle (0,-5)\)). Row $\displaystyle 4$ - No (the midpoint is \(\displaystyle (-1,2)\)). The connection is that the midpoint's coordinates are the averages of the endpoints' coordinates: \(\displaystyle x_\mathrm{M}=\frac{x_\mathrm{S}+x_\mathrm{T}}{2}\) and \(\displaystyle y_\mathrm{M}=\frac{y_\mathrm{S}+y_\mathrm{T}}{2}\).

Read the midpoint rule backwards.From Problem $\displaystyle 9$, if M is the midpoint of AB then each coordinate of M is the average of the corresponding coordinates of A and B:\[x_\mathrm{M} = \frac{x_\mathrm{A}+x_\mathrm{B}}{2}, \qquad y_\mathrm{M} = \frac{y_\mathrm{A}+y_\mathrm{B}}{2}. \]Here we are told the answer of each average and one of the two numbers being averaged, so we can solve for the missing one.With \(\displaystyle \mathrm{M}(-7,1)\), \(\displaystyle \mathrm{A}(3,-4)\) and \(\displaystyle \mathrm{B}(x,y)\):x-coordinate. \[\frac{3+x}{2} = -7 \ \Longrightarrow\ 3+x = -14 \ \Longrightarrow\ x = -17. \]y-coordinate. \[\frac{-4+y}{2} = 1 \ \Longrightarrow\ -4+y = 2 \ \Longrightarrow\ y = 6. \]So \(\displaystyle \mathrm{B}(-17,\ 6)\).Check by going forwards again. The midpoint of A\(\displaystyle (3,-4)\) and B\(\displaystyle (-17,6)\) is \[\left(\frac{3+(-17)}{2},\ \frac{-4+6}{2}\right) = \left(\frac{-14}{2},\ \frac{2}{2}\right) = (-7,\ 1) = \mathrm{M}. \ \checkmark \]A second check, using distances: \(\displaystyle \mathrm{AM} = \sqrt{(-7-3)^2+(1-(-4))^2} = \sqrt{100+25} = \sqrt{125}\), and \(\displaystyle \mathrm{MB} = \sqrt{(-17-(-7))^2+(6-1)^2} = \sqrt{100+25} = \sqrt{125}\). Equal, as they must be.There is also a quick way to see the answer without algebra: to get from A to M you move \(\displaystyle -7-3 = -10\) in x and \(\displaystyle 1-(-4) = +5\) in y. Since M is halfway, you must repeat exactly the same step to reach B: \(\displaystyle -7-10 = -17\) and \(\displaystyle 1+5 = 6\).Answer: \(\displaystyle \mathrm{B} = (-17,\ 6)\).