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

Split the journey into three equal steps instead of two.The idea. The midpoint formula works because going from A to B is one journey: the total change is \(\displaystyle \big(x_\mathrm{B}-x_\mathrm{A},\ y_\mathrm{B}-y_\mathrm{A}\big)\), and the midpoint is where you have completed half of that change in x and half of it in y. Trisection points are the same idea with thirds. P, the point closer to A, is reached after one third of the journey; Q is reached after two thirds. So:\[\text{compute the total step } (\Delta x,\ \Delta y),\ \text{take one third of it, and add that step on twice.} \]Doing it for A\(\displaystyle (4,7)\) and B\(\displaystyle (16,-2)\).Total step from A to B: \[\Delta x = 16 - 4 = 12, \qquad \Delta y = -2 - 7 = -9. \]One third of the step: \[\left(\frac{12}{3},\ \frac{-9}{3}\right) = (4,\ -3). \]Add it once to reach P, and once more to reach Q: \[\mathrm{P} = (4+4,\ 7-3) = (8,\ 4), \qquad \mathrm{Q} = (8+4,\ 4-3) = (12,\ 1). \]And adding it a third time gives \(\displaystyle (16, -2) = \mathrm{B}\), which confirms the arithmetic.Written as a formula, this says \[\mathrm{P} = \left(\frac{2x_\mathrm{A}+x_\mathrm{B}}{3},\ \frac{2y_\mathrm{A}+y_\mathrm{B}}{3}\right), \qquad \mathrm{Q} = \left(\frac{x_\mathrm{A}+2x_\mathrm{B}}{3},\ \frac{y_\mathrm{A}+2y_\mathrm{B}}{3}\right), \] which for our points gives \(\displaystyle \mathrm{P}=\left(\frac{8+16}{3},\frac{14-2}{3}\right)=(8,4)\) and \(\displaystyle \mathrm{Q}=\left(\frac{4+32}{3},\frac{7-4}{3}\right)=(12,1)\), as before.Checking with midpoints - the link the question asks for. If P and Q really cut AB into three equal parts, then P must be the midpoint of A and Q, and Q must be the midpoint of P and B. Test both with the averaging rule:\[\text{midpoint of A}(4,7)\text{ and Q}(12,1) = \left(\frac{4+12}{2},\ \frac{7+1}{2}\right) = (8,\ 4) = \mathrm{P}\ \checkmark \]\[\text{midpoint of P}(8,4)\text{ and B}(16,-2) = \left(\frac{8+16}{2},\ \frac{4+(-2)}{2}\right) = (12,\ 1) = \mathrm{Q}\ \checkmark \]Checking with distances. Each of the three pieces should be the same length: \[\mathrm{AP} = \sqrt{4^2+(-3)^2} = 5, \quad \mathrm{PQ} = \sqrt{4^2+(-3)^2} = 5, \quad \mathrm{QB} = \sqrt{4^2+(-3)^2} = 5, \] and the whole segment is \(\displaystyle \mathrm{AB} = \sqrt{12^2+(-9)^2} = \sqrt{144+81} = \sqrt{225} = 15 = 3\times 5\). The three parts are equal and they add up to AB, so P and Q really are the points of trisection, with P nearer A.Answer: take one third of the total step from A to B and add it on twice. For A\(\displaystyle (4,7)\) and B\(\displaystyle (16,-2)\) the total step is \(\displaystyle (12,-9)\), one third of it is \(\displaystyle (4,-3)\), so \(\displaystyle \mathrm{P} = (8,\ 4)\) and \(\displaystyle \mathrm{Q} = (12,\ 1)\), with \(\displaystyle \mathrm{AP}=\mathrm{PQ}=\mathrm{QB}=5\) and \(\displaystyle \mathrm{AB}=15\).

A circle is the set of points at one fixed distance from its centre - so compare distances from O.(i) Showing A, B, C lie on one circle centred at O.If three points are all the same distance from O, then a circle drawn with centre O and that distance as radius passes through all three. So compute OA, OB and OC with the distance formula. A useful trick: compare the squares of the distances. That avoids square roots altogether and avoids any rounding, and since distances are never negative, equal squares mean equal distances.\[\mathrm{OA}^2 = (1-0)^2 + (-8-0)^2 = 1 + 64 = 65, \] \[\mathrm{OB}^2 = (-4-0)^2 + (7-0)^2 = 16 + 49 = 65, \] \[\mathrm{OC}^2 = (-7-0)^2 + (-4-0)^2 = 49 + 16 = 65. \]All three squares are $\displaystyle 65$, so \[\mathrm{OA} = \mathrm{OB} = \mathrm{OC} = \sqrt{65} \approx 8.06. \]Since A, B and C are all exactly \(\displaystyle \sqrt{65}\) units from O, they all lie on the circle K with centre O\(\displaystyle (0,0)\) and radius \(\displaystyle \sqrt{65}\).\[\text{radius of K} = \sqrt{65} \approx 8.06\ \text{units}. \](Notice the three points are in three different quadrants - A in IV, B in II, C in III - which is a nice reminder that the distance formula does not care about signs; only the squares matter.)(ii) Where D and E lie.The rule is simply: a point is inside K if its distance from O is less than the radius, on K if it equals the radius, and outside if it is greater. Again compare squares against \(\displaystyle 65\).For D\(\displaystyle (-5,6)\): \[\mathrm{OD}^2 = (-5)^2 + 6^2 = 25 + 36 = 61. \] \(\displaystyle 61 < 65\), so \(\displaystyle \mathrm{OD} = \sqrt{61} \approx 7.81 < \sqrt{65} \approx 8.06\). D lies inside the circle (though only just - about a quarter of a unit in from the edge).For E\(\displaystyle (0,9)\): \[\mathrm{OE}^2 = 0^2 + 9^2 = 81. \] \(\displaystyle 81 > 65\), so \(\displaystyle \mathrm{OE} = 9 > \sqrt{65} \approx 8.06\). E lies outside the circle. (E is on the y-axis, $\displaystyle 9$ units above O, while the circle crosses the y-axis at \(\displaystyle (0,\sqrt{65})\), about $\displaystyle 8.06$ units up - so E is roughly $\displaystyle 0.94$ units beyond it.)Answer: (i) \(\displaystyle \mathrm{OA}^2=\mathrm{OB}^2=\mathrm{OC}^2=65\), so A, B and C are all \(\displaystyle \sqrt{65}\) units from the origin and therefore lie on the circle K with centre O and radius \(\displaystyle \sqrt{65}\approx 8.06\) units. (ii) \(\displaystyle \mathrm{OD}^2=61<65\), so D\(\displaystyle (-5,6)\) lies inside K; \(\displaystyle \mathrm{OE}^2=81>65\), so E\(\displaystyle (0,9)\) lies outside K.

Turn each midpoint into an equation, then add the equations up.Let the vertices be \(\displaystyle \mathrm{A}(x_1,y_1)\), \(\displaystyle \mathrm{B}(x_2,y_2)\), \(\displaystyle \mathrm{C}(x_3,y_3)\), and take the usual labelling: D is the midpoint of BC, E of CA and F of AB.From the midpoint rule \(\displaystyle x_\mathrm{M} = \frac{x_\mathrm{S}+x_\mathrm{T}}{2}\), it is neater to clear the fraction at once and write "sum of the two endpoints \(\displaystyle =\) twice the midpoint".The x-coordinates. \[x_2 + x_3 = 2 \times 5 = 10 \qquad (\text{from D}(5,1)) \] \[x_3 + x_1 = 2 \times 6 = 12 \qquad (\text{from E}(6,5)) \] \[x_1 + x_2 = 2 \times 0 = 0 \qquad (\text{from F}(0,3)) \]Add all three. Each of \(\displaystyle x_1, x_2, x_3\) appears exactly twice on the left: \[2(x_1+x_2+x_3) = 10+12+0 = 22 \ \Longrightarrow\ x_1+x_2+x_3 = 11. \]Now subtract each original equation from this total: \[x_1 = 11 - (x_2+x_3) = 11 - 10 = 1, \quad x_2 = 11 - (x_3+x_1) = 11 - 12 = -1, \quad x_3 = 11 - (x_1+x_2) = 11 - 0 = 11. \]The y-coordinates, exactly the same way. \[y_2+y_3 = 2, \qquad y_3+y_1 = 10, \qquad y_1+y_2 = 6. \] \[2(y_1+y_2+y_3) = 2+10+6 = 18 \ \Longrightarrow\ y_1+y_2+y_3 = 9, \] \[y_1 = 9-2 = 7, \qquad y_2 = 9-10 = -1, \qquad y_3 = 9-6 = 3. \]So \[\mathrm{A}(1,\,7), \qquad \mathrm{B}(-1,\,-1), \qquad \mathrm{C}(11,\,3). \]Check all three midpoints. \[\text{midpoint of BC} = \left(\frac{-1+11}{2},\ \frac{-1+3}{2}\right) = (5,1) = \mathrm{D}\ \checkmark \] \[\text{midpoint of CA} = \left(\frac{11+1}{2},\ \frac{3+7}{2}\right) = (6,5) = \mathrm{E}\ \checkmark \] \[\text{midpoint of AB} = \left(\frac{1+(-1)}{2},\ \frac{7+(-1)}{2}\right) = (0,3) = \mathrm{F}\ \checkmark \]A remark on the labelling. The question does not say which midpoint belongs to which side, and it does not need to. Working through the algebra above shows that each vertex comes out as \[(\text{sum of the other two midpoints}) - (\text{the opposite midpoint}), \] for instance \(\displaystyle \mathrm{A} = \mathrm{E}+\mathrm{F}-\mathrm{D} = (6+0-5,\ 5+3-1) = (1,7)\). Pairing the midpoints with the sides differently only swaps the names A, B, C around; the three vertices themselves always come out as the same three points.Answer: \(\displaystyle \mathrm{A}(1,\,7)\), \(\displaystyle \mathrm{B}(-1,\,-1)\) and \(\displaystyle \mathrm{C}(11,\,3)\).

Read the city plan as a coordinate grid.The two main roads cross at the centre of the city, one running North–South, the other East–West. Every other street runs parallel to one of these two, so the whole city is a rectangular grid: a family of N–S lines and a family of E–W lines, with neighbouring parallel lines \(\displaystyle 200\) m apart. Read "there are $\displaystyle 10$ streets in each direction" as ten lines running N–S and ten running E–W, the two main roads being among them.(i) Drawing the model.The scale is \(\displaystyle 1\) cm \(\displaystyle =200\) m. So a real gap of \(\displaystyle 200\) m must be drawn as \[\frac{200\ \text{m}}{200\ \text{m per cm}} = 1\ \text{cm}. \] Every pair of neighbouring streets is therefore \(\displaystyle 1\) cm apart on paper. Then: Ten equally spaced lines leave \(\displaystyle 10-1=9\) gaps between them, so the grid measures \(\displaystyle 9\) cm by \(\displaystyle 9\) cm on paper. Since \(\displaystyle 9\) cm stands for \(\displaystyle 9\times 200 = 1800\) m, the model represents a square of city \(\displaystyle 1800\) m \(\displaystyle \times\) \(\displaystyle 1800\) m — a useful check that your drawing fits on a notebook page.(ii) Naming the street intersections.Number the ten N–S streets \(\displaystyle 1, 2, 3, \dots, 10\) in order from west to east, and number the ten E–W streets \(\displaystyle 1, 2, 3, \dots, 10\) in order from south to north. Then the label \(\displaystyle (m, n)\) means:\[(m,\,n) \;=\; \text{the crossing of the } m\text{-th N–S street with the } n\text{-th E–W street}. \]For example, \(\displaystyle (2, 5)\) is the single point where the 2nd street in the N–S direction meets the 5th street in the E–W direction. This is exactly how coordinates work: the first entry tells you how far east you have gone, the second how far north.Two facts follow straight from this, and they are what the counting parts of the question turn on: CANNOT SOLVE: needs the rest of the stem — the sentence "Using this convention, find:" is where the printed sub-parts begin, and those sub-parts are missing from the text supplied, so there is nothing specific left to count.Answer. (i) A \(\displaystyle 9\) cm \(\displaystyle \times\) \(\displaystyle 9\) cm grid of \(\displaystyle 10\) vertical and \(\displaystyle 10\) horizontal lines drawn \(\displaystyle 1\) cm apart, with the two main roads drawn thicker. (ii) With the streets numbered \(\displaystyle 1\) to \(\displaystyle 10\) in each direction, \(\displaystyle (m,n)\) names the crossing of the \(\displaystyle m\)-th N–S street with the \(\displaystyle n\)-th E–W street; there are \(\displaystyle 100\) such crossings, and \(\displaystyle (m,n)\) and \(\displaystyle (n,m)\) are different crossings whenever \(\displaystyle m\neq n\). The specific items to be found are missing from the question text.

Compare distances with radii.Because the origin sits at the bottom-left corner, the screen is the rectangle \[0 \le x \le 800, \qquad 0 \le y \le 600, \] and its four edges are the lines \(\displaystyle x=0\) (left), \(\displaystyle x=800\) (right), \(\displaystyle y=0\) (bottom) and \(\displaystyle y=600\) (top).(i) Does either circle spill off the screen?A circle with centre \(\displaystyle (h,k)\) and radius \(\displaystyle r\) reaches exactly \(\displaystyle r\) pixels away from its centre in every direction. So it stays completely inside an edge precisely when the centre is more than \(\displaystyle r\) away from that edge. From the equations of the edges,\[\text{distance to left edge}=h,\quad \text{to right edge}=800-h,\quad \text{to bottom}=k,\quad \text{to top}=600-k. \]Circle A, centre \(\displaystyle A(100,150)\), radius \(\displaystyle 80\): \[h = 100,\quad 800-h = 700,\quad k = 150,\quad 600-k = 450. \] The smallest of these is \(\displaystyle 100\), and \(\displaystyle 100 > 80\). So circle A clears every edge.The same thing said with extreme points: the circle runs from \(\displaystyle x=100-80=20\) to \(\displaystyle x=100+80=180\), and from \(\displaystyle y=150-80=70\) to \(\displaystyle y=150+80=230\). Both ranges sit inside \(\displaystyle [0,800]\) and \(\displaystyle [0,600]\).Circle B, centre \(\displaystyle B(250,230)\), radius \(\displaystyle 100\): \[h = 250,\quad 800-h = 550,\quad k = 230,\quad 600-k = 370. \] The smallest is \(\displaystyle 230\), and \(\displaystyle 230 > 100\). So circle B clears every edge too. Its extreme points are \(\displaystyle x\) from \(\displaystyle 150\) to \(\displaystyle 350\) and \(\displaystyle y\) from \(\displaystyle 130\) to \(\displaystyle 330\) — again inside the screen.So no part of either circle lies outside the screen.(ii) Do the two circles meet?Whether two circles meet depends only on how far apart their centres are compared with their radii, so first find \(\displaystyle AB\).Go from \(\displaystyle A(100,150)\) horizontally across to the point \(\displaystyle (250,150)\), then vertically up to \(\displaystyle B(250,230)\). The first leg is along a horizontal line and the second along a vertical line, so they meet at a right angle, and the two legs are \[250-100 = 150 \quad\text{and}\quad 230-150 = 80 . \] By the Pythagoras theorem in that right triangle, \[AB^{2} = 150^{2} + 80^{2} = 22500 + 6400 = 28900, \] \[AB = \sqrt{28900} = 170 \text{ pixels.} \]Now compare with the radii \(\displaystyle r_A = 80\) and \(\displaystyle r_B = 100\): \[r_B - r_A = 20, \qquad r_A + r_B = 180, \qquad 20 < 170 < 180 . \] Between those two limits the circles must cross, and they cross at two points.A check you can do without that rule. Walk along the straight line from \(\displaystyle A\) towards \(\displaystyle B\). Circle B's near point on this line is at distance \(\displaystyle 170-100=70\) from \(\displaystyle A\), and \(\displaystyle 70 < 80\), so that point is inside circle A. Circle B's far point on this line is at distance \(\displaystyle 170+100 = 270\) from \(\displaystyle A\), and \(\displaystyle 270 > 80\), so that point is outside circle A. A circle with some points inside A and some outside A has to cross A's boundary — twice, once on each side. This agrees with the comparison above.Answer. (i) No — both icons are drawn completely on the screen; circle A stays at least \(\displaystyle 100-80=20\) pixels clear of the nearest edge and circle B at least \(\displaystyle 230-100=130\) pixels clear. (ii) Yes — the distance between the centres is \(\displaystyle AB = 170\) pixels, and since \(\displaystyle 20 < 170 < 180\), the two circular icons overlap, meeting at two points.

All four sides equal, then one right angle by the converse of the Pythagoras theorem.Plotting. \(\displaystyle A(2,1)\) has both coordinates positive, so it is in the first quadrant. \(\displaystyle B(-1,2)\) has \(\displaystyle x\) negative and \(\displaystyle y\) positive — second quadrant. \(\displaystyle C(-2,-1)\) has both negative — third quadrant. \(\displaystyle D(1,-2)\) has \(\displaystyle x\) positive and \(\displaystyle y\) negative — fourth quadrant. Joining \(\displaystyle A \to B \to C \to D \to A\) takes you once round the origin in the anticlockwise sense, so the four segments do not cross one another and \(\displaystyle ABCD\) really is a quadrilateral.Lengths from coordinates. For any two points, go across and then up (or down): the horizontal step and the vertical step form the legs of a right triangle whose hypotenuse is the segment itself. So \[(\text{length})^{2} = (\text{horizontal step})^{2} + (\text{vertical step})^{2}. \]

Side	horizontal step	vertical step	(length)\(\displaystyle ^2\)
\(\displaystyle AB\): \(\displaystyle (2,1)\to(-1,2)\)	\(\displaystyle |-1-2| = 3\)	\(\displaystyle |2-1| = 1\)	\(\displaystyle 9+1 = 10\)
\(\displaystyle BC\): \(\displaystyle (-1,2)\to(-2,-1)\)	\(\displaystyle |-2+1| = 1\)	\(\displaystyle |-1-2| = 3\)	\(\displaystyle 1+9 = 10\)
\(\displaystyle CD\): \(\displaystyle (-2,-1)\to(1,-2)\)	\(\displaystyle |1+2| = 3\)	\(\displaystyle |-2+1| = 1\)	\(\displaystyle 9+1 = 10\)
\(\displaystyle DA\): \(\displaystyle (1,-2)\to(2,1)\)	\(\displaystyle |2-1| = 1\)	\(\displaystyle |1+2| = 3\)	\(\displaystyle 1+9 = 10\)

So \[AB = BC = CD = DA = \sqrt{10}. \] All four sides are equal, so \(\displaystyle ABCD\) is at least a rhombus. That is not yet enough — a rhombus can be squashed — so we need an angle.Finding a right angle. The diagonal \(\displaystyle AC\) runs from \(\displaystyle (2,1)\) to \(\displaystyle (-2,-1)\): horizontal step \(\displaystyle 4\), vertical step \(\displaystyle 2\), so \[AC^{2} = 4^{2} + 2^{2} = 16 + 4 = 20 . \] Now look at triangle \(\displaystyle ABC\): \[AB^{2} + BC^{2} = 10 + 10 = 20 = AC^{2}. \] By the converse of the Pythagoras theorem, a triangle whose two shorter sides satisfy this relation is right-angled at the vertex between them. So \(\displaystyle \angle ABC = 90^{\circ}\).A rhombus with one right angle has all four angles right: opposite angles of a rhombus are equal and adjacent angles add to \(\displaystyle 180^{\circ}\), so \(\displaystyle \angle B = 90^{\circ}\) forces \(\displaystyle \angle A = \angle C = \angle D = 90^{\circ}\).Four equal sides and four right angles: \(\displaystyle ABCD\) is a square.Area. \[\text{Area} = (\text{side})^{2} = \left(\sqrt{10}\right)^{2} = 10 \text{ square units.} \]A second, independent check — through the diagonals.Midpoint of \(\displaystyle AC\): \[\left(\frac{2+(-2)}{2},\ \frac{1+(-1)}{2}\right) = (0,0). \] Midpoint of \(\displaystyle BD\): \[\left(\frac{-1+1}{2},\ \frac{2+(-2)}{2}\right) = (0,0). \] The two diagonals have the same midpoint, the origin, so they bisect each other — that alone makes \(\displaystyle ABCD\) a parallelogram. Their lengths are \[AC^{2} = 4^{2}+2^{2} = 20, \qquad BD^{2} = 2^{2}+4^{2} = 20, \] so \(\displaystyle AC = BD = \sqrt{20} = 2\sqrt5\): equal diagonals, which upgrades the parallelogram to a rectangle. A rectangle with equal sides is a square — the same conclusion as before.For a square, area \(\displaystyle =\tfrac12 \times d_1 \times d_2\) where \(\displaystyle d_1,d_2\) are the diagonals, so \[\text{Area} = \tfrac12 \times \sqrt{20} \times \sqrt{20} = \tfrac12 \times 20 = 10, \] matching the \(\displaystyle 10\) found from the side. Two different routes give the same number, which is a good sign the answer is right.A remark worth noticing. Each vertex is the same distance from the origin: \[2^2+1^2 = 5,\quad (-1)^2+2^2 = 5,\quad (-2)^2+(-1)^2 = 5,\quad 1^2+(-2)^2 = 5, \] so all four points lie on the circle of radius \(\displaystyle \sqrt5\) about \(\displaystyle O\), and \(\displaystyle A, C\) are opposite ends of one diameter while \(\displaystyle B, D\) are opposite ends of another. That is exactly the picture of a square inscribed in a circle, with the centre of the circle at the centre of the square.Answer. Yes, \(\displaystyle ABCD\) is a square — all four sides measure \(\displaystyle \sqrt{10}\) units and \(\displaystyle AB^2+BC^2=AC^2\) gives a right angle at \(\displaystyle B\) (equivalently, the diagonals \(\displaystyle AC\) and \(\displaystyle BD\) are equal, each \(\displaystyle 2\sqrt5\), and bisect each other at the origin). Its area is \(\displaystyle 10\) square units.