Orienting Yourself: The Use of Coordinates

(i) If $D_1R_1$ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

(ii) What are the coordinates of $D_1$?

(iii) If $R_1$ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

(iv) If $B_1(0, 1.5)$ and $B_2(0, 4)$ represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

(i) Sol. From Fig. 1.3, the door $D_1R_1$ lies along the x-axis, with $D_1 = (8, 0)$.

Distance of $D_1$ from the left wall (y-axis):

$= 8$ units

Distance of $D_1$ from the x-axis:

$= 0$ units

∴ The door is 8 units from the left wall and 0 units from the x-axis.

(ii) Sol. The coordinates of $D_1$ are $(8, 0)$.

(iii) Sol. The room door extends from $D_1 = (8, 0)$ to $R_1 = (11.5, 0)$.

Therefore, width of the door $= 11.5 - 8$ $=3.5$

Since 1 unit represents 1 foot,

Width of the door = 3.5 ft.

∴ The room door is 3.5 ft wide and is sufficiently wide for a person using a wheelchair to enter.

(iv) Sol. $B_1 = (0, 1.5)$ and $B_2 = (0, 4)$.

Width of the bathroom door:

$= 4 - 1.5 = 2.5$ units

Width of the room door (from part (iii)):

$= 3.5$ units

Since $2.5 < 3.5$,

∴ The bathroom door is narrower than the room door.

(i) Where will the fourth foot of the table be?

(ii) Is this a good spot for the table?

(iii) What is the width of the table? The length? Can you make out the height of the table?

(i) Sol. The three feet of the rectangular table are at $(8, 9), (11, 9), (11, 7)$.

Hence, the fourth foot will be at $(8, 7)$.

(ii) Sol. No, this is not a good spot for the table because it will be very close to the bed and the right wall, leaving less space for movement.

(iii) Sol. From the given coordinates,

Width of the table $=9-7=2$ units

Length of the table $=11-8=3$ units

The height of the table cannot be determined from the given graph because the graph shows only its position in two dimensions.

∴ Width = 2 units, Length = 3 units, and height cannot be determined.

(i) What are the coordinates of the four corners O, F, R and P of the bathroom?

(ii) What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the four corners.

(iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.

(i) Sol. From the figure, the four corners of the bathroom are:

(ii) Sol. The showering area SHWR is a trapezium. Its four corners are:

(iii) Sol. One possible arrangement is:

Washbasin (2 ft × 3 ft):

Toilet (3 ft × 2 ft):

(i) Reiaan's room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.

(ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

(i) Sol. From the figure, the coordinates of the corners are:

(ii) Sol. Finding the centre of the room, using the midpoint formula:

Centre = (3, -7.5).

The table is 5 ft long and 3 ft wide. From the centre, it extends half its length and half its width in each direction:

Half length (x-direction) = 2.5

Half width (y-direction) = 1.5

Coordinates of the table's feet:

Q2. Point W has x-coordinate equal to −5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

Sol. Since the line through $W(-5, y)$ is parallel to the y-axis, the x-coordinate remains constant. The coordinates of H are $(-5, y)$.

Quadrants:

If $y > 0$, H lies in Quadrant II.

If $y < 0$, H lies in Quadrant III.

(i) Two sides of RAMP that are perpendicular to each other.

(ii) One side of RAMP that is parallel to one of the axes.

(iii) Two points that are mirror images of each other in one axis. Which axis will this be?

Now plot the points and verify your predictions.

Sol. Given points are

(i) Sol. A and M have the same y-coordinate, so AM is parallel to the x-axis.

M and P have the same x-coordinate, so MP is parallel to the y-axis.

Since the x-axis is perpendicular to the y-axis,

∴ AM and MP are perpendicular sides.

(ii) Sol.$A=(0,-2)$ and $M=(-5,-2)$ have the same y-coordinate.

So AM is parallel to the x-axis.

$M=(-5,-2)$ and $P=(-5,2)$ have the same x-coordinate.

So MP is parallel to the y-axis.

(iii) Sol. The points $M = (-5, -2)$ and $P = (-5, 2)$ have the same x-coordinate and opposite y-coordinates.

Hence, M and P are mirror images of each other in the x-axis.

Sol.Given $Z=(5,-6)$.

Take $I=(5,0)$ and $N=(0,-6)$.

Since IZ is vertical and ZN is horizontal,

Using the distance formula,

∴ The lengths of the three sides are $IZ=6$ units, $ZN=5$ units and $IN=\sqrt{61}$ units.

Q7. Use your method (from Problem 6) to check if the points R(−5, −1), B(−2, −5) and C(4, −12) are on the same straight line. Now plot both sets of points and check your answers.

Sol. Given points are $R=(-5,-1),\; B=(-2,-5)\\ C=(4,-12)$

Using the distance formula,

$$\begin{array}{l} \textcolor{#1B5E20}{\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \\[0.6em] \displaystyle RB=\sqrt{(-2+5)^2+(-5+1)^2} \end{array}$$

$$=\sqrt{9+16}=5$$

$$BC=\sqrt{(4+2)^2+(-12+5)^2}$$

$$=\sqrt{36+49}=\sqrt{85}$$

$$RC=\sqrt{(4+5)^2+(-12+1)^2}$$

$$=\sqrt{81+121}=\sqrt{202}$$

Now,

$$RB+BC=5+\sqrt{85}$$

$$RC=\sqrt{202}$$

$$RB+BC\ne RC$$

∴ The points R, B and C are not collinear.

(i) A right-angled isosceles triangle.

(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

(i) Sol. Take the vertices $O=(0,0)$, $A=(4,0)$ and $B=(0,4)$.

Here, $OA=OB=4$ units and $\angle AOB=90^\circ$.

∴ $\triangle OAB$ is a right-angled isosceles triangle.

(ii) Sol. An isosceles triangle with vertices

Using the distance formula,

So $\text{OC} = \text{OD}$, and the triangle is isosceles, with C in Quadrant III and D in Quadrant IV.

Q10. Use the connection you found to find the coordinates of B given that M(−7, 1) is the midpoint of A(3, −4) and B(x, y).

Sol. Given, M = (−7, 1) is the midpoint of A = (3, −4) and B = (x, y).

Using the midpoint formula,

$$\textcolor{#1B5E20}{ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) }$$

$$\text{Therefore, } (-7, 1) = \left( \dfrac{3+x}{2}, \dfrac{-4+y}{2} \right)$$

Comparing the x-coordinates,

$$\begin{array}{l} \hspace{3.2em} \displaystyle \dfrac{3+x}{2} = -7 \\[0.8em] \hspace{3.2em} \displaystyle 3 + x = -14 \\[0.4em] \hspace{3.2em} \displaystyle x = -17 \end{array}$$

Comparing the y-coordinates,

$$\begin{array}{l} \hspace{2.4em} \displaystyle \dfrac{-4+y}{2} = 1 \\[0.8em] \hspace{2.4em} \displaystyle -4 + y = 2 \\[0.4em] \hspace{3.2em} \displaystyle y = 6 \end{array}$$

Hence, $B = (-17, 6)$.

Q11. Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A(4, 7) and B(16, −2).

Sol. Given, $A = (4, 7), \quad B = (16, -2)$.

Let $P = (x_1, y_1), \quad Q = (x_2, y_2)$.

Since P and Q trisect AB, $\text{AP} = \text{PQ} = \text{QB}$.

Therefore, P is the midpoint of AQ, and Q is the midpoint of PB.

Using the midpoint formula,

$$\textcolor{#1B5E20}{\displaystyle M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)}$$

Since P is the midpoint of AQ,

$$\begin{array}{l} \hspace{3em} \displaystyle x_1 = \frac{4+x_2}{2} \\[0.4em] \hspace{2.6em} \displaystyle 2x_1 = 4 + x_2 \\[0.4em] \hspace{2.6em} \displaystyle 2x_1 - x_2 = 4 \quad \text{...(1)} \end{array}$$

and

$$\begin{array}{l} \hspace{3em} \displaystyle y_1 = \frac{7+y_2}{2} \\[0.4em] \hspace{2.6em} \displaystyle 2y_1 = 7 + y_2 \\[0.4em] \hspace{2.6em} \displaystyle 2y_1 - y_2 = 7 \quad \text{...(2)} \end{array}$$

Since Q is the midpoint of PB,

$$\begin{array}{l} \hspace{3em} \displaystyle x_2 = \frac{x_1+16}{2} \\[0.4em] \hspace{2.6em} \displaystyle 2x_2 = x_1 + 16 \\[0.4em] \hspace{2.4em} \displaystyle -x_1 + 2x_2 = 16 \quad \text{...(3)} \end{array}$$

and

$$\begin{array}{l} \hspace{3em} \displaystyle y_2 = \frac{y_1-2}{2} \\[0.4em] \hspace{2.4em} \displaystyle 2y_2 = y_1 - 2 \\[0.4em] \hspace{2.2em} \displaystyle -y_1 + 2y_2 = -2 \quad \text{...(4)} \end{array}$$

$$\begin{array}{l} \text{For x-coordinates,}\\ \text{from (3), }\; x_1 = 2x_2 - 16 \end{array}$$

Substituting in (1),

$$\begin{array}{l} \hspace{3em} \displaystyle 2(2x_2 - 16) - x_2 = 4 \\[0.4em] \hspace{3em} \displaystyle 4x_2 - 32 - x_2 = 4 \\[0.4em] \hspace{3em} \displaystyle 3x_2 = 36 \\[0.4em] \hspace{3.4em} \displaystyle x_2 = 12 \end{array}$$

Substituting $x_2 = 12$ in (3),

$$\begin{array}{l} \hspace{2.6em} \displaystyle -x_1 + 2(12) = 16 \\[0.4em] \hspace{2.6em} \displaystyle -x_1 + 24 = 16 \\[0.4em] \hspace{3.4em} \displaystyle x_1 = 8 \end{array}$$

$$\text{For y-coordinates, from (4), } y_1 = 2y_2 + 2$$

Substituting in (2),

$$\begin{array}{l} \hspace{3em} \displaystyle 2(2y_2 + 2) - y_2 = 7 \\[0.4em] \hspace{3em} \displaystyle 4y_2 + 4 - y_2 = 7 \\[0.4em] \hspace{3em} \displaystyle 3y_2 = 3 \\[0.4em] \hspace{3.6em} \displaystyle y_2 = 1 \end{array}$$

Substituting $y_2 = 1$ in (4),

$$\begin{array}{l} \hspace{3em} \displaystyle -y_1 + 2(1) = -2 \\[0.4em] \hspace{3em} \displaystyle -y_1 + 2 = -2 \\[0.4em] \hspace{3.8em} \displaystyle y_1 = 4 \end{array}$$

Hence,

$$\begin{array}{l} \hspace{3em} \displaystyle x_1 = 8, \quad y_1 = 4 \\[0.4em] \hspace{3em} \displaystyle x_2 = 12, \quad y_2 = 1 \end{array}$$

$$\therefore P = (8, 4), \quad Q = (12, 1)$$

(ii) Given the points D(−5, 6) and E(0, 9), check whether D and E lie within the circle, on the circle, or outside the circle.

(i) Sol. Given, $\begin{aligned} A &= (1,-8), \qquad & B &= (-4,7) \\ C &= (-7,-4) \end{aligned}$

Centre of circle K is $O = (0, 0)$. Using the distance formula,

Thus, $\text{OA} = \text{OB} = \text{OC} = \sqrt{65}$.

Since A, B and C are at the same distance from O, they lie on the same circle with centre O, and the radius of circle K is $\sqrt{65}$ units.

(ii) Sol.

For point D,

Since $\sqrt{61} < \sqrt{65}$, D lies within the circle.

For point E,

Since $9 > \sqrt{65}$, E lies outside the circle.

Q13. The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C.

Sol. Given, $D = (5, 1), \quad E = (6, 5), \quad F = (0, 3)$.

Let

$\begin{aligned} A &= (x_1,y_1), \qquad & B &= (x_2,y_2) \\ C &= (x_3,y_3) \end{aligned}$.

Since D, E, F are the midpoints of BC, CA, AB respectively:

For D

Using the midpoint formula,

$$\begin{array}{l} \textcolor{#1B5E20}{\displaystyle M = \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)} \\[1.2em] \displaystyle (5, 1) = \left( \dfrac{x_2+x_3}{2}, \dfrac{y_2+y_3}{2} \right) \\[1em] \displaystyle \Rightarrow \; x_2+x_3 = 10 \quad \text{...(1)} \\[0.4em] \displaystyle \Rightarrow \; y_2+y_3 = 2 \quad \text{...(2)} \end{array}$$

For E,

$$\begin{array}{l} \displaystyle (6, 5) = \left( \dfrac{x_3+x_1}{2}, \dfrac{y_3+y_1}{2} \right) \\[1em] \displaystyle \Rightarrow \; x_3+x_1 = 12 \quad \text{...(3)} \\[0.4em] \displaystyle \Rightarrow \; y_3+y_1 = 10 \quad \text{...(4)} \end{array}$$

For F,

$$\begin{array}{l} \displaystyle (0, 3) = \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right) \\[1em] \displaystyle \Rightarrow \; x_1+x_2 = 0 \quad \text{...(5)} \\[0.4em] \displaystyle \Rightarrow \; y_1+y_2 = 6 \quad \text{...(6)} \end{array}$$

Adding (3) and (5),

$$x_3 + x_1 + x_1 + x_2 = 12$$

Using (1), $x_2 + x_3 = 10$,

Substituting in the above equation,

$$\begin{array}{l} \hspace{0.6em} \displaystyle 2x_1 + 10 = 12 \\[0.4em] \hspace{0.6em} \displaystyle 2x_1 = 2 \\[0.4em] \hspace{0.6em} \displaystyle x_1 = 1 \end{array}$$

$$\text{From (5), } x_1 + x_2 = 0$$

Substituting the value of $x_1$ in (5),

$$1 + x_2 = 0 \Rightarrow x_2 = -1$$

$$\text{From (1), } x_2 + x_3 = 10$$

Substituting the value of $x_2$ in (1),

$$-1 + x_3 = 10 \Rightarrow x_3 = 11$$

Similarly, adding (4) and (6),

$$y_3 + y_1 + y_1 + y_2 = 16$$

Using (2), $y_2 + y_3 = 2$,

$$\begin{array}{l} \hspace{0.6em} \displaystyle 2y_1 + 2 = 16 \\[0.4em] \hspace{0.6em} \displaystyle 2y_1 = 14 \\[0.4em] \hspace{0.6em} \displaystyle y_1 = 7 \end{array}$$

$$\text{From (6), } y_1 + y_2 = 6$$

Substituting the value of $y_1$ in (6),

$$7 + y_2 = 6 \Rightarrow y_2 = -1$$

$$\text{From (2), } y_2 + y_3 = 2$$

Substituting the value of $y_2$ in (2),

$$-1 + y_3 = 2 \Rightarrow y_3 = 3$$

Hence, the vertices of the triangle are:

$$A = (1, 7), \quad B = (-1, -1), \quad C = (11, 3)$$

(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

(ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: if the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find: (a) how many street intersections can be referred to as (4, 3); (b) how many street intersections can be referred to as (3, 4).

(i) Sol.

(ii) Sol. (a) A street intersection is formed by exactly one N–S street and one E–W street. So the 4th N–S street meets the 3rd E–W street at exactly one point.

∴ There is only one street intersection that can be referred to as (4, 3).

(b) Similarly, the 3rd N–S street meets the 4th E–W street at exactly one point.

∴ There is only one street intersection that can be referred to as (3, 4).

(i) whether any part of either circle lies outside the screen.

(ii) whether the two circles intersect each other.

(i) Sol. The screen is 800 pixels wide and 600 pixels high, with the origin at the bottom-left corner. Therefore, $0 \leq x \leq 800, \quad 0 \leq y \leq 600$.

For the first circle,

$A = (100, 150)$, $r_1 = 80$:

Since all these values lie within the dimensions of the screen, the first circle lies completely inside the screen.

For the second circle,

$B = (250, 230)$, $r_2 = 100$:

Since these values also lie within the dimensions of the screen, the second circle lies completely inside the screen.

∴ No part of either circle lies outside the screen.

(ii) Sol. Using the distance formula,

Sum of the radii:

$r_1+r_2 = 80+100 = 180$ pixels

Difference of the radii: $r_2-r_1 = 100-80 = 20$ pixels

Now, $20 < 170 < 180$.

Since the distance between the centres is greater than the difference of the radii and less than the sum of the radii, the two circles intersect at two points.

∴ The two circles intersect each other.

$\begin{aligned} &\text{Given,}\\ A &= (2,1), \qquad & B &= (-1,2)\\ C &= (-2,-1), \qquad & D &= (1,-2) \end{aligned}$

Using the distance formula,

Thus, $\text{AB} = \text{BC} = \text{CD} = \text{DA} = \sqrt{10}$.

Hence, ABCD is a rhombus.

Now, the diagonals are:

Thus, $\text{AC} = \text{BD}$.

Since ABCD is a rhombus whose diagonals are equal, ABCD is a square.