Orienting Yourself: The Use of Coordinates

Fig. 1.3 shows Reiaan's room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.

Referring to Fig. 1.3, answer the following questions:

(i) If D₁R₁ 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 D₁?

(iii) If R₁ 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 B₁ (0, 1.5) and B₂ (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Answer

(i) From Fig. 1.3, the door D₁R₁ lies along the x-axis. The end D₁ of the door is the point closer to the y-axis, and from the figure, D₁ is at +8 on the x-axis.

So, the distance of the door from the left wall (the y-axis) = 8 units (this is the distance of the nearest end D₁ from the y-axis).

Since the door lies on the x-axis itself, its distance from the x-axis = 0 unit.

Hence, the door is 8 units from the y-axis and 0 unit from the x-axis.

(ii) The point D₁ lies on the x-axis at +8.

Since any point on the x-axis has y-coordinate 0,

Hence, D₁ = (8, 0).

(iii) The door D₁R₁ extends from D₁(8, 0) to R₁(11.5, 0). Since both points lie on the x-axis,

Width of the door = 11.5 − 8 = 3.5 units

Assuming the unit here is feet, a width of 3.5 ft (≈ 107 cm) is a comfortable width for a room door.

The standard width of a wheelchair is roughly 2 to 2.5 ft. Since 3.5 ft is greater than this, a person in a wheelchair would be able to enter the room comfortably.

Hence, width of the door = 3.5 units, which is comfortable; a person in a wheelchair will be able to enter the room easily.

(iv) The bathroom door extends from B₁(0, 1.5) to B₂(0, 4). Since both points lie on the y-axis,

Width of bathroom door = 4 − 1.5 = 2.5 units

Width of room door = 3.5 units (from part (iii)).

Since 2.5 units < 3.5 units,

The bathroom door (2.5 units) is narrower than the room door (3.5 units).

What are the standard widths for a room door? Look around your home and in school.

Answer

The standard widths for room doors generally lie between 2.5 ft and 3 ft (about 75 cm to 90 cm). Main entrance doors of houses are usually wider, around 3 ft to 3.5 ft (about 90 cm to 105 cm). Bathroom doors are often slightly narrower, around 2 ft to 2.5 ft (about 60 cm to 75 cm).

Hence, the standard width of an internal room door is about 2.5 ft to 3 ft.

Are the doors in your school suitable for people in wheelchairs?

Answer

For a door to be wheelchair-accessible, its clear width should be at least 2.75 ft (about 81 cm to 90 cm), as a standard wheelchair is about 2 to 2.5 ft wide and needs extra room to manoeuvre.

If the doors in the school are 3 ft wide or more, then they are suitable for people in wheelchairs. If they are narrower than about 2.75 ft, then they would not be suitable, and would need to be widened.

Hence, doors of width 2.75 ft or more are suitable for wheelchairs; narrower doors are not.

What is the x-coordinate of a point on the y-axis?

Answer

The y-axis is the vertical line passing through the origin. Every point on the y-axis lies at a perpendicular distance of 0 from the y-axis itself.

Since the x-coordinate represents the perpendicular distance of a point from the y-axis,

Hence, the x-coordinate of every point on the y-axis is 0.

Is there a similar generalisation for a point on the x-axis?

Answer

The x-axis is the horizontal line passing through the origin. Every point on the x-axis lies at a perpendicular distance of 0 from the x-axis itself.

Since the y-coordinate represents the perpendicular distance of a point from the x-axis,

Hence, the y-coordinate of every point on the x-axis is 0. So, every point on the x-axis is of the form (x, 0).

Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.

Answer

Two points coincide when both their coordinates are equal.

P(x, y) and Q(y, x) coincide when:

⇒ x = y (first coordinates equal) and y = x (second coordinates equal)

Both conditions reduce to the single condition x = y.

For example, if x = y = 3, then P = (3, 3) and Q = (3, 3); they coincide.

Hence, point Q(y, x) coincides with P(x, y) if and only if x = y.

If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?

Answer

For two ordered pairs to be equal, both their corresponding coordinates must be equal. So, (x, y) = (y, x) requires:

⇒ x = y (first coordinates) and y = x (second coordinates)

Both reduce to x = y.

So:

• If x = y, then both pairs become (x, x), which are equal.

• If x ≠ y, then (x, y) and (y, x) have different first coordinates, so they are not equal.

Hence, the claim is true. (x, y) = (y, x) if and only if x = y, and (x, y) ≠ (y, x) whenever x ≠ y.

On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (–7, 0) to (13, 0) on the x-axis and from (0, -15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.) Using Fig. 1.5, answer the given questions.

1. Place Reiaan's rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7).
   (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?

2. If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

3. Look at Reiaan's bathroom.
   (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.

4. Other rooms in the house:
   (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.

Answer

1. The three given feet of the table are at (8, 9), (11, 9) and (11, 7).

(i) For a rectangular table, the four feet are at the four corners of a rectangle with sides parallel to the axes.

The known feet are:

• (8, 9) and (11, 9): same y-coordinate (9), so they share a horizontal side.
• (11, 9) and (11, 7): same x-coordinate (11), so they share a vertical side.

The fourth foot must share its x-coordinate with (8, 9) and its y-coordinate with (11, 7).

So, the fourth foot is at (8, 7).

Hence, the fourth foot of the table is at (8, 7).

(ii) The table occupies the rectangular region with x from 8 to 11 and y from 7 to 9. From Fig. 1.5, this region lies in the right-upper part of the bedroom, which is empty (the bed occupies x from 1 to 6 and y from 5 to 8, and the wardrobe occupies x from 3 to 7 and y from 0 to 2). So the table does not overlap with any other furniture and is also away from the door on the x-axis.

Hence, this is a good spot for the table.

(iii) Width of the table = |11 − 8| = 3 ft

Length of the table = |9 − 7| = 2 ft

Since Fig. 1.5 shows only the floor plan (a 2-D top view), the height of the table cannot be determined from the figure.

Hence, the width of the table is 3 ft and the length is 2 ft. The height of the table cannot be made out from the figure.

2. The bathroom door has its hinge at B₁(0, 1.5), and the other end at B₂(0, 4). When opened into the bedroom, the door swings about B₁ through the bedroom side (positive x).

Length of the door = 4 − 1.5 = 2.5 ft

When fully opened (rotated 90°) into the bedroom, the door's free end traces an arc of radius 2.5 ft from B₁(0, 1.5). The farthest point reached along the x-direction is (2.5, 1.5).

The wardrobe occupies x from 3 to 7 and y from 0 to 2. Since the door reaches only up to x = 2.5, which is less than 3, the door does not hit the wardrobe.

If the door is made wider (say 3 ft or more), the door's free end would reach x = 3 or beyond, and could potentially hit the wardrobe. To accommodate a wider door, one suggestion is to shift the wardrobe to the right (so that it begins beyond the door's swing) or to make the door open outwards (away from the bedroom).

Hence, the door does not hit the wardrobe at present, but if it is made wider than 3 ft, the wardrobe should be shifted to the right or the door should be made to open outwards.

3.

(i) From Fig. 1.5, the bathroom is 6 ft wide (along the x-axis) and 9 ft tall (along the y-axis). Its four corners are:

• O = (0, 0) (bottom-right of the bathroom, at the origin)

• F = (0, 9) (top-right corner of the bathroom, on the y-axis)

• R = (−6, 9) (top-left corner of the bathroom)

• P = (−6, 0) (bottom-left corner of the bathroom)

Hence, the coordinates are O(0, 0), F(0, 9), R(−6, 9) and P(−6, 0).

(ii) From Fig. 1.5, the showering area SHWR is a trapezium (a quadrilateral) with one slanted side from H to W and the other three sides aligned with the bathroom walls. Its corners are:

• S = (−6, 6) (on the left wall of the bathroom)

• H = (−3, 6) (interior corner where the boundary bends)

• W = (−2, 9) (on the top wall of the bathroom)

• R = (−6, 9) (top-left corner of the bathroom)

The sides SR and HW are not parallel, while SH and WR are both horizontal (parallel to the x-axis).

Hence, SHWR is a trapezium (with SH ∥ WR), with corners S(−6, 6), H(−3, 6), W(−2, 9) and R(−6, 9).

(iii) The bathroom region not occupied by the showering area is approximately the rectangle x from −6 to 0, y from 0 to 6. We can place the washbasin and toilet within this free area as follows.

Washbasin (3 ft × 2 ft): Place a 3 ft (along x) × 2 ft (along y) space in the upper-left region just below the showering area.

The four corners of the washbasin are:

(−6, 4), (−3, 4), (−3, 6), (−6, 6)

Toilet (2 ft × 3 ft): Place a 2 ft (along x) × 3 ft (along y) space along the left wall in the lower portion of the bathroom.

The four corners of the toilet are:

(−6, 0), (−4, 0), (−4, 3), (−6, 3)

Hence, the washbasin corners are (−6, 4), (−3, 4), (−3, 6), (−6, 6); and the toilet corners are (−6, 0), (−4, 0), (−4, 3), (−6, 3).

(i) The dining room is 18 ft × 15 ft. Its length (18 ft) extends from P(−6, 0) to A(12, 0), and its width (15 ft) extends downwards (in the negative y-direction).

The four corners of the dining room are:

• P = (−6, 0) (top-left)

• A = (12, 0) (top-right)

• (12, −15) (bottom-right)

• (−6, −15) (bottom-left)

Hence, the dining-room corners are (−6, 0), (12, 0), (12, −15) and (−6, −15).

(ii) The centre of the dining room is the midpoint of its diagonals.

Centre = $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) = \left(\dfrac{-6 + 12}{2}, \dfrac{0 + (-15)}{2}\right) = \left(\dfrac{6}{2}, \dfrac{-15}{2}\right) = (3, -7.5)$

The 5 ft × 3 ft dining table is placed precisely at the centre. Taking 5 ft along the x-direction and 3 ft along the y-direction, the table extends:

• x from 3 − 2.5 = 0.5 to 3 + 2.5 = 5.5

• y from −7.5 + 1.5 = −6 to −7.5 − 1.5 = −9

The four feet of the table are at:

(0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9)

Hence, the feet of the dining table are at (0.5, −6), (5.5, −6), (5.5, −9) and (0.5, −9).

In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?

Answer

Distance covered along the x-axis = (x-coordinate of D) − (x-coordinate of A) = 7 − 3 = 4 units.

Distance covered along the y-axis = (y-coordinate of A) − (y-coordinate of D) = 4 − 1 = 3 units (downwards).

Hence, the horizontal distance is 4 units and the vertical distance is 3 units.

Can these distances help you find the distance AD?

Answer

Yes. Together with the right angle at the foot of the perpendicular C(7, 4) (or C(3, 1)), the horizontal and vertical distances form the two legs of a right-angled triangle whose hypotenuse is AD.

By the Baudhāyana–Pythagoras Theorem,

AD² = (horizontal distance)² + (vertical distance)² = 4² + 3² = 16 + 9 = 25

⇒ AD = $\sqrt{25}$ = 5 units.

Hence, AD = 5 units.

What has remained the same and what has changed with this reflection?

Answer

When ΔADM is reflected in the y-axis to give ΔA'D'M':

What has remained the same:

• The lengths of the sides of the triangle: AD = A'D' = 5, DM = D'M' = $\sqrt{29}$, MA = M'A' = $\sqrt{40}$.

• The shape and size of the triangle (it is congruent to the original).

• The y-coordinates of all the points.

What has changed:

• The x-coordinates of all the vertices have changed sign (A(3, 4) → A'(−3, 4), D(7, 1) → D'(−7, 1), M(9, 6) → M'(−9, 6)).

• The triangle is now situated in a different region of the plane (Quadrant II instead of Quadrant I).

• The orientation (left-right) has been flipped (the triangle is the mirror image of the original).

Hence, the side lengths and shape are preserved, but the x-coordinates change sign and the orientation is flipped.

Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)?

Answer

The observations would be similar but not the same. When ΔADM is reflected in the x-axis:

• The lengths of the sides remain the same (the reflected triangle is congruent to the original).

• The x-coordinates of all the vertices remain unchanged.

• The y-coordinates of all the vertices change sign (A(3, 4) → A''(3, −4), D(7, 1) → D''(7, −1), M(9, 6) → M''(9, −6)).

• The triangle is flipped vertically (top-bottom) instead of horizontally.

Hence, similar observations hold: side lengths are preserved, but now the y-coordinates change sign and the orientation is flipped vertically.

What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Answer

The point of intersection of the x-axis and the y-axis is called the origin, denoted by O.

A point on the x-axis has y-coordinate 0, and a point on the y-axis has x-coordinate 0. The origin lies on both axes simultaneously, so both its coordinates are 0.

Hence, the x-coordinate is 0 and the y-coordinate is 0; that is, O = (0, 0).

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?

Answer

The line through W parallel to the y-axis is a vertical line. Every point on a vertical line has the same x-coordinate.

So, the x-coordinate of H is also −5, and its y-coordinate can be any real number y.

Coordinates of H = (−5, y), where y can be any real number.

The point H has x-coordinate −5 (negative).

• If y > 0 (positive), then H lies in Quadrant II (x negative, y positive).

• If y < 0 (negative), then H lies in Quadrant III (x negative, y negative).

• If y = 0, then H lies on the x-axis.

Hence, H = (−5, y); H can lie in Quadrant II (when y > 0) or Quadrant III (when y < 0).

Consider the points R (3, 0), A (0, –2), M (–5, –2) and P (–5, 2). If they are joined in the same order, predict:

(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.

Answer

The four sides of RAMP are RA, AM, MP and PR.

(i) Sides perpendicular to each other:

• Side AM goes from A(0, −2) to M(−5, −2): both have y-coordinate −2, so AM is horizontal.

• Side MP goes from M(−5, −2) to P(−5, 2): both have x-coordinate −5, so MP is vertical.

A horizontal line is perpendicular to a vertical line.

Hence, AM ⊥ MP.

(ii) Side parallel to an axis:

• AM has y-coordinate −2 throughout, so AM is parallel to the x-axis.

• MP has x-coordinate −5 throughout, so MP is parallel to the y-axis.

Hence, AM is parallel to the x-axis (and MP is parallel to the y-axis).

(iii) Mirror images in one axis:

Two points are mirror images of each other in the x-axis if their x-coordinates are equal and their y-coordinates are opposite.

• M = (−5, −2) and P = (−5, 2): same x-coordinate, opposite y-coordinates.

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

On plotting, all the predictions are verified.

Plot point Z (5, –6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.

(Comment: Answers may differ from person to person.)

Answer

Point Z(5, −6) lies in Quadrant IV.

We can choose I and N so that IZN is a right-angled triangle. One simple choice is to take I directly above Z on the x-axis, and N directly to the left of Z on the line y = −6.

Let I = (5, 0) and N = (0, −6). Then:

• IZ goes from (5, 0) to (5, −6) — vertical segment.

• ZN goes from (5, −6) to (0, −6) — horizontal segment.

• IZ and ZN meet at Z and are perpendicular, so the right angle is at Z.

Lengths of the sides:

IZ = |0 − (−6)| = 6 units

ZN = |5 − 0| = 5 units

$\text{NI} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(5-0)^2 + (0-(-6))^2} \\[1em] = \sqrt{25 + 36} \\[1em] = \sqrt{61} \text{ units}$

Hence, the sides of triangle IZN are IZ = 6 units, ZN = 5 units and NI = $\mathbf{\sqrt{61}}$ units. (Other valid choices of I and N will give different lengths.)

What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

Answer

Without negative numbers, we could only use non-negative values (0 and positive numbers) for the x and y-coordinates of any point.

This would mean that we could only mark points with x ≥ 0 and y ≥ 0, i.e., points only in the first quadrant (along with the positive x-axis, the positive y-axis, and the origin).

We would not be able to locate any point lying in the second, third or fourth quadrants — that is, points to the left of the y-axis or below the x-axis.

Hence, without negative numbers, we could locate only points in the first quadrant. So this system would not allow us to locate all the points of a 2-D plane.

Are the points M (–3, –4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.

Answer

Method (using the distance formula):

If three points lie on a straight line, then the distance between the two outer points equals the sum of the distances from each outer point to the middle point.

Calculate the distances:

We have the formula:

Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\text{MA} = \sqrt{(0-(-3))^2 + (0-(-4))^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[2em] \text{AG} = \sqrt{(6-0)^2 + (8-0)^2} \\[1em] = \sqrt{36 + 64} \\[1em] = \sqrt{100} \\[1em] = 10 \text{ units} \\[2em] \text{MG} = \sqrt{(6-(-3))^2 + (8-(-4))^2} \\[1em] = \sqrt{81 + 144} \\[1em] = \sqrt{225} \\[1em] = 15 \text{ units}$

Now check:

MA + AG = 5 + 10 = 15 = MG.

Since the sum of two of the distances equals the third, the three points lie on a single straight line, with A between M and G.

Hence, the points M, A and G are collinear (lie on the same straight line).

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.

Answer

Calculate the distances:

We have the formula:

Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\text{RB} = \sqrt{(-2-(-5))^2 + (-5-(-1))^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[2em] \text{BC} = \sqrt{(4-(-2))^2 + (-12-(-5))^2} \\[1em] = \sqrt{36 + 49} \\[1em] = \sqrt{85} \\[1em] \approx 9.22 \text{ units} \\[2em] \text{RC} = \sqrt{(4-(-5))^2 + (-12-(-1))^2} \\[1em] = \sqrt{81 + 121} \\[1em] = \sqrt{202} \\[1em] \approx 14.21 \text{ units}$

Now check:

RB + BC = 5 + $\sqrt{85}$ ≈ 5 + 9.22 = 14.22 units

RC ≈ 14.21 units

Since RB + BC ≠ RC exactly (one is $\sqrt{25} + \sqrt{85}$ = 5 + $\sqrt{85}$, and the other is $\sqrt{202}$, which are not equal), the three points are not collinear.

(In fact, (5 + $\sqrt{85}$)² = 25 + 10 $\sqrt{85}$ + 85 = 110 + 10 $\sqrt{85}$ ≈ 202.2, while 202 = $\sqrt{202}^2$. So the values are very close but not equal.)

From the figure, we can see that the points R, B and C do not lie on a single straight line — the line through R, B is slightly different from the line through B, C.

Hence, the points R(−5, −1), B(−2, −5) and C(4, −12) are not on the same straight line.

Using the origin as one vertex, plot the vertices of:

(i) A right-angled isosceles triangle.

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

Answer

(i) A right-angled isosceles triangle with one vertex at the origin:

A right-angled isosceles triangle has two equal sides meeting at the right angle. Take the two equal sides along the x-axis and y-axis.

Let the vertices be O(0, 0), A(4, 0) and B(0, 4).

Then OA = OB = 4 units (equal legs), and OA ⊥ OB (the right angle is at O).

The hypotenuse AB = $\sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$ units.

Hence, O(0, 0), A(4, 0) and B(0, 4) form a right-angled isosceles triangle with the right angle at O.

(ii) An isosceles triangle with one vertex at O, one in Quadrant III and one in Quadrant IV:

Choose the two non-origin vertices to be mirror images of each other in the y-axis, so they are equidistant from O.

Let the vertices be O(0, 0), P(−3, −4) and Q(3, −4).

Then, let's find the distance:

We have the formula,

Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\text{OP} = \sqrt{(-3 - 0)^2 + (-4 - 0)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[2em] \text{OQ} = \sqrt{(3 - 0)^2 + (-4 - 0)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}$

So OP = OQ = 5 units, which makes ΔOPQ isosceles.

P(−3, −4) lies in Quadrant III (x < 0, y < 0); Q(3, −4) lies in Quadrant IV (x > 0, y < 0).

Hence, O(0, 0), P(−3, −4) and Q(3, −4) form an isosceles triangle with one vertex at the origin, one in Quadrant III and one in Quadrant IV.

The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

Answer

For M to be the midpoint of segment ST, M must lie on segment ST and ST must be split into two equal halves at M.

Geometrically, this means SM = MT and S, M, T are collinear.

We check each case by computing distances SM, MT and ST.

Row 1: S(−3, 0), M(0, 0), T(3, 0)

SM = $\sqrt{(0-(-3))^2 + (0-0)^2} = \sqrt{9}$ = 3 units

MT = $\sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9}$ = 3 units

ST = $\sqrt{(3-(-3))^2 + 0^2} = \sqrt{36}$ = 6 units

SM = MT = 3 and SM + MT = ST, so S, M, T are collinear and M divides ST in two equal halves.

Yes, M is the midpoint of ST.

Row 2: S(2, 3), M(3, 4), T(4, 5)

SM = $\sqrt{(3-2)^2 + (4-3)^2} = \sqrt{1 + 1} = \sqrt{2}$ units

MT = $\sqrt{(4-3)^2 + (5-4)^2} = \sqrt{1 + 1} = \sqrt{2}$ units

ST = $\sqrt{(4-2)^2 + (5-3)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units

SM = MT = $\sqrt{2}$ and SM + MT = $2\sqrt{2}$ = ST, so S, M, T are collinear and SM = MT.

Yes, M is the midpoint of ST.

Row 3: S(0, 0), M(0, 5), T(0, −10)

All three points have x-coordinate 0, so they lie on the y-axis (collinear).

SM = |5 − 0| = 5 units

MT = |−10 − 5| = 15 units

Since SM ≠ MT, M does not split ST into equal halves.

No, M is not the midpoint of ST.

Row 4: S(−8, 7), M(0, −2), T(6, −3)

SM = $\sqrt{(0-(-8))^2 + (-2-7)^2} = \sqrt{64 + 81} = \sqrt{145}$ units

MT = $\sqrt{(6-0)^2 + (-3-(-2))^2} = \sqrt{36 + 1} = \sqrt{37}$ units

Since SM ≠ MT (and in fact, S, M, T are not even collinear; SM + MT ≠ ST), M is not the midpoint.

No, M is not the midpoint of ST.

The completed table:

Connection between the coordinates of M, S and T:

When M is the midpoint of ST, the x-coordinate of M is the average of the x-coordinates of S and T, and the y-coordinate of M is the average of the y-coordinates of S and T. That is, if S = (x₁, y₁) and T = (x₂, y₂), then

M = $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$

This is the midpoint formula.

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).

Answer

By the midpoint formula, if M is the midpoint of A and B, then:

M = $\left(\dfrac{x_A + x_B}{2}, \dfrac{y_A + y_B}{2}\right)$

Given M(−7, 1), A(3, −4) and B(x, y):

$\dfrac{3 + x}{2}$ = −7 and $\dfrac{-4 + y}{2}$ = 1

⇒ 3 + x = −14 and −4 + y = 2

⇒ x = −14 − 3 and y = 2 + 4

⇒ x = −17 and y = 6

Hence, B = (−17, 6).

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).

Answer

Since P and Q are points of trisection,

AP = PQ = QB

So P is closer to A, Q is closer to B, and the segment is divided as:

A $\quad$ P $\quad$ Q $\quad$ B

Now notice that P is the midpoint of AQ because:

AP = PQ

Similarly, Q is the midpoint of PB because:

PQ = QB

Let

P(x₁, y₁), Q(x₂, y₂)

Given:

A(4, 7), B(16, -2)

Since P is the midpoint of A and Q,

P = $\left(\dfrac{4 + x_2}{2}, \dfrac{7 + y_2}{2}\right)$

So,

x₁ = $\dfrac{4 + x_2}{2}$, y₁ = $\dfrac{7 + y_2}{2}$

Since Q is the midpoint of P and B,

Q = $\left(\dfrac{x_1 + 16}{2}, \dfrac{y_1 + (-2)}{2}\right)$

So,

$x_2 = \dfrac{x_1 + 16}{2}, \quad y_2 = \dfrac{y_1 - 2}{2}$

Now solve the x-coordinates:

$x_1 = \dfrac{4 + x_2}{2} \\[1em] \Rightarrow 2x_1 = 4 + x_2 \\[1em] \Rightarrow x_2 = 2x_1 - 4$

Also,

$x_2 = \dfrac{x_1 + 16}{2}$

Therefore,

$2x_1 - 4 = \dfrac{x_1 + 16}{2} \\[1em] \Rightarrow 4x_1 - 8 = x_1 + 16 \\[1em] \Rightarrow 3x_1 = 24 \\[1em] \Rightarrow x_1 = \dfrac{24}{3} \\[1em] \Rightarrow x_1 = 8$

So,

$x_2 = 2(8) - 4 = 12$

Now solve the y-coordinates:

$y_1 = \dfrac{7 + y_2}{2} \\[1em] \Rightarrow 2y_1 = 7 + y_2 \\[1em] \Rightarrow y_2 = 2y_1 - 7$

Also,

$y_2 = \dfrac{y_1 - 2}{2}$

Therefore,

$2y_1 - 7 = \dfrac{y_1 - 2}{2} \\[1em] \Rightarrow 4y_1 - 14 = y_1 - 2 \\[1em] \Rightarrow 3y_1 = 12 \\[1em] \Rightarrow y_1 = \dfrac{12}{3} \\[1em] \Rightarrow y_1 = 4$

So,

$y_2 = 2(4) - 7 = 1$

Hence, P = (8, 4) and Q = (12, 1).

(i) Given the points A (1, –8), B (–4, 7) and C (–7, – 4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?

(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 K.

Answer

(i) The three points lie on a circle centred at O(0, 0) if and only if their distances from O are all equal. This common distance is the radius.

$\text{OA} = \sqrt{(1 - 0)^2 + (-8 - 0)^2} \\[1em] = \sqrt{1 + 64} \\[1em] = \sqrt{65} \text{ units} \\[2em] \text{OB} = \sqrt{(-4 - 0)^2 + (7 - 0)^2} \\[1em] = \sqrt{16 + 49} \\[1em] = \sqrt{65} \text{ units} \\[2em] \text{OC} = \sqrt{(-7 - 0)^2 + (-4 - 0)^2} \\[1em] = \sqrt{49 + 16} \\[1em] = \sqrt{65} \text{ units}$

Since OA = OB = OC = $\sqrt{65}$, the points A, B and C all lie on the circle K with centre O and radius $\sqrt{65}$.

Hence, A, B and C lie on circle K, whose radius is $\sqrt{65}$ units.

(ii) For a point P, compare OP with the radius $\sqrt{65}$:

• If OP < $\sqrt{65}$, P lies inside the circle.

• If OP = $\sqrt{65}$, P lies on the circle.

• If OP > $\sqrt{65}$, P lies outside the circle.

For D(−5, 6):

$\text{OD} = \sqrt{(-5)^2 + 6^2} \\[1em] = \sqrt{25 + 36} \\[1em] = \sqrt{61} \text{ units}$

Since $\sqrt{61}$ < $\sqrt{65}$ (because 61 < 65), D lies inside circle K.

For E(0, 9):

$\text{OE} = \sqrt{0^2 + 9^2} \\[1em] = \sqrt{81} \\[1em] = 9 \text{ units}$

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

Hence, D lies within circle K, and E lies outside circle K.

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.

Answer

Let A = (a₁, a₂), B = (b₁, b₂) and C = (c₁, c₂).

Take D as the midpoint of BC, E as the midpoint of CA, and F as the midpoint of AB. Using the midpoint formula:

D = $\left(\dfrac{b_1 + c_1}{2}, \dfrac{b_2 + c_2}{2}\right)$ = (5, 1)

E = $\left(\dfrac{c_1 + a_1}{2}, \dfrac{c_2 + a_2}{2}\right)$ = (6, 5)

F = $\left(\dfrac{a_1 + b_1}{2}, \dfrac{a_2 + b_2}{2}\right)$ = (0, 3)

Working with x-coordinates:

$\dfrac{b_1 + c_1}{2}$ = 5

⇒ b₁ + c₁ = 2 x 5

⇒ b₁ + c₁ = 10 ……(1)

$\dfrac{c_1 + a_1}{2}$ = 6

⇒ c₁ + a₁ = 2 x 6

⇒ c₁ + a₁ = 12 ……(2)

$\dfrac{a_1 + b_1}{2}$ = 0

⇒ a₁ + b₁ = 2 x 0

⇒ a₁ + b₁ = 0 ……(3)

Adding (1), (2) and (3):

2(a₁ + b₁ + c₁) = 10 + 12 + 0

⇒ 2(a₁ + b₁ + c₁) = 22

⇒ a₁ + b₁ + c₁ = $\dfrac{22}{2}$

⇒ a₁ + b₁ + c₁ = 11 ……(4)

Subtracting (1) from (4):

(a₁ + b₁ + c₁) − (b₁ + c₁) = 11 − 10

⇒ a₁ = 1

Subtracting (2) from (4):

(a₁ + b₁ + c₁) − (c₁ + a₁) = 11 − 12

⇒ b₁ = −1

Subtracting (3) from (4):

(a₁ + b₁ + c₁) − (a₁ + b₁) = 11 − 0

⇒ c₁ = 11

Working with y-coordinates:

$\dfrac{b_2 + c_2}{2}$ = 1

⇒ b₂ + c₂ = 2 x 1

⇒ b₂ + c₂ = 2 ……(5)

$\dfrac{c_2 + a_2}{2}$ = 5

⇒ c₂ + a₂ = 2 x 5

⇒ c₂ + a₂ = 10 ……(6)

$\dfrac{a_2 + b_2}{2}$ = 3

⇒ a₂ + b₂ = 2 x 3

⇒ a₂ + b₂ = 6 ……(7)

Adding (5), (6) and (7):

2(a₂ + b₂ + c₂) = 2 + 10 + 6

⇒ 2(a₂ + b₂ + c₂) = 18

⇒ a₂ + b₂ + c₂ = $\dfrac{18}{2}$

⇒ a₂ + b₂ + c₂ = 9 ……(8)

Subtracting (5) from (8):

(a₂ + b₂ + c₂) − (b₂ + c₂) = 9 − 2

⇒ a₂ = 7

Subtracting (6) from (8):

(a₂ + b₂ + c₂) − (c₂ + a₂) = 9 − 10

⇒ b₂ = −1

Subtracting (7) from (8):

(a₂ + b₂ + c₂) − (a₂ + b₂) = 9 − 6

⇒ c₂ = 3

Verification:

Midpoint of BC = $\left(\dfrac{-1 + 11}{2}, \dfrac{-1 + 3}{2}\right) = (5, 1)$ = D

Midpoint of CA = $\left(\dfrac{11 + 1}{2}, \dfrac{3 + 7}{2}\right) = (6, 5)$ = E

Midpoint of AB = $\left(\dfrac{1 + (-1)}{2}, \dfrac{7 + (-1)}{2}\right) = (0, 3)$ = F

Hence, A = (1, 7), B = (−1, −1) and C = (11, 3).

A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.

(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).

Answer

(i) We draw two perpendicular main roads (one N–S, one E–W) crossing at the centre of the city. With the centre as the origin and the scale 1 cm = 200 m, the streets parallel to the N–S road are vertical lines 1 cm apart, and the streets parallel to the E–W road are horizontal lines 1 cm apart.

(ii) Each street intersection (i, j) is uniquely formed by the i^th N–S street meeting the j^th E–W street. So an ordered pair (i, j) refers to exactly one crossing in the model.

(a) (4, 3) refers to the meeting of the 4^th N–S street and the 3^rd E–W street, which is exactly one specific intersection.

Hence, exactly 1 intersection can be referred to as (4, 3).

(b) Similarly, (3, 4) refers to the meeting of the 3^rd N–S street and the 4^th E–W street, which is exactly one specific intersection (and a different one from (4, 3)).

Hence, exactly 1 intersection can be referred to as (3, 4).

This shows that the order of the coordinates matters: (4, 3) ≠ (3, 4).

A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine:

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

(ii) whether the two circles intersect each other.

Answer

The screen occupies the region 0 ≤ x ≤ 800 and 0 ≤ y ≤ 600.

(i) For Circle A (centre A(100, 150), radius 80):

x ranges from 100 − 80 = 20 to 100 + 80 = 180; both lie in [0, 800].

y ranges from 150 − 80 = 70 to 150 + 80 = 230; both lie in [0, 600].

So Circle A lies entirely inside the screen.

For Circle B (centre B(250, 230), radius 100):

x ranges from 250 − 100 = 150 to 250 + 100 = 350; both lie in [0, 800].

y ranges from 230 − 100 = 130 to 230 + 100 = 330; both lie in [0, 600].

So Circle B also lies entirely inside the screen.

Hence, no part of either circle lies outside the screen.

(ii) Distance between the two centres:

$\text{AB} = \sqrt{(250 - 100)^2 + (230 - 150)^2} \\[1em] = \sqrt{22500 + 6400} \\[1em] = \sqrt{28900} \\[1em] = 170 \text{ pixels}$

Sum of radii = 80 + 100 = 180 pixels.

Difference of radii = |100 − 80| = 20 pixels.

For two circles to intersect at two points, we need:

|r₁ − r₂| < AB < r₁ + r₂

⇒ 20 < 170 < 180

This condition is satisfied, so the two circles intersect each other at two points.

Hence, the two circles intersect each other (at two points).

Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?

Answer

The points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) are shown below:

To check whether ABCD is a square, we verify two conditions:

(1) all four sides are equal in length, and

(2) the two diagonals are equal in length.

Lengths of the sides:

$\text{AB} = \sqrt{(-1 - 2)^2 + (2 - 1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units} \\[2em] \text{BC} = \sqrt{(-2 - (-1))^2 + (-1 - 2)^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10} \text{ units} \\[2em] \text{CD} = \sqrt{(1 - (-2))^2 + (-2 - (-1))^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units} \\[2em] \text{DA} = \sqrt{(2 - 1)^2 + (1 - (-2))^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10} \text{ units}$

So AB = BC = CD = DA = $\sqrt{10}$ units. All four sides are equal.

Lengths of the diagonals:

$\text{AC} = \sqrt{(-2 - 2)^2 + (-1 - 1)^2} \\[1em] = \sqrt{16 + 4} \\[1em] = \sqrt{20} \text{ units} \\[2em] \text{BD} = \sqrt{(1 - (-1))^2 + (-2 - 2)^2} \\[1em] = \sqrt{4 + 16} \\[1em] = \sqrt{20} \text{ units}$

So AC = BD = $\sqrt{20}$ units. Both diagonals are equal.

Since all four sides are equal AND the two diagonals are equal, ABCD is a square.

Area of the square:

Area = (side)² = $(\sqrt{10})^2$ = 10 sq units.

Hence, ABCD is a square (all sides equal and diagonals equal), and its area is 10 sq units.