NCERT Solutions: Orienting Yourself: The Use of Coordinates

Exercise Set 1.1 (Page 5)

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. (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?

Answer: D₁ lies on the x-axis. Looking at the figure, D₁ appears at approximately $(8, 0)$ and R₁ is given as $(11.5, 0)$ .
Distance of the door from the left wall (y-axis) = x-coordinate of D₁ ≈ 8 ft (the door starts 8 ft from the y-axis).
Distance of the door from the x-axis = y-coordinate = 0 ft (the door lies on the x-axis, i.e., on the bottom wall of the room).

(ii)What are the coordinates of D₁?

Answer: The door D₁R₁ lies along the x-axis (bottom wall). D₁is the left end of the door.

Coordinates of D₁ = (8, 0)

The point lies on the x-axis, so its y-coordinate is 0.

(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?

Answer: D₁ = (8, 0) and R₁= (11.5, 0). Both points lie on the x-axis.

Width of door = x-coordinate of R₁ – x-coordinate of D₁
$= 11.5 - 8 = 3.5 ft$

Width of the door = 3.5 ft

A width of 3.5 ft (approximately 105 cm) is generally considered a comfortable width for a standard room door. However, for a person in a wheelchair, a minimum width of about 3 ft (90 cm) is required, and 3.5 ft provides adequate space. So yes, a person in a wheelchair should be able to enter the room fairly easily, though it may be slightly tight depending on the wheelchair model.

(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: B₁ = (0, 1.5) and B₂= (0, 4). Both points lie on the y-axis.

Width of bathroom door = y-coordinate of B₂ – y-coordinate of B₁
$= 4 - 1.5 = 2.5 ft$

Width of bathroom door = 2.5 ft

Since 2.5 ft < 3.5 ft, the bathroom door is narrower than the room door.

Think and Reflect (Page 5)

1. What are the standard widths for a room door?
The standard width for a room door is usually 75 cm to 90 cm (about 2.5 to 3 feet). In many homes, doors are around 80 cm or 90 cm wide.
2. Are the doors in your school suitable for people in wheelchairs?
Doors suitable for wheelchairs should be at least 90 cm wide. If school doors are wide enough and have easy access (no steps or narrow entries), then they are suitable. Otherwise, they may not be convenient for wheelchair users.

Think and Reflect (Page 7)

1. What is the x-coordinate of a point on the y-axis?
The x-coordinate is 0.

2. Is there a similar generalisation for a point on the x-axis?
Yes, the y-coordinate is 0.

3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.
Yes, they coincide only when x = y, because then both points become the same.

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

Exercise Set 1.2 (Page 7 )

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.
Q1. 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?

Answer:

(i) Fourth foot of the table:

Three feet are at (8, 9), (11, 9) and (11, 7). Since it is a rectangle, the fourth foot must complete the rectangle.
The pink dots represent the table
The fourth foot must share x = 8 (from the first point) and y = 7 (from the third point):
Fourth foot = (8, 7)
Fourth foot = (8, 7)

(ii) Is this a good spot for the table?
The table occupies x from 8 to 11 and y from 7 to 9. The bedroom is 12 ft × 10 ft (from O(0,0) to B(12,10)). The table is placed well within the room, away from the bathroom wall and the door. It is a reasonable spot for a study table – near the right side of the room but not blocking the wardrobe or the door.

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

Width = $∣ 11 - 8 ∣ = 3$ ft (along the x-direction)
Length = $∣ 9 - 7 ∣ = 2$ ft (along the y-direction)

Width = 3 ft, Length = 2 ft

The height of the table cannot be determined from a 2-D floor plan, since the floor map only shows horizontal (x) and vertical (y) positions – height is a third dimension (z) not represented here.

Q2. 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?

Answer:

B₁ = (0, 1.5) is the hinge point. The door swings into the bedroom. When open, the door sweeps an arc of radius equal to the door width (2.5 ft) from B₁.
The wardrobe W₁W₂W₃W₄ lies approximately between x = 3 to 7 and y = 0 to 2. The arc of the door from B₁(0, 1.5) with radius 2.5 ft extends to x ≈ 2.5, which is short of the wardrobe at x = 3.

The door will not hit the wardrobe at its current width of 2.5 ft.
If the door is made wider (say, 3 ft or more), the arc from B₁(0, 1.5) would extend further into the room and may reach the wardrobe. In that case, suggested changes include: making the door open outward (into the bathroom), or sliding the wardrobe farther from the bathroom wall.

Q3. 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.

Answer :

(i) Corners of the bathroom:
O = (0, 0), F = (0, 10), R = (-6, 10), P = (-6, 0)
The bathroom is a 6 ft × 10 ft rectangle (6 ft wide along x, 10 ft tall along y).

(ii) Showering area SHWR:
S = (-6, 6), H = (-6, 10), W = (0, 10), R = (0, 6)

(iii) Washbasin and Toilet spaces:
Marking a 3 ft × 2 ft washbasin space (for example, along the bottom of the bathroom):
Washbasin corners: (-3, 0), (0, 0), (0, 2), (-3, 2)
Marking a 2 ft × 3 ft toilet space (adjacent to it):
Toilet corners: (-6, 0), (-4, 0), (-4, 3), (-6, 3)

Q4. 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:

(i) Dining room:
P = (-6, 0) and A = (12, 0). The length from P to A is $12 - (- 6) = 18$ ft ✓. The width of the dining room is 15 ft, extending downward (in the negative y direction) from the x-axis.
Four corners of the dining room:
P = (-6, 0), A = (12, 0), (12, -15), (-6, -15)

(ii) Dining table at the centre:

Centre of the dining room:
Centre x = $\frac{- 6 + 12}{2} = 3$
Centre y = $\frac{0 + (- 15)}{2} = - 7.5$
The 5 ft × 3 ft table centred at (3, -7.5):
Half-length = 2.5 ft (along x), half-width = 1.5 ft (along y)
Feet at: $(3 - 2.5, - 7.5 - 1.5)$ , $(3 + 2.5, - 7.5 - 1.5)$ , $(3 + 2.5, - 7.5 + 1.5)$ , $(3 - 2.5, - 7.5 + 1.5)$
Feet of the dining table:
(0.5, -9), (5.5, -9), (5.5, -6), (0.5, -6)

Think and Reflect (Page 9)

1. 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?
Distance along x-axis = 7 – 3 = 4 units
Distance along y-axis = 1 – 4 = -3 units (3 units downward)

2. Can these distances help you find the distance AD?
Yes, these distances help to find AD using the distance formula:

AD = √[(7 – 3)² + (1 – 4)²]
= √[4² + (-3)²]
= √(16 + 9)
= √25 = 5 units

Think and Reflect (Page 11)

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

The shape, size, and side lengths of ΔADM remain the same.
The triangle is just flipped across the y-axis (mirror image).
The x-coordinates change sign:
A(3, 4) → A′(-3, 4), D(7, 1) → D′(-7, 1), M(9, 6) → M′(-9, 6)
The y-coordinates remain the same.

2. Would these observations be the same if ΔADM is reflected in the x-axis?

Yes, the shape and size will still remain the same.
But this time, the triangle will flip vertically.
The y-coordinates will change sign, while the x-coordinates remain the same.

End-of-Chapter Exercises (Page 12)

Q1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Answer: The point of intersection of the two coordinate axes is the origin O.
x-coordinate = 0, y-coordinate = 0. Coordinates = (0, 0).

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?

AnswerAny line parallel to the y-axis has a constant x-coordinate. Since H is on the line through W parallel to the y-axis, and W has x-coordinate = -5, every point on this line also has x-coordinate = -5.

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

Since the x-coordinate of H is -5 (negative), H can lie in:

Quadrant II – when y > 0 (i.e., H = (-5, positive))
Quadrant III – when y < 0 (i.e., H = (-5, negative))
On the x-axis when y = 0 (i.e., H = (-5, 0), which is not in any quadrant)

Q3. 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

(i) Two sides perpendicular to each other:
AM goes from A(0, -2) to M(-5, -2) – this is a horizontal segment (y is constant). MP goes from M(-5, -2) to P(-5, 2) – this is a vertical segment (x is constant). A horizontal line and a vertical line are always perpendicular

Sides AM and MP are perpendicular to each other.

(ii) One side parallel to an axis:
Side AM connects A(0, -2) and M(-5, -2). Both have y = -2, so AM is parallel to the x-axis.
Side AM is parallel to the x-axis.
Also, side MP connects M(-5, -2) and P(-5, 2), both with x = -5, so MP is parallel to the y-axis.

(iii) Two mirror image points:
Points A(0, -2) and P(-5, 2) are not mirror images. However, consider M(-5, -2) and P(-5, 2): they have the same x-coordinate and y-coordinates that are negatives of each other (-2 and +2). So M and P are mirror images in the x-axis.

Points M(-5, -2) and P(-5, 2) are mirror images of each other in the x-axis.

Q4. 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

Let I = (5, 0) on the x-axis, and N = (0, 0) = origin.
Triangle IZN has the right angle at I (since IZ is vertical and IN is horizontal).
IZ = distance from I(5, 0) to Z(5, -6) = $∣ - 6 - 0 ∣ = 6$ units
NI = distance from N(0, 0) to I(5, 0) = $∣ 5 - 0 ∣ = 5$ units
NZ (hypotenuse) = $\sqrt{5^{2} + 6^{2}} = \sqrt{25 + 36} = \sqrt{61}$ units

Three sides: IZ = 6 units, NI = 5 units, NZ = √61 units

Q5. 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 mark points to the right of and above the origin. The coordinate system would only cover Quadrant I (where both x ≥ 0 and y ≥ 0).
The axes would be like two number lines starting from 0 going only in the positive direction.
No, such a system would not allow us to locate all points on a 2-D plane – only those in Quadrant I (or on the positive parts of the axes). Points in Quadrants II, III, and IV would be inaccessible.
This is precisely why Brahmagupta’s formalisation of negative numbers was so crucial to the development of the four-quadrant Cartesian coordinate system.

*Q6. 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 (without plotting): Three points are collinear if the slope between any two pairs of points is the same.
Slope of MA = $\frac{0 - (- 4)}{0 - (- 3)} = \frac{4}{3}$
Slope of AG = $\frac{8 - 0}{6 - 0} = \frac{8}{6} = \frac{4}{3}$
Since slope of MA = slope of AG = $\frac{4}{3}$ , and point A is common, the three points are collinear.
Yes, M(-3, -4), A(0, 0) and G(6, 8) lie on the same straight line.

*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.

Answer:

Slope of RB = $\frac{- 5 - (- 1)}{- 2 - (- 5)} = \frac{- 4}{3}$
Slope of BC = $\frac{- 12 - (- 5)}{4 - (- 2)} = \frac{- 7}{6}$

Since $- \frac{4}{3} \neq - \frac{7}{6}$ , the slopes are different.
R(-5, -1), B(-2, -5) and C(4, -12) do NOT lie on the same straight line (they are non-collinear).

*Q8. 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) Right-angled isosceles triangle with origin as one vertex:
Let the right angle be at the origin O(0, 0). The two equal legs must be perpendicular.
Example: O(0, 0), A(4, 0), B(0, 4)

OA = OB = 4 units (equal legs along the axes), angle at O = 90°.(ii) Isosceles triangle with origin as vertex, one other vertex in Quadrant III and one in Quadrant IV:
Example: O(0, 0), P(-3, -4), Q(3, -4)
OP = OQ = $\sqrt{9 + 16} = 5$ units (equal sides), so it is isosceles. P is in Quadrant III, Q is in Quadrant IV.

*Q9. 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:

The midpoint formula: if M is the midpoint of ST where S = $(x_{1}, y_{1})$ and T = $(x_{2}, y_{2})$ , then:
$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
Row 1: S(-3, 0), M(0, 0), T(3, 0)
Midpoint of ST = $(\frac{- 3 + 3}{2}, \frac{0 + 0}{2}) = (0, 0) = M$
Yes, M is the midpoint of ST.
Row 2: S(2, 3), M(3, 4), T(4, 5)
Midpoint of ST = $(\frac{2 + 4}{2}, \frac{3 + 5}{2}) = (3, 4) = M$
Yes, M is the midpoint of ST.
Row 3: S(0, 0), M(0, 5), T(0, -10)
Midpoint of ST = $(\frac{0 + 0}{2}, \frac{0 + (- 10)}{2}) = (0, - 5) \neq M (0, 5)$
No, M is not the midpoint of ST.
Row 4: S(-8, 7), M(0, -2), T(6, -3)
Midpoint of ST = $(\frac{- 8 + 6}{2}, \frac{7 + (- 3)}{2}) = (- 1, 2) \neq M (0, - 2)$
No, M is not the midpoint of ST.
Connection: When M $(x, y)$ is the midpoint of S $(x_{1}, y_{1})$ and T $(x_{2}, y_{2})$ :
$x = \frac{x_{1} + x_{2}}{2} and y = \frac{y_{1} + y_{2}}{2}$
The coordinates of the midpoint are the averages of the respective coordinates of S and T.*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).

Answer:

Using the midpoint formula: M = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
For x-coordinate: $\frac{3 + x}{2} = - 7$
$\Rightarrow 3 + x = - 14 \Rightarrow x = - 17$
For y-coordinate: $\frac{- 4 + y}{2} = 1$
$\Rightarrow - 4 + y = 2 \Rightarrow y = 6$
Coordinates of 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).

Answer:

P and Q trisect AB, so AP = PQ = QB = $\frac{1}{3}$ AB.
Method: P is the midpoint of AQ, and Q is the midpoint of PB. We can use the following:
P divides AB in ratio 1:2 from A, and Q divides AB in ratio 2:1 from A.
A useful approach: P is the midpoint of A and the midpoint M of AB. M is the midpoint of AB.
Midpoint M of AB = $(\frac{4 + 16}{2}, \frac{7 + (- 2)}{2}) = (10, \frac{5}{2})$
P = midpoint of A(4, 7) and M $(10, \frac{5}{2})$ :
$P = (\frac{4 + 10}{2}, \frac{7 + \frac{5}{2}}{2}) = (7, \frac{\frac{19}{2}}{2}) = (7, \frac{19}{4})$
Q = midpoint of M $(10, \frac{5}{2})$ and B(16, -2):
$Q = (\frac{10 + 16}{2}, \frac{\frac{5}{2} + (- 2)}{2}) = (13, \frac{\frac{1}{2}}{2}) = (13, \frac{1}{4})$
P = $(7, \frac{19}{4})$ and Q = $(13, \frac{1}{4})$

*Q12.

(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) Showing A, B, C lie on circle K:
A point lies on a circle centred at origin with radius r if its distance from the origin equals r.
OA = $\sqrt{1^{2} + (- 8)^{2}} = \sqrt{1 + 64} = \sqrt{65}$
OB = $\sqrt{(- 4)^{2} + 7^{2}} = \sqrt{16 + 49} = \sqrt{65}$
OC = $\sqrt{(- 7)^{2} + (- 4)^{2}} = \sqrt{49 + 16} = \sqrt{65}$
Since OA = OB = OC = $\sqrt{65}$ , all three points are equidistant from the origin.
A, B and C all lie on circle K with centre O(0,0) and radius = √65 units.

(ii) Checking D and E:
OD = $\sqrt{(- 5)^{2} + 6^{2}} = \sqrt{25 + 36} = \sqrt{61}$
Since $\sqrt{61} < \sqrt{65}$ , D is inside the circle.
OE = $\sqrt{0^{2} + 9^{2}} = \sqrt{81} = 9$
Since $9 = \sqrt{81} > \sqrt{65} \approx 8.06$ , E is outside the circle.
D(-5, 6) lies inside circle K. | E(0, 9) lies outside circle K.

*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.

Answer:

Let A = $(x_{1}, y_{1})$ , B = $(x_{2}, y_{2})$ , C = $(x_{3}, y_{3})$ .
Let D be midpoint of BC, E be midpoint of AC, F be midpoint of AB.
D = midpoint of BC: $\frac{x_{2} + x_{3}}{2} = 5, \frac{y_{2} + y_{3}}{2} = 1$ → $x_{2} + x_{3} = 10, y_{2} + y_{3} = 2$   …(1)
E = midpoint of AC: $\frac{x_{1} + x_{3}}{2} = 6, \frac{y_{1} + y_{3}}{2} = 5$ → $x_{1} + x_{3} = 12, y_{1} + y_{3} = 10$   …(2)
F = midpoint of AB: $\frac{x_{1} + x_{2}}{2} = 0, \frac{y_{1} + y_{2}}{2} = 3$ → $x_{1} + x_{2} = 0, y_{1} + y_{2} = 6$   …(3)
Adding all three x-equations from (1),(2),(3): $2 (x_{1} + x_{2} + x_{3}) = 10 + 12 + 0 = 22$ , so $x_{1} + x_{2} + x_{3} = 11$ .
From (1): $x_{2} + x_{3} = 10$ → $x_{1} = 11 - 10 = 1$
From (2): $x_{1} + x_{3} = 12$ → $x_{2} = 11 - 12 = - 1$
From (3): $x_{1} + x_{2} = 0$ → $x_{3} = 11 - 0 = 11$
Similarly for y: $2 (y_{1} + y_{2} + y_{3}) = 2 + 10 + 6 = 18$ , so $y_{1} + y_{2} + y_{3} = 9$ .
From (1): $y_{2} + y_{3} = 2$ → $y_{1} = 9 - 2 = 7$
From (2): $y_{1} + y_{3} = 10$ → $y_{2} = 9 - 10 = - 1$
From (3): $y_{1} + y_{2} = 6$ → $y_{3} = 9 - 6 = 3$
A = (1, 7), B = (-1, -1), C = (11, 3)

Q14. 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 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) Drawing the model:

(ii)(a) Intersections referred to as (4, 3):

The notation (4, 3) refers to “4th N-S street and 3rd E-W street.” With 10 streets in each direction, there is exactly one 4th N-S street and one 3rd E-W street.

There is exactly 1 street intersection referred to as (4, 3).

(ii)(b) Intersections referred to as (3, 4):

Similarly, (3, 4) refers to the 3rd N-S street and 4th E-W street – again uniquely determined.
There is exactly 1 street intersection referred to as (3, 4).

Q15. 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 spans x from 0 to 800 and y from 0 to 600.

(i) Checking if circles lie outside the screen:

Circle A (centre (100, 150), radius 80):
Left boundary: $100 - 80 = 20 \geq 0$ ✓
Right boundary: $100 + 80 = 180 \leq 800$ ✓
Bottom boundary: $150 - 80 = 70 \geq 0$ ✓
Top boundary: $150 + 80 = 230 \leq 600$ ✓

Circle A lies entirely within the screen.

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

Left: $250 - 100 = 150 \geq 0$ ✓, Right: $250 + 100 = 350 \leq 800$ ✓
Bottom: $230 - 100 = 130 \geq 0$ ✓, Top: $230 + 100 = 330 \leq 600$ ✓

Circle B lies entirely within the screen.

(ii) Do the two circles intersect?
Distance between centres AB = $\sqrt{(250 - 100)^{2} + (230 - 150)^{2}}$
$= \sqrt{15 0^{2} + 8 0^{2}} = \sqrt{22500 + 6400} = \sqrt{28900} = 170 pixels$
Sum of radii = 80 + 100 = 180 pixels. Difference of radii = |100 – 80| = 20 pixels.
Two circles intersect if: difference of radii < distance between centres < sum of radii.
20 < 170 < 180 ✓
Yes, the two circles intersect each other.

Q 16. 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

To check if ABCD is a square, we verify that all four sides are equal and the diagonals are equal.
AB = $\sqrt{(- 1 - 2)^{2} + (2 - 1)^{2}} = \sqrt{9 + 1} = \sqrt{10}$
BC = $\sqrt{(- 2 - (- 1))^{2} + (- 1 - 2)^{2}} = \sqrt{1 + 9} = \sqrt{10}$
CD = $\sqrt{(1 - (- 2))^{2} + (- 2 - (- 1))^{2}} = \sqrt{9 + 1} = \sqrt{10}$
DA = $\sqrt{(2 - 1)^{2} + (1 - (- 2))^{2}} = \sqrt{1 + 9} = \sqrt{10}$

All four sides are equal. Now check diagonals:
AC = $\sqrt{(- 2 - 2)^{2} + (- 1 - 1)^{2}} = \sqrt{16 + 4} = \sqrt{20}$
BD = $\sqrt{(1 - (- 1))^{2} + (- 2 - 2)^{2}} = \sqrt{4 + 16} = \sqrt{20}$
Both diagonals are equal. Also, diagonal² = AB² + BC² → $20 = 10 + 10$ ✓ (diagonals and sides satisfy Pythagoras, confirming right angles).
Yes, ABCD is a square. All sides = √10, diagonals = √20.
Area = side² = $(\sqrt{10})^{2} = 10$ sq. units
Area of square ABCD = 10 square units