NCERT Solutions for Class 9 Maths Chapter 1 Orienting Yourself the Use of Coordinates (PDF) – 2027

Because Fig. 1.1 is a map of the floor only, and a window is not on the floor. A point on the floor needs just two numbers: how far along the room and how far across it. A window is up on a wall, at some height above the floor. Height is a third measurement, and a flat floor map has no way to show it.

(Doors can be marked, because a doorway reaches down to the floor and leaves a gap in the wall line of the map.)

In Fig. 1.3 the door D₁R₁ lies on the bottom wall of the room, which is the x-axis. D₁ is at the +8 mark and R₁ is (11.5, 0).

D₁ is the pin on the x-axis at the +8 mark.

D₁ and R₁ both lie on the x-axis, so the width is just the difference of the x-coordinates.

Yes, it is a comfortable width. Room doors in Indian homes are usually 2.5 ft to 3 ft (750–900 mm) wide, so 3.5 ft is generous.

Yes, a wheelchair user can enter easily. India’s Harmonised Guidelines for barrier-free buildings ask for a clear door opening of at least 900 mm (about 3 ft). The door is 42 in, about 1070 mm, which is more than enough.

B₁ and B₂ both lie on the y-axis, so their separation is the difference of the y-coordinates.

The bathroom door is narrower, by 1 ft.

usual widths in Indian homes and schools (your own measurements may differ a little):

Doors are usually 2.0–2.1 m (about 6.5–7 ft) tall.

A door suits a wheelchair only if it passes all three checks:

1. Clear width of at least 900 mm (India’s Harmonised Guidelines for Barrier Free Built Environment).
2. No raised step at the door; where the floor level changes, a gentle ramp (about 1 : 12).
3. Turning space of about 1500 mm × 1500 mm near the door.

In many schools the classroom doors pass the width check, but toilet doors (often about 750 mm) fail it, so those doors are not yet suitable.

Both coordinates are non-zero, and their two signs together decide which of the four quadrants the point is in.

S (3, −5) and Q (−5, 3) are fixed by the question. P and R may be chosen freely — here P (4, 2) and R (−4, −3).

A point on the y-axis is at zero distance from the y-axis, so its x-coordinate is 0. Every such point has the form (0, y), for example H (0, 4) in Fig. 1.2.

Yes, with the roles of the two axes swapped.

A point on the x-axis is at zero distance from the x-axis, so its y-coordinate is 0; for example B (4.5, 0) in Fig. 1.2.

Yes, but only when x = y.

For example P (4, 4) and Q (4, 4) are the same point. But P (4, 7) and Q (7, 4) are two different points.

Yes, the claim is true. Two ordered pairs are equal only when their first entries are equal and their second entries are equal.

(i) The fourth foot is at (8, 7).

(ii) Yes, it is a good spot. The table covers 8 ≤ x ≤ 11 and 7 ≤ y ≤ 9. It does not overlap the bed (x only up to 6.5), the wardrobe (y only up to 2) or the room door, which is on the bottom wall (y = 0).

(iii)

No, it will not hit the wardrobe — but the clearance is only 0.5 ft.

If the door is made wider (3 ft or more, it would hit the wardrobe), suggested changes:

• move the wardrobe 0.5–1 ft to the right;
• hinge the door at B₂ (0, 4): the nearest wardrobe corner W₄ (3, 2) is then √13 ≈ 3.6 ft away;
• make the door open into the bathroom, or use a sliding door.

(i) The bathroom lies to the left of the y-axis:

(ii) SHWR is a trapezium (exactly one pair of parallel sides).

(iii) Answers may differ. One arrangement, clear of the shower and the doorway:

(i) PA = |12 − (−6)| = 18 ft, the given length. The bedroom and bathroom are above PA, so the dining room lies below the x-axis, 15 ft deep.

(ii) The centre of a rectangle is the midpoint of a diagonal.

None of AD, DM, MA is parallel to an axis, so we cannot subtract coordinates directly. Instead, make each side the hypotenuse of a right triangle whose legs are parallel to the axes.

In Fig. 1.7 these two shifts appear as the segments CD = 4 and AC = 3, where C = (3, 1).

Yes. The two shifts are the legs of a right triangle whose hypotenuse is AD.

Reflection in the y-axis sends (x, y) to (−x, y).

So: the image triangle is congruent to ΔADM — same size and shape — but it is in a different place and is flipped over.

Yes — every observation carries over, with the roles of x and y interchanged.

The triangle moves from Quadrant I to Quadrant IV. Side lengths, angles and area stay the same; now the x-coordinates stay the same and the y-coordinates change sign; and the image is again flipped over.

The two axes meet at the origin.

H is on the vertical line through W, so it has the same x-coordinate, −5; its y-coordinate can be any number.

So H can lie only in Quadrant II or Quadrant III — never in I or IV, because those need a positive x-coordinate.

(i) AM and MP are perpendicular.

(ii) AM is parallel to the x-axis (both ends have y = −2). MP is parallel to the y-axis, so either answer is acceptable; the other two sides, RA and PR, are parallel to neither.

(iii) M (−5, −2) and P (−5, 2) are mirror images in the x-axis.

Z (5, −6) is in Quadrant IV: 5 to the right of O, then 6 down. The easiest way to guarantee a right angle at Z is to take one other vertex directly above it and one directly beside it.

No. Without negative numbers you could label only one quarter of the plane.

A number without a sign tells only how far, not in which direction. So (3, 2) and (−3, 2) would get the same label, and the system could not locate all points of the plane.

Yes, M, A and G are collinear, with A lying between M and G.

Method — use distances. Three points are on one line exactly when the longest of the three distances equals the sum of the other two.

Another way: the shift from M to A is (3, 4) and from A to G is (6, 8) = 2 × (3, 4). The direction does not change, so the points lie on one line.

No — R, B and C are not collinear, although they come remarkably close.

Distance method (from Problem 6):

On the graph: M, A, G plot on a straight line. R, B, C look almost straight: B is only about 0.2 unit (about 2 mm on your graph at 1 cm = 1 unit) off the line RC, so the bend is easy to miss. The exact test settles it.

Answers may differ. Two convenient choices:

(i) Right-angled isosceles triangle: O (0, 0), (4, 0), (0, 4).

(ii) Isosceles triangle with one vertex in Quadrant III and one in Quadrant IV: O (0, 0), U (−3, −4), V (3, −4).

Test each row by averaging the coordinates of S and T and comparing with M.

The connection: each coordinate of M is the average of the corresponding coordinates of S and T.

Apply the averaging rule one coordinate at a time and solve for the unknown.

Check: midpoint of A (3, −4) and B (−17, 6) = ((3 − 17)/2, (−4 + 6)/2) = (−7, 1) = M ✓

Method: the midpoint is half of the shift from A to B; P and Q are one-third and two-thirds of the shift. Also, P is the midpoint of AQ and Q is the midpoint of PB.

For A (4, 7) and B (16, −2):

Check using midpoints:

(i) Find the distance of each point from O.

(ii) Compare each distance with the radius. It is cleaner to compare the squares and avoid roots altogether:

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

(i) Draw 10 vertical lines (the N–S streets) and 10 horizontal lines (the E–W streets), each set spaced 1 cm apart, since 200 m at the scale 1 cm = 200 m is exactly 1 cm. Number the N–S streets 1 to 10 from left to right and the E–W streets 1 to 10 from bottom to top. The grid has 10 × 10 = 100 crossings.

(ii) (a) Exactly one. (b) Exactly one — but a different one.

With the origin at the bottom-left corner, the visible screen is exactly the set of points with 0 ≤ x ≤ 800 and 0 ≤ y ≤ 600.

(i) No — both circles lie completely on the screen. A circle stays inside the screen when its centre is at least one radius away from every edge.

(ii) Yes, the two circles intersect (they cut each other at two points).

Yes, ABCD is a square, and its area is 10 square units.

Area = side² = (√10)² = 10 square units.