CBSE Class 9 Mathematics Ganita Manjari Chapter 01 Orienting Yourself The Use of Coordinates MCQs with Answers Set 01

Answer: (b) Rectangle; Quadrant III

Explanation:
1. The four vertices lie one in each quadrant, placed symmetrically about both axes.
2. The horizontal sides have length \( 2a \) and the vertical sides have length \( 2b \), so opposite sides are equal.
3. The sides are parallel to the axes, so every corner is a right angle. The figure is a rectangle. It becomes a square only in the special case \( a = b \).
4. The vertex \( (-a, -b) \) has a negative x-coordinate and a negative y-coordinate, so it lies in Quadrant III.

Answer: (c) City B is the reflection of City A through the origin (0, 0).

Explanation:
1. City A \( (-d, k) \) lies in Quadrant II and City B \( (d, -k) \) lies in Quadrant IV, so (d) is wrong.
2. Reflection in the x-axis would give \( (-d, -k) \), and reflection in the y-axis would give \( (d, k) \). Neither matches B.
3. Going from A to B, both coordinates change sign: \( (x, y) \to (-x, -y) \). This is reflection through the origin.

Answer: (d) The desk is positioned completely inside the bathroom space.

Explanation:
1. The desk is rectangular. The fourth corner takes its x-value from (8, 9) and its y-value from (11, 7), giving (8, 7). So (a) is correct.
2. The desk runs from x = 8 to x = 11, which is 3 units parallel to the x-axis. It runs from y = 7 to y = 9, which is 2 units parallel to the y-axis. So (c) is correct.
3. The area is \( 3 \times 2 = 6 \) square feet, so (b) is correct.
4. The desk lies between x = 8 and x = 11, well inside the bedroom (x = 0 to 12). In Reiaan's floor plan the bathroom is to the left of the y-axis (x < 0). So statement (d) is the incorrect one.

Answer: (b) (−4, 7)

Explanation:
1. The distance from the x-axis is the size of the y-coordinate, so \( |y| = 7 \).
2. The distance from the y-axis is the size of the x-coordinate, so \( |x| = 4 \).
3. In Quadrant II, x is negative and y is positive, so x = −4 and y = 7.
4. Therefore P = (−4, 7).

Answer: (c) If and only if \( x = y \)

Explanation:
1. Two points are the same only when their x-coordinates are equal and their y-coordinates are equal.
2. For (x, y) = (y, x), we need x = y and y = x. Both conditions say the same thing: x = y.
3. For example, (3, 3) and (3, 3) coincide, but (2, 5) and (5, 2) are different points.

Answer: (b) (0, −6.5) and distance is 6.5 units

Explanation:
1. Every point on the y-axis has x-coordinate 0.
2. Being 6.5 units below the x-axis means y = −6.5. So the point is (0, −6.5).
3. Its distance from the origin is \( \sqrt{0^2 + (-6.5)^2} = 6.5 \) units.
4. Option (c) is wrong because a distance can never be negative.

Answer: (b) Yes, because the door width is 3.5 feet.

Explanation:
1. Both points lie on the x-axis, so the width is the difference of their x-coordinates.
2. Width = \( |11.5 - 8| = 3.5 \) feet.
3. Since 3.5 ft is more than the required 3.2 ft, the door complies with the guideline.

Answer: (a) \( 5 + \sqrt{29} + \sqrt{40} \) units

Explanation:
1. \( AD = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
2. \( DM = \sqrt{(9-7)^2 + (6-1)^2} = \sqrt{4 + 25} = \sqrt{29} \)
3. \( MA = \sqrt{(9-3)^2 + (6-4)^2} = \sqrt{36 + 4} = \sqrt{40} \)
4. Perimeter = \( 5 + \sqrt{29} + \sqrt{40} \) units. (Note that \( \sqrt{40} \) can also be written as \( 2\sqrt{10} \).)

Answer: (c) The signs of the x-coordinates of its vertices

Explanation:
1. Reflection in the y-axis changes (x, y) to (−x, y).
2. So A′ = (−3, 4), D′ = (−7, 1) and M′ = (−9, 6). Only the signs of the x-coordinates change.
3. A reflection is like flipping the triangle over. Its shape and size stay the same, so side lengths, perimeter and area do not change.

Answer: (b) (0, 4)

Explanation:
1. Midpoint M = \( \left( \frac{-3 + 3}{2}, \frac{0 + 0}{2} \right) = (0, 0) \).
2. Shifting up by 4 units adds 4 to every y-coordinate: S′ = (−3, 4) and T′ = (3, 4).
3. New midpoint M′ = \( \left( \frac{-3 + 3}{2}, \frac{4 + 4}{2} \right) = (0, 4) \). The midpoint moves up by 4 units along with the segment.

Answer: (a) \( |-5 - 5| = 10 \) units along the x-axis, since the y-coordinates are equal

Explanation:
1. Both points have y = −2, so segment PQ is horizontal (parallel to the x-axis).
2. For a horizontal segment, distance = difference of the x-coordinates = \( |-5 - 5| = 10 \) units.
3. Option (b) is also wrong on its own terms, because \( |-2 - (-2)| = 0 \), not 4.

Answer: (b) Quadrant III; 10 units

Explanation:
1. Both coordinates are negative, so the hub is in Quadrant III.
2. Distance from origin = \( \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) units.
3. 14 units (6 + 8) would be the distance travelled along the avenues, not the crow-fly distance.

Answer: (a) \( \sqrt{65} \) units

Explanation:
1. The radius is the distance from the centre O(0, 0) to any point on the circle.
2. \( OA = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65} \)
3. Check: \( OB = \sqrt{16 + 49} = \sqrt{65} \) and \( OC = \sqrt{49 + 16} = \sqrt{65} \). All three distances match.
4. Option (d) is \( r^2 \), not r.

Answer: (b) x_min = 20, x_max = 180

Explanation:
1. The circle reaches one radius to the left and one radius to the right of its centre.
2. x_min = 100 − 80 = 20 and x_max = 100 + 80 = 180.
3. Option (c) gives the y-range (150 − 80 = 70 to 150 + 80 = 230), not the x-range.

Answer: (b) Square of area 10 sq units

Explanation:
1. \( AB^2 = (-1-2)^2 + (2-1)^2 = 9 + 1 = 10 \). In the same way, \( BC^2 = CD^2 = DA^2 = 10 \). All four sides are equal to \( \sqrt{10} \).
2. Diagonals: \( AC^2 = (-4)^2 + (-2)^2 = 20 \) and \( BD^2 = 2^2 + (-4)^2 = 20 \). The diagonals are equal.
3. Equal sides and equal diagonals mean ABCD is a square.
4. Area = side² = \( (\sqrt{10})^2 = 10 \) sq units.

Answer: (a) 48 sq units

Explanation:
1. Keeping 2 units away from every wall, the rug runs from x = 2 to x = 10 and from y = 2 to y = 8.
2. Length = 10 − 2 = 8 units and breadth = 8 − 2 = 6 units.
3. Area = \( 8 \times 6 = 48 \) sq units.
4. Remember that 2 units are removed from both ends of each side, so each dimension shrinks by 4 units.

Answer: (b) (3, −5)

Explanation:
1. Distance from the y-axis gives |x| = 3, and distance from the x-axis gives |y| = 5.
2. In Quadrant IV, x is positive and y is negative.
3. So K = (3, −5). Option (d) mixes up the two distances.

Answer: (a) It can only represent points lying in Quadrant I and on the positive axes.

Explanation:
1. Quadrants II, III and IV all need at least one negative coordinate.
2. With only non-negative numbers, the only points available have x ≥ 0 and y ≥ 0. That is Quadrant I together with the positive axes and the origin.
3. The origin (0, 0) can still be defined, and distances can still be measured, so (b) and (c) are wrong.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. P(−4, 0) has y = 0, so it lies on the x-axis. Its x-value is negative, so it is on the negative side. Its distance from the y-axis is |−4| = 4 units. The Assertion is true.
2. The Reason correctly states that points on the x-axis have y = 0 and that the abscissa gives the signed distance from the y-axis. The Reason is true.
3. The Reason is exactly the rule used to place P, so it correctly explains the Assertion.

Answer: (d) A is false, but R is true.

Explanation:
1. P(0, −5) has x = 0, so it lies on the negative y-axis, not inside Quadrant IV. The Assertion is false.
2. The Reason is a correct rule: points on an axis do not belong to any quadrant. The Reason is true.