CBSE Class 9 Maths Ganita Manjari Part 1 Ch 01 Orienting Yourself The Use of Coordinates Short Answer Questions with Answers

2 Mark Questions

Question 1. A point A is located on the x-axis at a distance of 7 units to the left of the origin, and point B is located on the y-axis at a distance of 24 units above the origin. Write the coordinates of A and B, and find the straight-line distance AB.
Answer:
A lies on the x-axis, 7 units to the left of the origin, so \( A(-7, 0) \).
B lies on the y-axis, 24 units above the origin, so \( B(0, 24) \).
\( AB = \sqrt{(0 - (-7))^2 + (24 - 0)^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} \)
\( \therefore AB = 25 \) units
In simple words: A and B form a right triangle with the origin, with legs 7 and 24, so the slanting side AB is 25 units.

Exam Tip: A point on the x-axis always has y = 0, and "to the left" means the x-coordinate is negative.

Question 2. Explain why Brahmagupta's formalization of zero and negative numbers was a necessary prerequisite for the development of the modern four-quadrant Cartesian coordinate system.
Answer:
1. The Cartesian plane is built from two perpendicular number lines that meet at the origin (0, 0). The origin is simply the number zero on both axes, so the whole system needs zero as its reference point.
2. Points to the left of the y-axis have negative x-coordinates, and points below the x-axis have negative y-coordinates. Quadrants II, III and IV all need at least one negative coordinate.
3. Brahmagupta (in his Brāhmasphuṭasiddhānta, 628 CE) treated zero and negative numbers as real numbers and gave rules for adding, subtracting and multiplying them.
4. Without zero there would be no fixed origin, and without negative numbers only Quadrant I could be described. So his work was needed before a complete four-quadrant system could exist.
In simple words: The origin is zero and three of the four quadrants need negative numbers, so the coordinate plane cannot work without them.

Exam Tip: Link each idea to a part of the plane: zero gives the origin, and negative numbers give Quadrants II, III and IV.

Question 3. Point M(x, y) is the midpoint of a line segment joining S(−8, 7) and T(6, −3). Calculate the values of x and y using coordinate principles.
Answer:
Midpoint formula: \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
\( x = \frac{-8 + 6}{2} = \frac{-2}{2} = -1 \)
\( y = \frac{7 + (-3)}{2} = \frac{4}{2} = 2 \)
\( \therefore x = -1,\ y = 2 \), so M is (−1, 2).
In simple words: The midpoint is the average of the two x-values and the average of the two y-values.

Exam Tip: Take care with signs. Adding −3 to 7 gives 4, not 10.

Question 4. A rectangular room floor has corner coordinates O(0, 0), A(12, 0), B(12, 10) and C(0, 10). A circular rug of radius 3 feet is placed with its centre at (6, 5). Determine whether any part of the rug extends outside the room boundaries.
Answer:
The room covers \( 0 \le x \le 12 \) and \( 0 \le y \le 10 \).
The rug extends 3 ft on every side of its centre (6, 5).
x-range of rug: 6 − 3 = 3 to 6 + 3 = 9. Since \( 0 \le 3 \) and \( 9 \le 12 \), it fits left to right.
y-range of rug: 5 − 3 = 2 to 5 + 3 = 8. Since \( 0 \le 2 \) and \( 8 \le 10 \), it fits bottom to top.
\( \therefore \) No part of the rug extends outside the room. There is a gap of 3 ft from the side walls and 2 ft from the other two walls.
In simple words: The rug's leftmost, rightmost, lowest and highest points all stay inside the room.

Exam Tip: To check whether a circle fits inside a rectangle, compare centre ± radius with the rectangle's boundaries in both directions.

Question 5. Without plotting on graph paper, determine whether the point P(−4.5, 0) and point Q(0, 3.8) lie in any quadrant or on coordinate axes. Justify your reasoning.
Answer:
P(−4.5, 0): The y-coordinate is 0, so P lies on the x-axis. Since x = −4.5 is negative, P is on the negative x-axis.
Q(0, 3.8): The x-coordinate is 0, so Q lies on the y-axis. Since y = 3.8 is positive, Q is on the positive y-axis.
\( \therefore \) Neither P nor Q lies in any quadrant. Points on the axes separate the quadrants and do not belong to any of them.
In simple words: If either coordinate is zero, the point sits on an axis, not inside a quadrant.

Exam Tip: y = 0 means the point is on the x-axis, and x = 0 means it is on the y-axis. Students often mix these two up.

3 Mark Questions

Question 6. Three points are given as M(−3, −4), A(0, 0) and G(6, 8).
(i) Suggest and derive a mathematical method using distances to check if M, A and G lie on the same straight line (are collinear) without plotting them.
(ii) Execute your method to conclude whether they are collinear.
Answer:
(i) Method: Find the three distances MA, AG and MG. If the point in the middle lies on the straight path, then (shorter distance) + (other shorter distance) = (longest distance). If the points form a triangle instead, the sum of two sides is always greater than the third side.
(ii) Execution:
\( MA = \sqrt{(0+3)^2 + (0+4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
\( AG = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \)
\( MG = \sqrt{(6+3)^2 + (8+4)^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \)
\( MA + AG = 5 + 10 = 15 = MG \)
\( \therefore \) M, A and G are collinear, with A lying between M and G.
In simple words: Going from M to A and then A to G is exactly as long as going straight from M to G, so there is no bend and the points are on one line.

Exam Tip: Always add the two smaller distances and compare them with the largest one, not with any random pair.

Question 7. In Reiaan's bathroom layout, the bathroom occupies a rectangular region with vertices P(−6, 0), O(0, 0), F(0, 10) and R(−6, 10).
(i) A washbasin area of size 3 ft × 2 ft is placed against the left wall (x = −6) in the bottom corner starting from y = 0. Write the coordinates of its four corners.
(ii) A toilet space of size 2 ft × 3 ft is placed along the y-axis starting above the door hinge B₂(0, 4) up to y = 7. Write the coordinates of its four corners.
Answer:
(i) The washbasin starts at the bottom-left corner (−6, 0). Taking 3 ft away from the left wall (along x) and 2 ft up the wall (along y), it spans x = −6 to x = −3 and y = 0 to y = 2.
Corners: (−6, 0), (−3, 0), (−3, 2), (−6, 2)
(ii) The toilet space runs along the y-axis from y = 4 to y = 7, which is 3 ft. It extends 2 ft into the bathroom (to the left), so it spans x = −2 to x = 0.
Corners: (0, 4), (−2, 4), (−2, 7), (0, 7)
In simple words: Start from the given corner or wall, then move the given lengths along x and y to find the other three corners.

Exam Tip: The bathroom lies at x < 0, so anything placed inside it must move in the negative x-direction from the y-axis.

Question 8. Points P and Q are points of trisection of the line segment AB, where P is closer to A and Q is closer to B. Given A(4, 7) and B(16, −2):
(i) Find the midpoint of AB (call it M).
(ii) Using midpoint concepts, calculate the precise coordinates of trisection points P and Q.
Answer:
(i) \( M = \left( \frac{4 + 16}{2}, \frac{7 + (-2)}{2} \right) = (10,\ 2.5) \)
(ii) Since AP = PQ = QB:
• M is also the midpoint of PQ, so if P = (x, y), then Q = (2 × 10 − x, 2 × 2.5 − y) = (20 − x, 5 − y).
• P is the midpoint of AQ, so \( x = \frac{4 + (20 - x)}{2} \Rightarrow 2x = 24 - x \Rightarrow x = 8 \)
and \( y = \frac{7 + (5 - y)}{2} \Rightarrow 2y = 12 - y \Rightarrow y = 4 \)
• So P = (8, 4) and Q = (20 − 8, 5 − 4) = (12, 1).
\( \therefore P(8, 4) \) and \( Q(12, 1) \)
Check: Each step from A → P → Q → B moves +4 in x and −3 in y, so all three parts are equal, each of length 5 units.
In simple words: Trisection cuts AB into three equal pieces, so P is one-third of the way and Q is two-thirds of the way from A to B.

Exam Tip: The midpoint of AM is a quarter-point, not a trisection point. Trisection points are at one-third and two-thirds of the segment.

Question 9. The points R(3, 0), A(0, −2), M(−5, −2) and P(−5, 2) are joined in order to form quadrilateral RAMP.
(i) Identify two sides that are perpendicular to each other.
(ii) Identify one side that is parallel to the x-axis.
(iii) Identify two vertices that are mirror images of each other across an axis, and name the axis.
Answer:
(i) Side AM joins (0, −2) and (−5, −2). Both points have y = −2, so AM is horizontal. Side MP joins (−5, −2) and (−5, 2). Both points have x = −5, so MP is vertical. A horizontal line and a vertical line meet at 90°, so AM ⊥ MP.
(ii) Side AM is parallel to the x-axis, because both its endpoints have the same y-coordinate (−2).
(iii) M(−5, −2) and P(−5, 2) have the same x-coordinate and opposite y-coordinates. So M and P are mirror images of each other in the x-axis.
In simple words: Same y means a horizontal side, same x means a vertical side, and flipping the sign of y reflects a point in the x-axis.

Exam Tip: Reflection in the x-axis changes the sign of y. Reflection in the y-axis changes the sign of x.

Question 10. Two circular computer icons are given: Icon 1 is centred at A(100, 150) with radius 80 pixels, and Icon 2 is centred at B(250, 230) with radius 100 pixels.
(i) Calculate the exact distance between centres A and B.
(ii) Determine analytically whether the two circular icons overlap/intersect each other.
Answer:
(i) \( AB = \sqrt{(250 - 100)^2 + (230 - 150)^2} = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900} = 170 \) pixels
(ii) Sum of radii = 80 + 100 = 180 pixels. Difference of radii = 100 − 80 = 20 pixels.
Two circles overlap when (difference of radii) < (distance between centres) < (sum of radii).
Here 20 < 170 < 180.
\( \therefore \) The icons overlap and their edges cross at two points. Along the line joining the centres, they overlap by 180 − 170 = 10 pixels.
In simple words: The centres are closer together than the two radii added up, so the circles run into each other.

Exam Tip: If the distance between centres equals the sum of the radii, the circles only touch. If it is more, they are completely separate.

Question 11. Construct an isosceles triangle on a Cartesian plane such that one vertex is at the origin O(0, 0), the second vertex lies in Quadrant III, and the third vertex lies in Quadrant IV. Provide explicit coordinates for all three vertices and prove analytically that two side lengths are equal.
Answer:
Choose O(0, 0), P(−4, −3) in Quadrant III and Q(4, −3) in Quadrant IV. P and Q are mirror images in the y-axis.
\( OP = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units
\( OQ = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units
\( PQ = |4 - (-4)| = 8 \) units (P and Q have the same y-coordinate)
\( \therefore OP = OQ = 5 \) units, so triangle OPQ is isosceles.
In simple words: Picking two points that are mirror images across the y-axis makes them equally far from the origin, which gives two equal sides.

Exam Tip: Other correct answers are possible, such as (−a, −b) and (a, −b) for any positive a and b. Always show the distance calculation as proof.