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CHAPTER 1 — THE USE OF COORDINATES
Class IX Mathematics — Coordinate Geometry
Preeti Kushwah Classes — Unit Test (Set 2: Standard)
General Instructions:
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Draw neat figures wherever required. Use graph paper for plotting questions.
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Draw neat figures wherever required. Use graph paper for plotting questions.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
The perpendicular distance of point (−7, 4) from the y-axis is ___.
Q2.1
If both coordinates of a point are negative, it lies in Quadrant ___.
Q3.1
The mirror image of point (3, −5) in the x-axis is ___.
Q4.1
True or False: Points (1, −1) and (−1, 1) lie in the same quadrant.
Q5.1
A point on the y-axis is 7 units above the origin. Its coordinates are ___.
Q6.1
The ordinate of every point on the x-axis is ___.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
The point (a, b) lies in Quadrant III. What can you say about the signs of a and b? Give two examples of such points.
Q8.2
Find the mirror image of each point in the y-axis: A(3, 5), B(−2, 7), C(−4, −3), D(6, −1)
Q9.2
Without plotting, determine whether the points (3, 5), (3, −2), (3, 0), (3, 7) lie on a line parallel to the x-axis or the y-axis. Justify your answer.
Q10.2
Write coordinates of a point which is equidistant from both axes and lies in Quadrant IV. Give two such points.
Q11.2
The base of an isosceles triangle lies on the x-axis with vertices at (−3, 0) and (3, 0). The third vertex is on the y-axis. Give two possible positions for the third vertex.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Plot P(−3, 4), Q(3, 4), R(3, −4), S(−3, −4). Join PQRS.
(a) Name the figure.
(b) Find the length of diagonal PR.
(c) Write the coordinates of the point of intersection of the diagonals.
(a) Name the figure.
(b) Find the length of diagonal PR.
(c) Write the coordinates of the point of intersection of the diagonals.
Q13.3
For each condition, identify which quadrant(s) the point (x, y) can lie in:
(a) x > 0 and y < 0
(b) x < 0 and y > 0
(c) x = y (where x ≠ 0)
(d) x = −y (where x > 0)
(a) x > 0 and y < 0
(b) x < 0 and y > 0
(c) x = y (where x ≠ 0)
(d) x = −y (where x > 0)
Q14.3
A triangle has vertices A(−2, 3), B(4, 3), C(1, −1). Plot the triangle.
(a) Find the length of AB.
(b) Find the perpendicular height from C to AB.
(c) Calculate the area of the triangle.
(a) Find the length of AB.
(b) Find the perpendicular height from C to AB.
(c) Calculate the area of the triangle.
Q15.3
Three vertices of a parallelogram are A(1, 2), B(4, 2), C(5, 5). Find the fourth vertex D. (Hint: In a parallelogram, opposite sides are equal and parallel.)
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
A rectangular garden on a coordinate grid has corners at (−5, −3), (5, −3), (5, 4), (−5, 4). A circular fountain has its centre at the origin.
(a) Plot the garden and mark the centre of the fountain. [1]
(b) Find the length and breadth of the garden. [1]
(c) Find the area of the garden. [1]
(d) If the fountain has radius 2 units, does any part go outside the garden? Justify. [1]
(e) A bench is placed at the reflection of (3, 2) in the y-axis. What are its coordinates? [1]
(a) Plot the garden and mark the centre of the fountain. [1]
(b) Find the length and breadth of the garden. [1]
(c) Find the area of the garden. [1]
(d) If the fountain has radius 2 units, does any part go outside the garden? Justify. [1]
(e) A bench is placed at the reflection of (3, 2) in the y-axis. What are its coordinates? [1]
Q17.5
Plot P(0, 0), Q(6, 0), R(6, 4), S(4, 7), T(2, 7), U(0, 4). Join PQRSTU in order.
(a) Name the figure formed. [1]
(b) Which sides are parallel to the x-axis? [1]
(c) Which sides are parallel to the y-axis? [1]
(d) Find the length of PQ and ST. [1]
(e) Is the hexagon symmetric about any vertical line? If yes, write its equation. [1]
(a) Name the figure formed. [1]
(b) Which sides are parallel to the x-axis? [1]
(c) Which sides are parallel to the y-axis? [1]
(d) Find the length of PQ and ST. [1]
(e) Is the hexagon symmetric about any vertical line? If yes, write its equation. [1]
Bonus Question (Optional) [2 Marks]
Q18.2
★ A point moves such that it is always equidistant from both coordinate axes. What path does it trace? Write the relationship between x and y. In how many quadrants can such points exist?
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Answer Key & Detailed Solutions
Q1. [1 Mark]
7 units.
Perpendicular distance from y-axis = |x-coordinate| = |−7| = 7.
Perpendicular distance from y-axis = |x-coordinate| = |−7| = 7.
Q2. [1 Mark]
Quadrant III.
x < 0 and y < 0 → Quadrant III.
x < 0 and y < 0 → Quadrant III.
Q3. [1 Mark]
(3, 5).
Reflection in x-axis: (x, y) → (x, −y). So (3, −5) → (3, 5).
Reflection in x-axis: (x, y) → (x, −y). So (3, −5) → (3, 5).
Q4. [1 Mark]
False.
(1, −1) lies in Quadrant IV. (−1, 1) lies in Quadrant II. Different quadrants.
(1, −1) lies in Quadrant IV. (−1, 1) lies in Quadrant II. Different quadrants.
Q5. [1 Mark]
(0, 7).
On the y-axis, x = 0. Seven units above means y = 7.
On the y-axis, x = 0. Seven units above means y = 7.
Q6. [1 Mark]
0.
Every point on the x-axis has coordinates (x, 0).
Every point on the x-axis has coordinates (x, 0).
Q7. [2 Marks]
In Quadrant III, both coordinates are negative: a < 0 and b < 0. [1 mark]
Examples: (−2, −3) and (−5, −1). [1 mark]
Examples: (−2, −3) and (−5, −1). [1 mark]
Q8. [2 Marks]
Reflection in y-axis: (x, y) → (−x, y).
A(3, 5) → A′(−3, 5) [½ mark]
B(−2, 7) → B′(2, 7) [½ mark]
C(−4, −3) → C′(4, −3) [½ mark]
D(6, −1) → D′(−6, −1) [½ mark]
A(3, 5) → A′(−3, 5) [½ mark]
B(−2, 7) → B′(2, 7) [½ mark]
C(−4, −3) → C′(4, −3) [½ mark]
D(6, −1) → D′(−6, −1) [½ mark]
Q9. [2 Marks]
All four points have x-coordinate = 3. Since the x-coordinate is constant, these points lie on a vertical line (line x = 3), which is parallel to the y-axis. [1 mark for observation, 1 mark for conclusion]
Q10. [2 Marks]
In Quadrant IV: x > 0, y < 0. Equidistant from both axes means |x| = |y|, so x = −y (since y is negative). [1 mark]
Examples: (1, −1) and (5, −5). [1 mark]
Examples: (1, −1) and (5, −5). [1 mark]
Q11. [2 Marks]
Third vertex is on y-axis: (0, k) where k ≠ 0. Since base vertices (−3, 0) and (3, 0) are symmetric about the y-axis, any point on the y-axis makes the triangle isosceles. [1 mark]
Two possible positions: (0, 4) above x-axis, and (0, −4) below x-axis. [1 mark]
Two possible positions: (0, 4) above x-axis, and (0, −4) below x-axis. [1 mark]
Q12. [3 Marks]
(a) Rectangle (PQ = 6 horizontal, QR = 8 vertical). [1 mark]
(b) PR = √((3−(−3))² + (−4−4)²) = √(36 + 64) = √100 = 10 units. [1 mark]
(c) Intersection of diagonals = midpoint of PR = ((−3+3)/2, (4+(−4))/2) = (0, 0) = origin. [1 mark]
(b) PR = √((3−(−3))² + (−4−4)²) = √(36 + 64) = √100 = 10 units. [1 mark]
(c) Intersection of diagonals = midpoint of PR = ((−3+3)/2, (4+(−4))/2) = (0, 0) = origin. [1 mark]
Q13. [3 Marks]
(a) x > 0, y < 0 → Quadrant IV. [¾ mark]
(b) x < 0, y > 0 → Quadrant II. [¾ mark]
(c) x = y (non-zero): both positive → QI, or both negative → QIII. [¾ mark]
(d) x = −y with x > 0: x positive, y negative → QIV. [¾ mark]
(b) x < 0, y > 0 → Quadrant II. [¾ mark]
(c) x = y (non-zero): both positive → QI, or both negative → QIII. [¾ mark]
(d) x = −y with x > 0: x positive, y negative → QIV. [¾ mark]
Q14. [3 Marks]
(a) AB = |4 − (−2)| = 6 units (horizontal line at y = 3). [1 mark]
(b) Height from C(1, −1) to line AB (y = 3): perpendicular distance = |−1 − 3| = 4 units. [1 mark]
(c) Area = ½ × base × height = ½ × 6 × 4 = 12 sq units. [1 mark]
(b) Height from C(1, −1) to line AB (y = 3): perpendicular distance = |−1 − 3| = 4 units. [1 mark]
(c) Area = ½ × base × height = ½ × 6 × 4 = 12 sq units. [1 mark]
Q15. [3 Marks]
In parallelogram ABCD: AB ∥ DC and AD ∥ BC.
Vector AB = (4−1, 2−2) = (3, 0). Since DC must equal AB:
D = C − AB = (5−3, 5−0) = (2, 5). [2 marks]
Verify: AD = (2−1, 5−2) = (1, 3), BC = (5−4, 5−2) = (1, 3). AD = BC ✓ [1 mark]
Vector AB = (4−1, 2−2) = (3, 0). Since DC must equal AB:
D = C − AB = (5−3, 5−0) = (2, 5). [2 marks]
Verify: AD = (2−1, 5−2) = (1, 3), BC = (5−4, 5−2) = (1, 3). AD = BC ✓ [1 mark]
Q16. [5 Marks]
(a) Plot rectangle with 4 corners and mark O(0,0) as fountain centre. [1 mark]
(b) Length = 5 − (−5) = 10 units. Breadth = 4 − (−3) = 7 units. [1 mark]
(c) Area = 10 × 7 = 70 sq units. [1 mark]
(d) Nearest boundary to origin: y = −3 (distance = 3 units). Since 3 > 2 (radius), the fountain stays entirely inside the garden. No part goes outside. [1 mark]
(e) Reflection of (3, 2) in y-axis: (−3, 2). [1 mark]
(b) Length = 5 − (−5) = 10 units. Breadth = 4 − (−3) = 7 units. [1 mark]
(c) Area = 10 × 7 = 70 sq units. [1 mark]
(d) Nearest boundary to origin: y = −3 (distance = 3 units). Since 3 > 2 (radius), the fountain stays entirely inside the garden. No part goes outside. [1 mark]
(e) Reflection of (3, 2) in y-axis: (−3, 2). [1 mark]
Q17. [5 Marks]
(a) Hexagon (6-sided polygon). [1 mark]
(b) PQ lies on y = 0 and ST lies on y = 7. Both parallel to x-axis. [1 mark]
(c) QR lies on x = 6 and UP lies on x = 0. Both parallel to y-axis. [1 mark]
(d) PQ = |6 − 0| = 6 units. ST = |4 − 2| = 2 units. [1 mark]
(e) Yes, symmetric about the line x = 3 (vertical line through the middle). Check: P(0,0)↔Q(6,0), U(0,4)↔R(6,4), T(2,7)↔S(4,7) — all symmetric about x = 3. [1 mark]
(b) PQ lies on y = 0 and ST lies on y = 7. Both parallel to x-axis. [1 mark]
(c) QR lies on x = 6 and UP lies on x = 0. Both parallel to y-axis. [1 mark]
(d) PQ = |6 − 0| = 6 units. ST = |4 − 2| = 2 units. [1 mark]
(e) Yes, symmetric about the line x = 3 (vertical line through the middle). Check: P(0,0)↔Q(6,0), U(0,4)↔R(6,4), T(2,7)↔S(4,7) — all symmetric about x = 3. [1 mark]
Q18. Bonus [2 Marks]
Points equidistant from both axes: |x| = |y|.
This gives two lines: y = x and y = −x. [1 mark]
These lines pass through the origin at 45° angles, forming an X shape. Points can exist in all 4 quadrants:
QI: y = x (e.g., (2,2)); QII: y = −x (e.g., (−3,3)); QIII: y = x (e.g., (−2,−2)); QIV: y = −x (e.g., (3,−3)). [1 mark]
This gives two lines: y = x and y = −x. [1 mark]
These lines pass through the origin at 45° angles, forming an X shape. Points can exist in all 4 quadrants:
QI: y = x (e.g., (2,2)); QII: y = −x (e.g., (−3,3)); QIII: y = x (e.g., (−2,−2)); QIV: y = −x (e.g., (3,−3)). [1 mark]