Chapter 1 practice
Practice: Orienting Yourself: The Use of Coordinates
A question bank of 50 exam-style questions on this chapter only: 20 one-mark, 12 two-mark, 10 three-mark, 8 four-mark. Every question is followed by its answer and a full explanation. Use the buttons to list only the questions of one mark value.
Sources are shown on each question. "NCERT Exemplar" questions are adapted from NCERT's official Exemplar Problems. "Board pattern" questions follow standard CBSE exam questions on the same skills. "Written for this site" questions cover this chapter's own topics. Every answer has been worked out and checked.
The point (−3, 5) lies in the
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(b) second quadrant. x is negative (left) and y is positive (up), which is the pattern (−, +).
The signs of the x-coordinate and y-coordinate of a point in the second quadrant are, in that order,
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(c) −, +. Quadrant II is left of the y-axis (x negative) and above the x-axis (y positive).
The point (0, −7) lies
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(c) on the y-axis. Its x-coordinate is 0, so it has not moved left or right. It sits on the y-axis, 7 units below O.
The point (−10, 0) lies
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(a). y = 0 puts it on the x-axis, and x = −10 puts it on the left (negative) side.
The y-coordinate of every point on the x-axis is
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(a) 0. Points on the x-axis have not moved up or down at all.
The x-coordinate of points on the x-axis can be
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(d) any number. A point can be anywhere along the x-axis, left or right, so its x-coordinate can be any number. Only its y-coordinate is fixed at 0.
A point whose coordinates are both negative lies in
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(c) Quadrant III, the (−, −) quadrant: left and down.
The points (1, −1), (2, −2), (4, −5) and (−3, −4)
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(d). The first three are (+, −), so Quadrant IV. But (−3, −4) is (−, −), so Quadrant III.
The points (−5, 2) and (2, −5) lie in
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(c). (−5, 2) is (−, +), Quadrant II. (2, −5) is (+, −), Quadrant IV. Swapping the numbers moved the point to a different quadrant.
The perpendicular distance of a point P from the x-axis is 5 units. Then P has
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(d). Distance from the x-axis is the size of the y-coordinate. P could be 5 above (y = 5) or 5 below (y = −5).
O (0, 0), A (3, 0), B (3, 4) and C (0, 4) are plotted and joined in order. The figure is a
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(b) rectangle. The sides are 3, 4, 3, 4 and all corners are right angles. It is not a square because 3 ≠ 4.
The x-coordinate of a point is positive in
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(b). x is positive to the right of the y-axis, which covers Quadrant I (top right) and Quadrant IV (bottom right).
Points whose two coordinates have different signs lie in
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(d). Different signs means (−, +) or (+, −), which are Quadrants II and IV.
The point on the y-axis at a distance of 5 units from O in the negative direction is
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(c) (0, −5). On the y-axis x = 0; the negative direction is downward.
The perpendicular distance of the point P (3, 4) from the y-axis is
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(a) 3. The distance from the y-axis is the x-coordinate. (Distractor (c) 5 is the distance from the origin.)
The distance of the point (−6, 8) from the origin is
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(b) 10. √(6² + 8²) = √(36 + 64) = √100 = 10.
The reflection of (4, −7) in the x-axis is
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(b) (4, 7). Reflecting in the x-axis flips the sign of y and keeps x.
The midpoint of the segment joining (2, 6) and (8, −2) is
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(a) (5, 2). Average the x's: (2 + 8) ÷ 2 = 5. Average the y's: (6 − 2) ÷ 2 = 2.
Write the coordinates of the point on the x-axis that is 3 units to the left of the origin.
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(−3, 0). Left means negative x, and on the x-axis y = 0.
True or false: (3, 4) and (4, 3) are the same point.
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False. (x, y) = (y, x) only when x = y. Here 3 ≠ 4, so they are different points.
Without plotting, say where each point lies (which quadrant, or which axis): (i) (−2, −7) (ii) (5, 0) (iii) (−1, 8) (iv) (0, −3)
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- (i) (−, −): Quadrant III
- (ii) y = 0, x positive: positive x-axis
- (iii) (−, +): Quadrant II
- (iv) x = 0, y negative: negative y-axis
P (−1, 1), Q (3, −4), R (1, −1), S (−2, −3) and T (−4, 4) are plotted. Which of them lie in Quadrant IV?
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Quadrant IV is (+, −). Q (3, −4) and R (1, −1) fit. P and T are (−, +), Quadrant II. S is (−, −), Quadrant III.
Q and R
Which of the points P (5, 1), Q (8, 0), R (0, 4), S (0, 5) and O (0, 0) lie on the x-axis?
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A point is on the x-axis when its y-coordinate is 0. That is true for Q (8, 0) and O (0, 0). (O is on both axes.) R and S are on the y-axis. P is in Quadrant I.
Q and O
Find the distance between A (−3, 2) and B (5, 2).
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Both have y = 2, so AB is horizontal. Distance = |5 − (−3)| = |5 + 3| = 8 units.
Find the distance between the points (2, 3) and (4, 1).
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Across: 4 − 2 = 2. Down: 3 − 1 = 2. Distance = √(2² + 2²) = √8 = 2√2 units (about 2.83), since √8 = √4 × √2 = 2√2.
Find the point on the x-axis that is equidistant from A (2, −5) and B (−2, 9).
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- A point on the x-axis is (x, 0).
- Its distance² from A: (x − 2)² + (0 + 5)² = x² − 4x + 4 + 25. From B: (x + 2)² + (0 − 9)² = x² + 4x + 4 + 81.
- Set them equal: −4x + 29 = 4x + 85, so −8x = 56, so x = −7.
(−7, 0)
The point P (a, 3) is 5 units from the origin. Find the possible values of a.
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a² + 3² = 5², so a² + 9 = 25, so a² = 16, so a = 4 or a = −4. Both (4, 3) and (−4, 3) are 5 units from O.
Find the midpoint of the segment joining (−4, 7) and (6, −3).
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((−4 + 6) ÷ 2, (7 + (−3)) ÷ 2) = (2 ÷ 2, 4 ÷ 2) = (1, 2).
M (2, −1) is the midpoint of AB, where A is (5, 3). Find B.
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Let B = (x, y). (5 + x) ÷ 2 = 2 gives x = −1. (3 + y) ÷ 2 = −1 gives y = −5.
B = (−1, −5)
Quick check: from A to M is −3 across and −4 down; from M to B is again −3 and −4.
A triangle has vertices (1, 2), (4, 2) and (1, 6). Reflect it in the y-axis. Write the new vertices and the length of its longest side.
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Reflecting in the y-axis flips the sign of x: (−1, 2), (−4, 2), (−1, 6).
The sides are 3 (across), 4 (up) and √(3² + 4²) = 5. Reflection keeps lengths, so the longest side is still 5 units.
"The point (0, −2) lies in Quadrant IV." Is this true or false? Give a reason.
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False. Its x-coordinate is 0, so it lies on the y-axis (2 units below O). Points on an axis are not in any quadrant.
Find the perimeter of the rectangle with vertices (−2, −1), (4, −1), (4, 3) and (−2, 3).
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Length: 4 − (−2) = 6. Width: 3 − (−1) = 4. Perimeter = 2 × (6 + 4) = 20 units.
Three vertices of a rectangle are (3, 2), (−4, 2) and (−4, 5). Find the fourth vertex and the area of the rectangle.
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(3, 2) and (−4, 2) form a horizontal side. (−4, 2) and (−4, 5) form a vertical side. The missing corner is straight above (3, 2), level with (−4, 5): (3, 5).
Sides: 3 − (−4) = 7 and 5 − 2 = 3. Area = 7 × 3 = 21 square units.
Show that (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.
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Call them A, B, C, D in order.
- AB = √(3² + 5²) = √34
- BC = √(5² + 3²) = √34
- CD = √(3² + 5²) = √34
- DA = √(5² + 3²) = √34
- Diagonals: AC = √(2² + 8²) = √68 and BD = √(8² + 2²) = √68
All four sides are equal and the diagonals are equal, so ABCD is a square.
Are the points A (1, 5), B (2, 3) and C (−2, −11) on one straight line? Justify.
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Steps method: A to B is 1 across and 2 down (2 down per step). B to C is 4 back and 14 down (3.5 down per step). The steepness changes, so the path bends at B.
Distance check: AB = √5 ≈ 2.24, BC = √212 ≈ 14.56, AC = √265 ≈ 16.28. AB + BC ≈ 16.80, which is not equal to AC.
No, they are not collinear.
Find the values of y for which the distance between P (2, −3) and Q (10, y) is 10 units.
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- Across: 10 − 2 = 8. Up or down: y − (−3) = y + 3.
- 8² + (y + 3)² = 10², so 64 + (y + 3)² = 100, so (y + 3)² = 36.
- y + 3 = 6 or y + 3 = −6.
y = 3 or y = −9
Show that A (5, −2), B (6, 4) and C (7, −2) are the vertices of an isosceles triangle.
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AB = √(1² + 6²) = √37. BC = √(1² + 6²) = √37. AC = |7 − 5| = 2.
AB = BC, so two sides are equal: ABC is isosceles.
Is the triangle with vertices (0, 0), (3, 0) and (3, 4) right-angled? Find its perimeter and area.
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Sides: 3 (along the x-axis), 4 (straight up) and √(3² + 4²) = 5. Since 3² + 4² = 9 + 16 = 25 = 5², yes, it is right-angled at (3, 0).
Perimeter = 3 + 4 + 5 = 12 units. Area = ½ × 3 × 4 = 6 square units.
A (−2, 3), B (4, 3) and C (4, −5). Find AB, BC and AC, and check the Baudhāyana-Pythagoras theorem.
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AB = 4 − (−2) = 6 (horizontal). BC = 3 − (−5) = 8 (vertical). AC = √(6² + 8²) = √100 = 10.
Check: 6² + 8² = 36 + 64 = 100 = 10². The theorem holds, with the right angle at B.
Triangle ABC has A (2, 1), B (6, 1), C (4, 5). Reflect it in the x-axis. Write the image vertices, name the quadrant of the image, and show that AB keeps its length.
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Flip the sign of y: A′ (2, −1), B′ (6, −1), C′ (4, −5). All three are (+, −), so the image is in Quadrant IV.
AB = 6 − 2 = 4 and A′B′ = 6 − 2 = 4. The length is unchanged, as reflections always keep lengths.
A (1, 2), B (5, 2) and C (5, 6) are three corners of a square ABCD. Find D, the area, and the length of a diagonal.
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D is above A and level with C: D (1, 6). Side = 5 − 1 = 4, so area = 16 square units.
Diagonal AC = √(4² + 4²) = √32 = 4√2, about 5.66 units.
Find the point on the y-axis that is equidistant from A (6, 5) and B (−4, 3).
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- A point on the y-axis is (0, y).
- Distance² to A: 6² + (y − 5)² = 36 + y² − 10y + 25. To B: 4² + (y − 3)² = 16 + y² − 6y + 9.
- Equal: 61 − 10y = 25 − 6y, so 36 = 4y, so y = 9.
(0, 9)
A park is drawn on a grid where 1 unit = 10 m. The gate G is at (0, 0), a fountain F at (6, 8), a bench B at (6, 0) and a café C at (−3, 4).
(i) How far is the fountain from the gate, in metres? (ii) Which is closer to the gate: the café or the bench? (iii) How far is the fountain from the café? (iv) In which quadrant is the café?
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(i) GF = √(6² + 8²) = 10 units = 100 m.
(ii) GC = √(3² + 4²) = 5 units = 50 m. GB = 6 units = 60 m. The café is closer.
(iii) FC: across 9, down 4. √(81 + 16) = √97 ≈ 9.85 units, about 98.5 m.
(iv) (−3, 4) is (−, +): Quadrant II.
Show that A (−1, 2), B (3, 5) and C (6, 1) form a right-angled isosceles triangle, and find its area.
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- AB = √(4² + 3²) = √25 = 5
- BC = √(3² + 4²) = √25 = 5
- AC = √(7² + 1²) = √50
AB = BC, so it is isosceles. And AB² + BC² = 25 + 25 = 50 = AC², so it is right-angled at B.
Area = ½ × AB × BC = ½ × 5 × 5 = 12.5 square units.
Triangle ABC has A (0, 0), B (8, 0) and C (4, 6). Find the midpoints of its three sides, and show that each side of the triangle formed by the midpoints is exactly half of one side of ABC.
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Midpoints: of AB is P (4, 0); of BC is Q (6, 3); of CA is R (2, 3).
Sides of ABC: AB = 8, BC = √(4² + 6²) = √52, CA = √(4² + 6²) = √52.
Sides of PQR: PQ = √(2² + 3²) = √13, QR = 6 − 2 = 4, RP = √(2² + 3²) = √13.
Now √52 = √(4 × 13) = 2√13, so √13 is half of √52. And 4 is half of 8. Each side of PQR is half of a side of ABC. (QR is half of AB, PQ is half of CA, RP is half of BC.)
Show that A (1, −2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.
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Method 1, opposite sides: AB = √(2² + 8²) = √68, CD = √(2² + 8²) = √68. BC = √(2² + 4²) = √20, DA = √(2² + 4²) = √20. Both pairs of opposite sides are equal.
Method 2, diagonals: midpoint of AC = ((1 + 5) ÷ 2, (−2 + 10) ÷ 2) = (3, 4). Midpoint of BD = ((3 + 3) ÷ 2, (6 + 2) ÷ 2) = (3, 4). The diagonals cut each other in half.
Either way, ABCD is a parallelogram.
A circle has centre O (0, 0) and passes through P (5, 12). (i) Find its radius. (ii) Does (−13, 0) lie on the circle? (iii) Is (10, 10) inside, on or outside it? (iv) Name a point on the circle in Quadrant III.
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(i) OP = √(5² + 12²) = √169 = 13.
(ii) Distance of (−13, 0) from O is 13, so yes.
(iii) √(10² + 10²) = √200 ≈ 14.1, more than 13, so outside.
(iv) For example (−5, −12): √(25 + 144) = 13, and it is (−, −).
The vertices of PQRS are P (−3, −1), Q (2, −1), R (2, 3), S (−3, 3). (i) Name the shape. (ii) Find its area. (iii) Reflect P and R in the y-axis. (iv) Find the length of the diagonal PR.
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(i) PQ is horizontal (length 5), QR is vertical (length 4), and the opposite sides match: a rectangle.
(ii) Area = 5 × 4 = 20 square units.
(iii) P′ (3, −1) and R′ (−2, 3).
(iv) PR = √(5² + 4²) = √41, about 6.40 units.
Points P and Q split the segment from A (−3, 4) to B (9, −5) into three equal parts, with P nearer A. Find P and Q, and check your answer using midpoints.
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Whole journey: across 9 − (−3) = 12, down 4 − (−5) = 9. One third: across 4, down 3.
P = (−3 + 4, 4 − 3) = (1, 1) and Q = (1 + 4, 1 − 3) = (5, −2).
Check: P should be the midpoint of A and Q: ((−3 + 5) ÷ 2, (4 − 2) ÷ 2) = (1, 1). Correct. Q should be the midpoint of P and B: ((1 + 9) ÷ 2, (1 − 5) ÷ 2) = (5, −2). Correct.
A delivery drone starts from a depot at the origin. Each unit is 1 km. A house H is at (3, 4), a school S at (−5, 12) and a shop K at (8, −6).
(i) How far is the house from the depot? (ii) Which place is farthest from the depot? (iii) How far is it from the house to the shop? (iv) The drone can fly at most 12 km from the depot. Which places can it reach?
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(i) √(3² + 4²) = 5 km.
(ii) School: √(25 + 144) = 13 km. Shop: √(64 + 36) = 10 km. The school is farthest.
(iii) House to shop: across 5, down 10. √(25 + 100) = √125 ≈ 11.18 km.
(iv) House (5 km) and shop (10 km) are within 12 km. The school (13 km) is not.