Exercise 7.1 — Coordinate Geometry

1. 13 marksNCERT Cl-10 Maths, Ex 7.1, Q1

   Find the distance between the following pairs of points: (i) (2, 3) and (4, 1); (ii) (−5, 7) and (−1, 3); (iii) (a, b) and (−a, −b).

   Hint. Straight substitution into √((x₂−x₁)² + (y₂−y₁)²). Keep the subtraction order the same in both brackets.

   (i) (2, 3) and (4, 1). x₂ − x₁ = 4 − 2 = 2 and y₂ − y₁ = 1 − 3 = −2 d = √(2² + (−2)²) = √(4 + 4) = √8

   ✦ d = 2√2 units (≈ 2.83)

   (ii) (−5, 7) and (−1, 3). x₂ − x₁ = −1 − (−5) = 4 and y₂ − y₁ = 3 − 7 = −4 d = √(4² + (−4)²) = √32

   ✦ d = 4√2 units (≈ 5.66)

   (iii) (a, b) and (−a, −b). x₂ − x₁ = −a − a = −2a and y₂ − y₁ = −b − b = −2b d = √(4a² + 4b²) = √(4(a² + b²))

   ✦ d = 2√(a² + b²) units

   That last one has a neat reading: (−a, −b) is the reflection of (a, b) through the origin, so the two points sit at equal distances on opposite sides of O. The distance between them is therefore twice the distance of either from the origin, which is exactly 2√(a² + b²).

   Where students slip. Writing √(x₂ − x₁)² + (y₂ − y₁)² with the root covering only the first bracket. The square root must sit over the whole sum.

   Another way. The signs never matter, because both differences get squared. If you find negatives error-prone, take |x₂ − x₁| and |y₂ − y₁| and the answer is unchanged.

2. 22 marksNCERT Cl-10 Maths, Ex 7.1, Q2

   Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2 (town B lies 36 km east and 15 km north of town A)?

   Hint. The second part is the first part with units attached — that is the whole point of the question.

   Step 1 — Apply the distance formula, or its origin form OP = √(x² + y²). d = √(36² + 15²) = √(1296 + 225) = √1521

   Step 2 — Recognise the square root. 1521 = 39², since 39 × 39 = 1521. d = 39

   Step 3 — Now read the towns problem. Putting town A at the origin, "36 km east and 15 km north" places town B at exactly (36, 15) with 1 unit = 1 km. So the calculation just done is the answer.

   ✦ Answer: the distance between the points is 39 units, and the distance between towns A and B is 39 km.

   Note that 39 is much less than 36 + 15 = 51 — you save 12 km by going straight rather than east-then-north.

   Where students slip. Leaving the answer as √1521 because the number looks unfriendly. Try the perfect squares near it: 40² = 1600 is a little too big, 39² = 1521 is exact.

   Another way. Spot the Pythagorean triple. Dividing by 3 gives 12 and 5, and (5, 12, 13) is a standard triple, so the hypotenuse is 3 × 13 = 39 with no arithmetic at all.

3. 33 marksNCERT Cl-10 Maths, Ex 7.1, Q3

   Determine whether the points (1, 5), (2, 3) and (−2, −11) are collinear.

   Hint. Find all three distances. Collinear means the two smaller ones add up exactly to the largest.

   Call the points A(1, 5), B(2, 3) and C(−2, −11).

   Step 1 — Compute the three distances. AB = √((2−1)² + (3−5)²) = √(1 + 4) = √5 ≈ 2.24 BC = √((−2−2)² + (−11−3)²) = √(16 + 196) = √212 ≈ 14.56 AC = √((−2−1)² + (−11−5)²) = √(9 + 256) = √265 ≈ 16.28

   Step 2 — Test the largest against the sum of the other two. AB + BC ≈ 2.24 + 14.56 = 16.80 AC ≈ 16.28

   Step 3 — Compare. 16.80 ≠ 16.28, and the difference is far too big to be rounding. So no point lies on the segment joining the other two.

   ✦ Answer: the points are not collinear.

   Because AB + BC is greater than AC, the three points form a genuine (very thin) triangle.

   Where students slip. Deciding "not collinear" from a rough sketch. The three points really do lie close to a line — you need the arithmetic, and you need enough decimal places to tell 16.80 from 16.28.

   Another way. Compare slopes: slope AB = (3−5)/(2−1) = −2, slope BC = (−11−3)/(−2−2) = 3.5. Different slopes, so not collinear. This is quicker, though the distance method is the one the textbook models.

4. 43 marksNCERT Cl-10 Maths, Ex 7.1, Q4

   Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.

   Hint. Isosceles needs exactly two equal sides. Compute all three and compare.

   Let A(5, −2), B(6, 4) and C(7, −2).

   Step 1 — Find the three side lengths. AB = √((6−5)² + (4+2)²) = √(1 + 36) = √37 BC = √((7−6)² + (−2−4)²) = √(1 + 36) = √37 AC = √((7−5)² + (−2+2)²) = √(4 + 0) = 2

   Step 2 — Compare. AB = BC = √37, and AC = 2 is different.

   Step 3 — Confirm it is a genuine triangle. √37 ≈ 6.08, and 6.08 + 2 = 8.08 > 6.08, so the triangle inequality holds and the three points are not collinear.

   ✦ Answer: yes — two sides are equal, so the points form an isosceles triangle.

   A and C share the same y-coordinate, which is why AC came out as a plain horizontal length of 2.

   Where students slip. Concluding "equilateral" after finding AB = BC and not bothering with AC. Compute all three sides every time; the third is often the one that decides the name.

   Another way. Notice the symmetry first: A and C are level at y = −2, and B sits at x = 6, exactly halfway between x = 5 and x = 7. B is therefore on the perpendicular bisector of AC, which forces BA = BC without any calculation.

5. 54 marksNCERT Cl-10 Maths, Ex 7.1, Q5 (Fig. 7.8)

   Four friends sit in a classroom at A(3, 4), B(6, 7), C(9, 4) and D(6, 1), read off the grid of rows and columns in the figure. Champa says ABCD is a square; Chameli disagrees. Using the distance formula, decide who is right.

   Hint. Four sides is not enough — you need both diagonals too, since equal sides alone would only make it a rhombus.

   Step 1 — Find the four sides, taking the vertices in order. AB = √((6−3)² + (7−4)²) = √(9 + 9) = 3√2 BC = √((9−6)² + (4−7)²) = √(9 + 9) = 3√2 CD = √((6−9)² + (1−4)²) = √(9 + 9) = 3√2 DA = √((3−6)² + (4−1)²) = √(9 + 9) = 3√2

   All four sides are equal. That is not yet enough, because a rhombus also has four equal sides — so the answer is still open at this stage.

   Step 2 — Find the two diagonals. AC = √((9−3)² + (4−4)²) = √36 = 6 BD = √((6−6)² + (1−7)²) = √36 = 6

   The diagonals are equal too, which is exactly what separates a square from a rhombus.

   Step 3 — Apply the test. A quadrilateral with all four sides equal and both diagonals equal is a square.

   ✦ Answer: Champa is correct — ABCD is a square.

   The figure makes it look tilted, which is what prompts the disagreement, but a square rotated 45° is still a square.

   Where students slip. Reading the coordinates off the grid the wrong way round — the horizontal axis is labelled *Columns* and the vertical one *Rows*, so A is at column 3, row 4, giving (3, 4) and not (4, 3).

   Another way. Equal sides plus one right angle also settles it. Here DA² + AB² = 18 + 18 = 36 = BD², so by the converse of Pythagoras ∠A = 90°, and a rhombus with a right angle is a square.

6. 65 marksNCERT Cl-10 Maths, Ex 7.1, Q6

   Name the type of quadrilateral formed, if any, by the following points, giving reasons: (i) (−1, −2), (1, 0), (−1, 2), (−3, 0); (ii) (−3, 5), (3, 1), (0, 3), (−1, −4); (iii) (4, 5), (7, 6), (4, 3), (1, 2).

   Hint. Sides and diagonals for each. The phrase "if any" in the question is a warning that one of the three sets does not form a quadrilateral at all.

   (i) A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0). AB = √(4 + 4) = 2√2 BC = √(4 + 4) = 2√2 CD = √(4 + 4) = 2√2 DA = √(4 + 4) = 2√2 AC = √(0 + 16) = 4 BD = √(16 + 0) = 4

   All four sides equal and both diagonals equal.

   ✦ A square

   (ii) A(−3, 5), B(3, 1), C(0, 3), D(−1, −4).

   Before measuring sides, check the points are in general position. Take A, C and B: slope of AC = (3 − 5)/(0 + 3) = −2/3 slope of CB = (1 − 3)/(3 − 0) = −2/3

   The slopes are equal and the two segments share the point C, so A, C and B lie on one straight line. Three of the four vertices are collinear, so joining them produces no enclosed four-sided figure.

   ✦ No quadrilateral is formed

   (iii) A(4, 5), B(7, 6), C(4, 3), D(1, 2). AB = √(9 + 1) = √10 BC = √(9 + 9) = 3√2 CD = √(9 + 1) = √10 DA = √(9 + 9) = 3√2 AC = √(0 + 4) = 2 BD = √(36 + 16) = √52 = 2√13

   Opposite sides are equal in pairs (AB = CD and BC = DA), but the diagonals are very different — 2 against about 7.21.

   ✦ A parallelogram

   Where students slip. Grinding through the side lengths in part (ii) and announcing some shape from them. The four lengths there are all different, which should prompt the collinearity check rather than a guess — and the "if any" in the question is the hint that this case exists.

   Another way. For (iii) you can skip the sides: the midpoint of AC is (4, 4) and the midpoint of BD is (4, 4) as well. Diagonals that bisect each other is precisely the definition of a parallelogram, and the unequal diagonal lengths rule out rectangle.

7. 73 marksNCERT Cl-10 Maths, Ex 7.1, Q7

   Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).

   Hint. A point on the x-axis has the form (x, 0) — that is one unknown, not two. Set the squared distances equal so the roots disappear.

   Step 1 — Use the form of the point. Any point on the x-axis is (x, 0). Call it P.

   Step 2 — Write both squared distances. Squaring is safe here because distances are non-negative, so equal distances and equal squared distances say the same thing.

   PA² = (x − 2)² + (0 + 5)² = (x − 2)² + 25 PB² = (x + 2)² + (0 − 9)² = (x + 2)² + 81

   Step 3 — Set them equal and expand. x² − 4x + 4 + 25 = x² + 4x + 4 + 81 −4x + 29 = 4x + 85

   The x² terms cancel, which is why the equation turns out linear.

   Step 4 — Solve. −8x = 56, so x = −7

   Check. Distance to (2, −5): √(81 + 25) = √106. Distance to (−2, 9): √(25 + 81) = √106 ✓

   ✦ Answer: the point is (−7, 0).

   Where students slip. Keeping the square roots and squaring only one side later. Square both sides at the very start — every root vanishes and the algebra stays linear.

   Another way. The set of points equidistant from A and B is the perpendicular bisector of AB. You could find its equation and intersect it with y = 0; the squared-distance route gets there in fewer lines.

8. 83 marksNCERT Cl-10 Maths, Ex 7.1, Q8

   Find the values of y for which the distance between P(2, −3) and Q(10, y) is 10 units.

   Hint. Square the distance condition. You will get a squared bracket equal to a number, which has two solutions.

   Step 1 — Write the squared distance and set it to 10². (10 − 2)² + (y + 3)² = 100

   Step 2 — Simplify. 64 + (y + 3)² = 100 (y + 3)² = 36

   Step 3 — Take both square roots. y + 3 = 6 or y + 3 = −6

   Both are legitimate, because Q may sit above or below P and the squaring loses that distinction. Geometrically there are two points on the vertical line x = 10 that are 10 units from P.

   y = 3 or y = −9

   Check. For y = 3: √(64 + 36) = 10 ✓. For y = −9: √(64 + 36) = 10 ✓

   ✦ Answer: y = 3 or y = −9.

   Where students slip. Reporting only y = 3. The question says "the values of y", plural — a squared bracket always offers two roots, and here nothing in the problem rules either out.

   Another way. Read it geometrically: P and Q are 8 apart horizontally and the straight-line distance is 10, so the vertical gap must satisfy 8-?-10, giving 6 by the (6, 8, 10) triple. Q is then 6 above or 6 below y = −3.

9. 94 marksNCERT Cl-10 Maths, Ex 7.1, Q9

   If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find the values of x. Also find the distances QR and PR.

   Hint. Set QP² = QR² to get x. Two values of x will appear, and the last part has to be answered for each.

   Step 1 — Compute QP², which involves no unknown. QP² = (5 − 0)² + (−3 − 1)² = 25 + 16 = 41

   Step 2 — Write QR² and set it equal. QR² = (x − 0)² + (6 − 1)² = x² + 25 x² + 25 = 41, so x² = 16

   Step 3 — Take both roots. x = 4 or x = −4

   Step 4 — Find QR. It is the same for both, since QR² was 41 either way. QR = √41 ≈ 6.40

   Step 5 — Find PR for each value separately, because R moves. For x = 4: PR = √((4 − 5)² + (6 + 3)²) = √(1 + 81) = √82 ≈ 9.06 For x = −4: PR = √((−4 − 5)² + (6 + 3)²) = √(81 + 81) = √162 = 9√2 ≈ 12.73

   ✦ Answer: x = 4 or x = −4; QR = √41 in both cases; PR = √82 when x = 4, and PR = 9√2 when x = −4.

   Where students slip. Giving a single value of PR. QR happens to be independent of the sign of x, but PR is not — the two answers are genuinely different and both must be stated.

   Another way. Since QR² = x² + 25 depends on x only through x², any solution comes in a ± pair automatically. Spotting that early tells you to expect two values of x before you solve anything.

10. 103 marksNCERT Cl-10 Maths, Ex 7.1, Q10

    Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (−3, 4).

    Hint. Set the two squared distances equal. Both x² and y² will cancel, leaving a linear relation.

    Step 1 — Let P(x, y) be equidistant from A(3, 6) and B(−3, 4), and square both distances. PA² = (x − 3)² + (y − 6)² PB² = (x + 3)² + (y − 4)²

    Step 2 — Set them equal and expand. x² − 6x + 9 + y² − 12y + 36 = x² + 6x + 9 + y² − 8y + 16

    Step 3 — Cancel x² and y². They appear identically on both sides, so they subtract away and what is left is linear — which is why the answer turns out to be a straight line. −6x − 12y + 45 = 6x − 8y + 25

    Step 4 — Collect terms. −6x − 6x − 12y + 8y + 45 − 25 = 0 −12x − 4y + 20 = 0

    Step 5 — Divide through by −4. 3x + y − 5 = 0

    Check with a convenient point on this line, say (1, 2): distance to (3, 6) is √(4 + 16) = √20; distance to (−3, 4) is √(16 + 4) = √20 ✓

    ✦ Answer: 3x + y = 5

    This relation is a straight line — the perpendicular bisector of AB, which is exactly the set of points equidistant from A and B.

    Where students slip. Expanding (y − 6)² as y² − 36 or (x + 3)² as x² + 9. Every one of these brackets has a middle term, and dropping it destroys the whole result.

    Another way. The answer must be the perpendicular bisector of AB, so you can build it directly: the midpoint of AB is (0, 5), the slope of AB is (4 − 6)/(−3 − 3) = 1/3, so the bisector has slope −3 and equation y − 5 = −3x, i.e. 3x + y = 5.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (chapter 7 now runs to two exercises — 7.1 and 7.2; the area of a triangle from coordinates, with old exercises 7.3 and 7.4, is no longer part of this chapter, and the summary lists only four results). Questions are referenced from the NCERT textbook for identification.