These are the complete exercise 7.1 class 10 maths NCERT solutions for Coordinate Geometry. All ten questions from Exercise 7.1 are solved below, step by step, with the working explained — not just the final answers. Each question is reproduced exactly as it appears in the textbook, and every solution shows the distance formula being set up, substituted and simplified.
One formula drives the whole exercise: the distance formula \( PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \), which is Pythagoras’ theorem in coordinates. Once you see that, the questions sort into three groups — direct distances (Q1-Q2), figure checks built from distances (Q3-Q6) and equidistance equations (Q7-Q10).
Every question below is taken word for word from the official NCERT Class 10 Mathematics textbook, and the working here is original. Open the Chapter 7 PDF on the NCERT website to cross-check the exercise text, the worked examples and Fig. 7.8 against the printed source as you solve.
For the chapter’s theory in one place, the Class 10 Maths notes page collects every chapter’s concepts and formulas.
Exercise 7.1 Class 10 Maths NCERT Solutions: The Distance Formula in Action
Exercise 7.1 is the first practice set of Chapter 7. It takes the distance formula developed in Section 7.2 and applies it in three ways: finding plain distances (Q1-Q2), judging figures through distances (Q3-Q6) and solving equidistance problems (Q7-Q10). The table below maps each question to the skill it tests.
| Question | Skill it tests |
|---|---|
| Q1 | Direct application of the distance formula to two given points |
| Q2 | Distance from the origin, plus the town A-B word problem (answer in km) |
| Q3 | Collinearity test using only three distances |
| Q4 | Isosceles triangle check — do two sides match? |
| Q5 | Square verification from a plotted figure (sides and diagonals) |
| Q6 | Naming quadrilaterals: square, no quadrilateral, parallelogram |
| Q7 | Finding an equidistant point on the x-axis, using the form (x, 0) |
| Q8 | Distance formula used backwards to find an unknown coordinate |
| Q9 | Equidistance condition plus two further distance calculations |
| Q10 | Deriving a relation (a straight line) from an equidistance condition |
The first two questions are direct substitutions, Q3-Q6 are figure checks where the working itself is the reasoning, and Q7-Q10 test setting up an equidistance equation. A full-marks answer writes the formula first, then the substitution with brackets kept, then the simplified result — never just the final number.
Everything in this exercise rests on a few ideas. Master these before attempting the questions.
- Distance formula: for P(x₁, y₁) and Q(x₂, y₂), \( PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \) (NCERT, p. 102). It is Pythagoras’ theorem: the two differences are the legs of a right triangle, and PQ is the hypotenuse.
- Distance from the origin: \( OP = \sqrt{x^2 + y^2} \), which is the distance formula with (0, 0) as the first point.
- Collinearity: three points are on one straight line exactly when the largest of the three distances equals the sum of the other two (Example 3, NCERT, p. 104).
- Isosceles triangle: at least two sides equal — so compute all three side lengths.
- Square: all four sides equal AND both diagonals equal (Example 2, NCERT, p. 103). Four equal sides alone only prove a rhombus.
- Parallelogram: both pairs of opposite sides equal.
- Points on the axes: a point on the x-axis has the form (x, 0); a point on the y-axis has the form (0, y).
- Equidistance: AP = BP has exactly the same solutions as \( AP^2 = BP^2 \), because distances are non-negative. Squaring removes the square roots and makes the algebra manageable.
Question 1: Find the distance between the following pairs of points :
- (i) (2, 3), (4, 1)
- (ii) (-5, 7), (-1, 3)
- (iii) (a, b), (-a, -b)
Answer: The distance formula \( PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \) is Pythagoras’ theorem written in coordinates: the difference of the x-coordinates and the difference of the y-coordinates are the two legs of a right triangle, and the distance between the points is the hypotenuse. Apply it to each pair separately, keeping every bracket.
Part (i): Take (2, 3) as \((x_1, y_1)\) and (4, 1) as \((x_2, y_2)\).
Subtract inside the brackets, then square:
\[ d = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \]
Part (ii): For (-5, 7) and (-1, 3), subtracting a negative needs its own bracket: \(-1 – (-5) = 4\).
\[ d = \sqrt{(-1-(-5))^2 + (3-7)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2} \]
Part (iii): With variables the pattern is the same: (a, b) and (-a, -b) differ by \(2a\) in x and \(2b\) in y.
\[ d = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2} \]
Common error: Dropping the bracket around a negative coordinate, for example writing \(-1 – 5\) instead of \(-1 – (-5)\). That changes \(4^2\) into \((-6)^2\) and gives a wrong distance.
Since both differences are squared, you may subtract in either order: \((x_2-x_1)^2 = (x_1-x_2)^2\). Use this as a self-check — if reversing the order changes your answer, the slip is in the signs, not in the formula.
Question 2: 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.
Answer: When one point is the origin, the distance formula collapses to \( OP = \sqrt{x^2 + y^2} \), the distance of (x, y) from (0, 0) (NCERT, p. 102). In Section 7.2, town B is located 36 km east and 15 km north of town A, so A is placed at (0, 0) and B at (36, 15) — exactly the two points in this question.
The distance between the towns is therefore the distance you are about to compute, and the unit is kilometres.
\[ OP = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \]
So the distance between the points is 39 units, and the distance between towns A and B is 39 km.
Common error: Stopping at 1521 by forgetting the final square root. The formula ends with a root, so the answer must be a number whose square is 1521.
15-36-39 is a Pythagorean triple (3 × 5, 3 × 12, 3 × 13). Spotting that the two legs are 15 and 36 tells you the hypotenuse is 39 — a built-in check on the arithmetic.
Question 3: Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Answer: Three points are collinear exactly when the largest of the three distances equals the sum of the other two. This is how the chapter’s Example 3 verifies that three desks are seated in a line (NCERT, p. 104). So compute all three distances first, then compare.
Step 1: Name the points A(1, 5), B(2, 3) and C(-2, -11).
\[ AB = \sqrt{(2-1)^2 + (3-5)^2} = \sqrt{1+4} = \sqrt{5} \approx 2.24 \]
\[ BC = \sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{16+196} = \sqrt{212} = 2\sqrt{53} \approx 14.56 \]
\[ AC = \sqrt{(-2-1)^2 + (-11-5)^2} = \sqrt{9+256} = \sqrt{265} \approx 16.28 \]
Step 2: Compare.
The largest distance is AC \(\approx 16.28\), but \(AB + BC \approx 2.24 + 14.56 = 16.80\).
Since the largest distance is not equal to the sum of the other two, the points are not collinear — they form a triangle.
Common error: Checking only two distances. Collinearity is decided by comparing the largest distance with the sum of the other two, which needs all three.
Sketch the points roughly. B(2, 3) is close to A(1, 5), while C(-2, -11) is far below the x-axis — clearly off any line through A and B, confirming the distance result.
Question 4: Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Answer: A triangle is isosceles when at least two of its sides are equal. The reliable method is to compute all three side lengths and compare them. Let A(5, -2), B(6, 4) and C(7, -2).
\[ AB = \sqrt{(6-5)^2 + (4-(-2))^2} = \sqrt{1 + 36} = \sqrt{37} \]
\[ BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1 + 36} = \sqrt{37} \]
\[ AC = \sqrt{(7-5)^2 + (-2-(-2))^2} = \sqrt{4 + 0} = 2 \]
Since AB = BC = \(\sqrt{37}\), two sides are equal, so the triangle is isosceles. The equal sides meet at B(6, 4), and the base AC = 2 units lies on the horizontal line y = -2.
Common error: Assuming equal sides must look horizontal or vertical. Here the equal sides AB and BC are tilted; only the base AC is horizontal.
The base AC being horizontal is a quick visual check: its length is simply the difference of the x-coordinates, 7 – 5 = 2, which matches the distance found by the formula.
Question 5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Answer: A quadrilateral is a square only if two conditions hold together: all four sides equal, and both diagonals equal. The chapter’s Example 2 uses exactly this double test (NCERT, p. 103). Read the coordinates from the figure: A(3, 4), B(6, 7), C(9, 4) and D(6, 1).
Step 1: Check the four sides.
\[ AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{9+9} = \sqrt{18} \]
\[ BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{9+9} = \sqrt{18}, \quad CD = \sqrt{18}, \quad DA = \sqrt{18} \]
Step 2: Check the diagonals.
AC runs horizontally from x = 3 to x = 9, so AC = 6; BD runs vertically from y = 7 to y = 1, so BD = 6.
All four sides are equal \((\sqrt{18})\) and the diagonals are equal (6 = 6). Both conditions hold, so ABCD is a square and Champa is correct. Chameli’s doubt is about orientation, not shape: a square rotated by 45° still has four equal sides and equal diagonals — the test for a square is purely a test of lengths.
Common error: Calling the figure a rhombus because you stopped after the four equal sides. The diagonal check is what separates a rhombus from a square.
Check with Pythagoras: if the sides are all \(\sqrt{18}\), a square’s diagonal must be \(\sqrt{18 + 18} = \sqrt{36} = 6\). The diagonal you measured from the graph, 6, matches — that is the proof in one line.
Question 6: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
- (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)
Answer: To name a quadrilateral reliably, compute the four side lengths and, where needed, the two diagonals. The phrase “if any” is a warning: it is possible that no quadrilateral exists, as happens when three points are collinear.
Part (i): Let A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0).
Every side is the same diagonal move: \(AB = BC = CD = DA = \sqrt{8}\).
The diagonals are AC = 4 and BD = 4.
All sides equal and diagonals equal, so the quadrilateral is a square.
Part (ii): Let A(-3, 5), B(3, 1), C(0, 3), D(-1, -4).
Compute the three distances among A, B and C:
\[ AB = \sqrt{(3-(-3))^2 + (1-5)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13} \]
\[ BC = \sqrt{(0-3)^2 + (3-1)^2} = \sqrt{13}, \quad AC = \sqrt{(0-(-3))^2 + (3-5)^2} = \sqrt{13} \]
Here \(BC + AC = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = AB\), so A, B and C are collinear — C lies between A and B. Three collinear vertices cannot form a quadrilateral, so no quadrilateral is formed.
Part (iii): Let A(4, 5), B(7, 6), C(4, 3), D(1, 2).
Compute the sides:
\[ AB = \sqrt{(7-4)^2 + (6-5)^2} = \sqrt{10}, \quad BC = \sqrt{(4-7)^2 + (3-6)^2} = \sqrt{18} \]
\[ CD = \sqrt{(1-4)^2 + (2-3)^2} = \sqrt{10}, \quad DA = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{18} \]
Opposite sides are equal: AB = CD and BC = DA. A quadrilateral with both pairs of opposite sides equal is a parallelogram.
Common error: Forcing a name in part (ii). The words “if any” exist precisely because the answer can be “no quadrilateral”.
Always sketch the four points before computing. In part (ii) the sketch shows A, B and C lying on one straight line, so no quadrilateral — the distances then confirm what the sketch suggests.
Question 7: Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Answer: Every point on the x-axis has y-coordinate 0, so let the required point be P(x, 0). Equidistant means PA = PB; squaring both sides is safe because distances are non-negative, so PA² = PB². This is the same idea as Example 5 of the chapter, but on the x-axis instead of the y-axis (NCERT, p. 105).
\[ (x-2)^2 + (0-(-5))^2 = (x-(-2))^2 + (0-9)^2 \]
\[ (x-2)^2 + 25 = (x+2)^2 + 81 \]
Expand: \(x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81\). Cancel \(x^2 + 4\) from both sides: \(-4x + 25 = 4x + 81\), so \(-8x = 56\) and \(x = -7\).
The required point is (-7, 0). Check: \(PA = \sqrt{(-7-2)^2 + 5^2} = \sqrt{106}\) and \(PB = \sqrt{(-7+2)^2 + 9^2} = \sqrt{106}\).
Common error: Writing the point as (0, x). That is a point on the y-axis, not the x-axis — remember which coordinate is forced to be zero.
Verification is the whole trick here: substitute x = -7 into both distance expressions. Arriving at the same number, \(\sqrt{106}\), proves the point is truly equidistant.
Question 8: Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer: Here the distance is given and a coordinate is unknown, so start from the distance formula and square both sides to remove the root. This always produces a quadratic in the missing coordinate — and a quadratic usually gives two values.
\[ PQ = \sqrt{(10-2)^2 + (y-(-3))^2} = \sqrt{64 + (y+3)^2} = 10 \]
Square both sides: \(64 + (y+3)^2 = 100\), so \((y+3)^2 = 36\). Taking square roots gives two possibilities:
\[ y + 3 = 6 \quad \text{or} \quad y + 3 = -6 \]
\[ y = 3 \quad \text{or} \quad y = -9 \]
Both values work: (10, 3) and (10, -9) are each 10 units from P(2, -3). Geometrically, the vertical line x = 10 crosses the circle of radius 10 around P at two points, one above and one below.
Common error: Dropping y = -9. Squaring hides the sign, and the question’s plural word “values” is the clue that two answers are expected.
Substitute each value back: from P(2, -3) to (10, 3) the differences are 8 and 6, and \(\sqrt{64+36} = 10\); to (10, -9) the differences are 8 and -6, again giving 10. Both check out.
Question 9: 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.
Answer: Start with the one distance you can compute directly, QP. Because Q is equidistant from P and R, set QR equal to QP and solve for x. Then use the values of x to answer the two distance questions the problem also asks for.
\[ QP = \sqrt{(5-0)^2 + (-3-1)^2} = \sqrt{25+16} = \sqrt{41} \]
\[ QR = \sqrt{(x-0)^2 + (6-1)^2} = \sqrt{x^2 + 25} = \sqrt{41} \]
Squaring gives \(x^2 + 25 = 41\), so \(x^2 = 16\) and x = 4 or x = -4.
Distances: In both cases \(QR = \sqrt{41}\), since that is exactly what the equidistance equation stated.
For PR, use each value of x in turn:
\[ x = 4: \quad PR = \sqrt{(4-5)^2 + (6-(-3))^2} = \sqrt{1+81} = \sqrt{82} \]
\[ x = -4: \quad PR = \sqrt{(-4-5)^2 + (6-(-3))^2} = \sqrt{81+81} = \sqrt{162} = 9\sqrt{2} \]
Common error: Stopping after finding x. The question explicitly asks for QR and PR as well, and PR has two different values depending on which x you choose.
Write the answer as a complete sentence: “x = ±4; QR = \(\sqrt{41}\) in both cases; PR = \(\sqrt{82}\) when x = 4 and \(9\sqrt{2}\) when x = -4.” That structure can be read in one glance by an examiner.
Question 10: Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Answer: Let P(x, y) be equidistant from A(3, 6) and B(-3, 4). Equidistance means PA = PB; square both sides and substitute the distance formula for each side — the same method as Example 4 of the chapter (NCERT, p. 105).
\[ (x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2 \]
Expand every square carefully:
\[ x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16 \]
Cancelling \(x^2\) and \(y^2\) from both sides leaves \(-6x – 12y + 45 = 6x – 8y + 25\). Take all terms to one side: \(-12x – 4y = -20\). Dividing by -4 gives the required relation:
\[ 3x + y = 5 \quad (\text{or } 3x + y – 5 = 0) \]
Meaning of the answer: the graph of \(3x + y = 5\) is a straight line, and it is the perpendicular bisector of AB. Every point on it is equidistant from A and B — the class of points you were asked to describe.
Common error: Sign slips when expanding \((x+3)^2\) or \((y-4)^2\). One wrong sign breaks the cancellation of \(x^2\) and \(y^2\), and the relation comes out wrong.
Test the relation with an easy point on the line, say (1, 2): \(3(1) + 2 = 5\). Its distances to A(3, 6) and B(-3, 4) are both \(\sqrt{20}\), confirming the line is correct.
How to Use the Distance Formula in Four Steps (a Fresh Worked Example)
Every question in this exercise is the same four steps repeated. Memorise the flow, not the individual answers.
- Subtract the x-coordinates: \(x_2 – x_1\). The order does not matter, because the next step squares it.
- Subtract the y-coordinates: \(y_2 – y_1\).
- Square both differences.
- Add the squares and take the positive square root.
A fresh worked example: find the distance between (-4, 1) and (2, -7).
\[ \sqrt{(2-(-4))^2 + (-7-1)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = \sqrt{100} = 10 \]
The distance is 10 units. The answer is always non-negative, which is why only the positive square root is taken — the distance formula is defined to give the non-negative root.
Mnemonic for exam pressure: “Square the differences, add, then root.”
Which distance check goes with which figure?
Questions 3 to 6 all ask you to recognise a figure from distances. This table collects the four checks in one place.
| Figure / condition being tested | What to verify with distances | Where it appears |
|---|---|---|
| Collinearity of three points | Largest distance = sum of the other two | Q3, Q6(ii) |
| Isosceles triangle | At least two sides equal | Q4 |
| Square | All four sides equal AND both diagonals equal | Q5, Q6(i) |
| Parallelogram | Both pairs of opposite sides equal | Q6(iii) |
| Rhombus (how to tell it from a square) | Four equal sides; the extra diagonal check separates it from a square | Prevents the Q5 error |
All of these checks rest on the same right-triangle idea behind the distance formula — the Pythagoras theorem you studied in Chapter 6 Triangles. In Chapter 8 Introduction to Trigonometry, you will use those same right triangles to define trigonometric ratios.
Common Mistakes in Exercise 7.1 and How to Fix Them
These are the errors students actually make on this exercise, each with the correct rule and a way to catch it while checking.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Forgetting the square root (writing 1521 or 100 instead of 39 or 10) | The formula ends with \(\sqrt{\ldots}\): take the positive square root of the sum. | Ask: does my answer make sense as a distance compared with the coordinates? |
| Dropping brackets with negatives, e.g. \(-1 – 5\) instead of \(-1 – (-5)\) | Keep each difference inside its own bracket; subtracting a negative means adding. | Rewrite the substitution and check each bracket sign by sign. |
| Testing collinearity with the wrong pair of distances | Compare the LARGEST distance with the sum of the other two. | Make a rough sketch; three collinear points clearly sit on one line. |
| Calling a rhombus a square in Q5 and Q6(i) | A square needs four equal sides AND equal diagonals. | Always compute both diagonals before naming a square. |
| Dropping the negative solution in Q8 and Q9 | Squaring creates two solutions; the word “values” (plural) signals both. | Substitute each value back into the distance formula. |
| Writing (0, x) for a point on the x-axis | x-axis points are (x, 0); y-axis points are (0, y). | Name the axis first — whichever coordinate must be zero goes in that position. |
If a concept still feels shaky, the Class 10 notes hub and the full CBSE notes index group every chapter’s theory in one place.
Frequently Asked Questions on Exercise 7.1
Why do we ignore the negative square root in the distance formula?
A distance is a length, and a length is never negative. The formula is defined to return the non-negative root: \(\sqrt{9}\) means 3, not -3. The signs were already used up inside the squared differences, so the final root must be taken positive.
How do I check whether three points are collinear using only distances?
Compute all three distances AB, BC and AC. If the largest of the three equals the sum of the other two, the points lie on one straight line. If not, they form a triangle. Q3 and the chapter’s Example 3 (NCERT, p. 104) both work this way.
What does it mean for a point to be equidistant from two points, and how do I set up the equation?
Equidistant means the point is at the same distance from both, so AP = BP. Because distances are non-negative, AP = BP is exactly the same as \(AP^2 = BP^2\), which removes the square roots and leaves a solvable equation. Questions 7, 9 and 10 all use this setup.
Why does Question 8 have two answers for y?
Because squaring creates two possibilities: \((y+3)^2 = 36\) gives \(y + 3 = 6\) or \(y + 3 = -6\). Geometrically, the vertical line x = 10 has two points at distance 10 from P(2, -3) — one above P and one below it.
Does the order of the two points matter when I apply the distance formula?
No. Both differences are squared, so \((x_2-x_1)^2 = (x_1-x_2)^2\), and the textbook itself gives both versions of the formula. Just keep the two coordinates of a point together inside their own bracket.
Which formula do I need next for Exercise 7.2?
Exercise 7.2 is built on the section formula of Section 7.3 (NCERT, p. 107): a point P dividing AB internally in the ratio m₁ : m₂ has coordinates \( \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \). The worked solutions for all its questions are on the Exercise 7.2 page.
Reference: NCERT Class 10 Mathematics textbook, chapter Coordinate Geometry.
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