CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 7 Coordinate Geometry
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Boards begin February 2027
Pre-boards are in December
Coordinate Geometry is in Coordinate Geometry, which carries 6 of the 80 marks in the 2026-27 CBSE Maths paper
35 NCERT Exemplar questions from Chapter 7
Exercise 7.1
Exemplar Ex 7.1 Q11 mark
The distance of the point P(2, 3) from the x-axis is
Step 1The distance of any point (x, y) from the x-axis equals |y|.
Step-by-step solution
The distance of any point (x, y) from the x-axis equals |y|.
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
Step 1[1 mark] The side lengths are 4 (from (0,4) to (0,0)), 3 (from (0,0) to (3,0)) and 5 (from (0,4) to (3,0), since √(3²+4²)=5); perimeter = 4 + 3 + 5 = 12 — option (B).
Step-by-step solution
[1 mark] The side lengths are 4 (from (0,4) to (0,0)), 3 (from (0,0) to (3,0)) and 5 (from (0,4) to (3,0), since √(3²+4²)=5); perimeter = 4 + 3 + 5 = 12 — option (B).
The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at
Step 1[1 mark] A point (0, y) on the y-axis equidistant from A and B satisfies 1²+(y−5)² = 4²+(y−6)²; expanding, 1+y²−10y+25 = 16+y²−12y+36, so 2y = 26 and y = 13, giving (0, 13) — option (A).
Step-by-step solution
[1 mark] A point (0, y) on the y-axis equidistant from A and B satisfies 1²+(y−5)² = 4²+(y−6)²; expanding, 1+y²−10y+25 = 16+y²−12y+36, so 2y = 26 and y = 13, giving (0, 13) — option (A).
A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, −5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
Step 1[1 mark] Let P = (0, b) on the y-axis and Q = (a, 0) on the x-axis; the mid-point is (a/2, b/2) = (2, −5), so a = 4 and b = −10, giving P(0, −10) and Q(4, 0) — option (D).
Step-by-step solution
[1 mark] Let P = (0, b) on the y-axis and Q = (a, 0) on the x-axis; the mid-point is (a/2, b/2) = (2, −5), so a = 4 and b = −10, giving P(0, −10) and Q(4, 0) — option (D).
State whether the following statement is true or false, justifying your answer: A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.
Step 1Radius = distance from centre (origin) to P(5, 0) = 5.
Step-by-step solution
Radius = distance from centre (origin) to P(5, 0) = 5.
State whether the following statement is true or false, justifying your answer: Point P(5, −3) is one of the two points of trisection of the line segment joining the points A(7, −2) and B(1, −5).
Step 1The trisection point closer to A divides AB in 1:2.
Step-by-step solution
The trisection point closer to A divides AB in 1:2.
State whether the following statement is true or false and justify: ΔABC with vertices A (−2, 0), B (2, 0) and C (0, 2) is similar to ΔDEF with vertices D (−4, 0), E (4, 0) and F (0, 4).
Step 1[1 mark] The sides of ΔABC are AB = 4, BC = 2√2, CA = 2√2; the sides of ΔDEF are DE = 8, EF = 4√2, FD = 4√2.
Step-by-step solution
[1 mark] The sides of ΔABC are AB = 4, BC = 2√2, CA = 2√2; the sides of ΔDEF are DE = 8, EF = 4√2, FD = 4√2.
[1 mark] Each side of ΔDEF is exactly twice the corresponding side of ΔABC, so the triangles are similar (SSS) — the statement is True.
Answer:True — all three pairs of sides are in the ratio 1 : 2.
State whether the following statement is true or false and justify: Point P (−4, 2) lies on the line segment joining the points A (−4, 6) and B (−4, −6).
Step 1[1 mark] A, B and P all have x-coordinate −4, so they are collinear on the vertical line x = −4.
Step-by-step solution
[1 mark] A, B and P all have x-coordinate −4, so they are collinear on the vertical line x = −4.
[1 mark] Since 2 lies between −6 and 6, P is between A and B — the statement is True.
State whether the following statement is true or false and justify: Point P (0, 2) is the point of intersection of the y-axis and the perpendicular bisector of the line segment joining the points A (−1, 1) and B (3, 3).
Step 1[1 mark] A point on the perpendicular bisector is equidistant from A and B; PA = √((−1)²+(1−2)²) = √2 and PB = √(3²+(3−2)²) = √10.
Step-by-step solution
[1 mark] A point on the perpendicular bisector is equidistant from A and B; PA = √((−1)²+(1−2)²) = √2 and PB = √(3²+(3−2)²) = √10.
[1 mark] Since PA ≠ PB, P is not on the perpendicular bisector — the statement is False.
State whether the following statement is true or false and justify: Points A (4, 3), B (6, 4), C (5, −6) and D (−3, 5) are the vertices of a parallelogram.
Step 1[1 mark] A quadrilateral is a parallelogram iff its diagonals bisect each other, i.e. mid-point of AC = mid-point of BD.
Step-by-step solution
[1 mark] A quadrilateral is a parallelogram iff its diagonals bisect each other, i.e. mid-point of AC = mid-point of BD.
[1 mark] Mid-point of AC = (4.5, −1.5) but mid-point of BD = (1.5, 4.5); they differ, so the statement is False.
Answer:False — the diagonals AC and BD do not bisect each other.
State whether the following statement is true or false and justify: The point A (2, 7) lies on the perpendicular bisector of the line segment joining the points P (6, 5) and Q (0, −4).
Step 1[1 mark] A lies on the perpendicular bisector iff AP = AQ; AP = √((6−2)²+(5−7)²) = √20 and AQ = √((0−2)²+(−4−7)²) = √125.
Step-by-step solution
[1 mark] A lies on the perpendicular bisector iff AP = AQ; AP = √((6−2)²+(5−7)²) = √20 and AQ = √((0−2)²+(−4−7)²) = √125.
[1 mark] Since AP ≠ AQ, A is not on the perpendicular bisector — the statement is False.
State whether the following statement is true or false and justify: The points A (−1, −2), B (4, 3), C (2, 5) and D (−3, 0) in that order form a rectangle.
Step 1[1 mark] Mid-point of AC = (0.5, 1.5) = mid-point of BD, so the diagonals bisect each other (it is a parallelogram).
Step-by-step solution
[1 mark] Mid-point of AC = (0.5, 1.5) = mid-point of BD, so the diagonals bisect each other (it is a parallelogram).
[1 mark] Also AC = √(3²+7²) = √58 and BD = √(7²+3²) = √58 are equal diagonals, so it is a rectangle — the statement is True.
Answer:True — diagonals bisect each other and are equal (√58).
Find a point which is equidistant from the points A (−5, 4) and B (−1, 6). How many such points are there?
Step 1[1 mark] A point equidistant from A and B lies on the perpendicular bisector of AB; the mid-point of AB is ((−5−1)/2, (4+6)/2) = (−3, 5), which is one such point.
Step-by-step solution
[1 mark] A point equidistant from A and B lies on the perpendicular bisector of AB; the mid-point of AB is ((−5−1)/2, (4+6)/2) = (−3, 5), which is one such point.
[1 mark] Setting PA² = PB² for P(x,y): (x+5)²+(y−4)² = (x+1)²+(y−6)² simplifies to 8x + 4y + 4 = 0, i.e. 2x + y + 1 = 0.
[1 mark] Every point on the line 2x + y + 1 = 0 is equidistant from A and B, so there are infinitely many such points (e.g. (−3, 5)).
Answer:(−3, 5) is one such point; infinitely many lie on 2x + y + 1 = 0.
Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A (−5, −2) and B (4, −2). Name the type of triangle formed by the points Q, A and B.
Step 1[1 mark] A and B have equal y-coordinates, so the perpendicular bisector of AB is the vertical line x = (−5+4)/2 = −1/2.
Step-by-step solution
[1 mark] A and B have equal y-coordinates, so the perpendicular bisector of AB is the vertical line x = (−5+4)/2 = −1/2.
[1 mark] On the x-axis (y = 0), this gives Q = (−1/2, 0).
[1 mark] Since Q is on the perpendicular bisector, QA = QB, so ΔQAB is an isosceles triangle.
Find the ratio in which the line 2x + 3y − 5 = 0 divides the line segment joining the points (8, −9) and (2, 1). Also find the coordinates of the point of division.
Step 1[1 mark] Let the line divide the segment in ratio k : 1; the point of division is ((2k+8)/(k+1), (k−9)/(k+1)).
Step-by-step solution
[1 mark] Let the line divide the segment in ratio k : 1; the point of division is ((2k+8)/(k+1), (k−9)/(k+1)).
If (−4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
Step 1Side AB = 8 (both endpoints have y = 3).
Step-by-step solution
Side AB = 8 (both endpoints have y = 3).
Let C(x, y). For equilateral: AC = BC = 8.
AC² = BC² gives (x+4)² = (x−4)² → 16x = 0 → x = 0 (C lies on perpendicular bisector of AB).
Substitute x = 0 in AC² = 64: 16 + (y−3)² = 64 → (y−3)² = 48 → y = 3 ± 4√3.
AB lies on the line y = 3. For origin to be interior, C must be BELOW AB → y < 3 → y = 3 − 4√3.
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Case-based4 marks
Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter's school and then reaches the office. House is at H(2, 4), bank at K(5, 8), school at S(13, 14) and office at O(13, 26). All coordinates are in km. (i) Find HK and KS and SO. (ii) Find the direct distance HO from house to office. (iii) What is the extra distance Ayush travels compared to going directly?
Step 1HK = √[(5−2)² + (8−4)²] = √(9 + 16) = 5 km.
Step-by-step solution
HK = √[(5−2)² + (8−4)²] = √(9 + 16) = 5 km.
KS = √[(13−5)² + (14−8)²] = √(64 + 36) = 10 km.
SO = √[0 + (26−14)²] = 12 km. Total path = 5 + 10 + 12 = 27 km.
Direct HO = √[11² + 22²] = √(121 + 484) = √605 = 11√5 km ≈ 24.6 km.
Extra distance = 27 − 11√5 km ≈ 2.4 km.
Answer:(i) HK=5, KS=10, SO=12 km. (ii) HO = 11√5 km. (iii) Extra = (27 − 11√5) km ≈ 2.4 km.
Assertion (A): The midpoint of the line segment joining A(−2, 8) and B(−6, −4) is (−4, 2).
Reason (R): The midpoint of a segment joining (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2).
Step 1Reason R states the standard midpoint formula — true.
Step-by-step solution
Reason R states the standard midpoint formula — true.
Apply R to A(−2, 8) and B(−6, −4): ((−2 + (−6))/2, (8 + (−4))/2) = (−8/2, 4/2) = (−4, 2). So A is true.
R is exactly the formula used to derive A → R is the correct explanation of A.
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Questions from the NCERT Class 10 Mathematics Exemplar Problems (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.