Notes Formulae Read NCERT Chapter 7 (jemh107) Practice Check my answer NCERT Exemplar Solutions 22 NCERT textbook questions from Chapter 7
Exercise 7.1 NCERT Ex 7.1 Q1(i) 2 marks Find the distance between the pair of points (2, 3) and (4, 1).
Stuck? Show a hint Step 1 Apply distance formula d = √[(x₂−x₁)² + (y₂−y₁)²].
NCERT Ex 7.1 Q1(iii) 2 marks Find the distance between the points (a, b) and (−a, −b).
Stuck? Show a hint Step 1 Apply d = √[(x₂−x₁)² + (y₂−y₁)²].
Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Stuck? Show a hint Step 1 AB = √[(2−1)² + (3−5)²] = √(1+4) = √5 ≈ 2.236.
Step-by-step solution AB = √[(2−1)² + (3−5)²] = √(1+4) = √5 ≈ 2.236. BC = √[(−2−2)² + (−11−3)²] = √(16+196) = √212 ≈ 14.560. AC = √[(−2−1)² + (−11−5)²] = √(9+256) = √265 ≈ 16.279. Check AB + BC = √5 + √212 ≈ 16.796 ≠ AC ≈ 16.279. No sum of two sides equals the third → points are NOT collinear. Answer: The points are NOT collinear.
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Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
Stuck? Show a hint Step 1 AB = √[(6−5)² + (4+2)²] = √(1+36) = √37.
Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).
Stuck? Show a hint Step 1 Point on x-axis: P(x, 0).
Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.
Stuck? Show a hint Step 1 PQ² = (10−2)² + (y+3)² = 64 + (y+3)².
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (−3, 4).
Stuck? Show a hint Step 1 AP = BP ⇒ AP² = BP².
Exercise 7.2 Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Stuck? Show a hint Step 1 P = ((m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂)) with m₁=2, m₂=3.
Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Stuck? Show a hint Step 1 P (ratio 1:2): x = (1(−2) + 2(4))/3 = 6/3 = 2; y = (1(−3) + 2(−1))/3 = −5/3.
Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6).
Stuck? Show a hint Step 1 Let ratio be k : 1.
Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Stuck? Show a hint Step 1 Let ratio be k : 1.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Stuck? Show a hint Step 1 Use diagonals bisect: mid(AC) = mid(BD).
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
Stuck? Show a hint Step 1 Centre = midpoint of diameter ⇒ ((x+1)/2, (y+4)/2) = (2, −3).
Examples and more questions Do the points (3, 2), (−2, −3) and (2, 3) form a triangle? If so, name the type of triangle formed.
Stuck? Show a hint Step 1 Apply distance formula d = √[(x₂−x₁)² + (y₂−y₁)²] to each pair.
Step-by-step solution Apply distance formula d = √[(x₂−x₁)² + (y₂−y₁)²] to each pair. PQ = √[(3+2)² + (2+3)²] = √(25+25) = √50. QR = √[(−2−2)² + (−3−3)²] = √(16+36) = √52. PR = √[(3−2)² + (2−3)²] = √(1+1) = √2. Check triangle inequality: PQ + PR > QR (since √50 + √2 > √52). Hence a triangle is formed. Test right angle: PQ² + PR² = 50 + 2 = 52 = QR², i.e. QR² = PQ² + PR², so ∠P = 90°. Therefore △PQR is a right-angled triangle. Answer: Yes — a right-angled triangle with the right angle at P.
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Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.
Stuck? Show a hint Step 1 Label A(1, 7), B(4, 2), C(−1, −1), D(−4, 4).
Step-by-step solution Label A(1, 7), B(4, 2), C(−1, −1), D(−4, 4). AB = √[(1−4)² + (7−2)²] = √(9+25) = √34. BC = √[(4−(−1))² + (2−(−1))²] = √(25+9) = √34. CD = √[(−1−(−4))² + (−1−4)²] = √(9+25) = √34. DA = √[(−4−1)² + (4−7)²] = √(25+9) = √34. So AB = BC = CD = DA. Diagonals: AC = √[(1−(−1))² + (7−(−1))²] = √(4+64) = √68; BD = √[(4−(−4))² + (2−4)²] = √(64+4) = √68. AC = BD. All sides equal and diagonals equal → ABCD is a square. Answer: Yes — ABCD is a square (all sides = √34 and both diagonals = √68).
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Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
Stuck? Show a hint Step 1 AP = BP ⇒ AP² = BP² (squaring removes the radical).
Step-by-step solution AP = BP ⇒ AP² = BP² (squaring removes the radical). (x−7)² + (y−1)² = (x−3)² + (y−5)². Expand: x² − 14x + 49 + y² − 2y + 1 = x² − 6x + 9 + y² − 10y + 25. Cancel x² and y² on both sides: −14x − 2y + 50 = −6x − 10y + 34. Rearrange: −8x + 8y = −16 → x − y = 2. Hence x − y = 2 is the perpendicular bisector of segment AB. Answer: x − y = 2.
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Find a point on the y-axis which is equidistant from the points A(6, 5) and B(−4, 3).
Stuck? Show a hint Step 1 Any point on the y-axis has the form (0, y).
Find the coordinates of the point which divides the line segment joining (4, −3) and (8, 5) in the ratio 3 : 1 internally.
Stuck? Show a hint Step 1 Use section formula P = ((m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂)).
In what ratio does the point (−4, 6) divide the line segment joining the points A(−6, 10) and B(3, −8)?
Stuck? Show a hint Step 1 Let (−4, 6) divide AB in ratio k : 1 with A(−6, 10), B(3, −8).
Find the coordinates of the points of trisection of the line segment joining A(2, −2) and B(−7, 4).
Stuck? Show a hint Step 1 P divides AB in ratio 1:2. P = ((1(−7) + 2(2))/3, (1(4) + 2(−2))/3) = (−3/3, 0/3) = (−1, 0).
Step-by-step solution P divides AB in ratio 1:2. P = ((1(−7) + 2(2))/3, (1(4) + 2(−2))/3) = (−3/3, 0/3) = (−1, 0). Q divides AB in ratio 2:1. Q = ((2(−7) + 1(2))/3, (2(4) + 1(−2))/3) = (−12/3, 6/3) = (−4, 2). Check: AP = √(9+4) = √13, PQ = √(9+4) = √13, QB = √(9+4) = √13. ✓ Answer: P(−1, 0) and Q(−4, 2).
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Find the ratio in which the y-axis divides the line segment joining the points (5, −6) and (−1, −4). Also find the point of intersection.
Stuck? Show a hint Step 1 Let ratio be k : 1. By section formula, x = (−k + 5)/(k+1).
If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
Stuck? Show a hint Step 1 Diagonals of a parallelogram bisect each other ⇒ mid(AC) = mid(BD).
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-based 4 marks In a classroom, Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. (i) Find AB, BC and AC. (ii) Are A, B, C seated in a straight line? Justify using the distance formula.
Stuck? Show a hint Step 1 AB = √[(6−3)² + (4−1)²] = √(9+9) = √18 = 3√2.
Step-by-step solution AB = √[(6−3)² + (4−1)²] = √(9+9) = √18 = 3√2. BC = √[(8−6)² + (6−4)²] = √(4+4) = √8 = 2√2. AC = √[(8−3)² + (6−1)²] = √(25+25) = √50 = 5√2. Sum AB + BC = 3√2 + 2√2 = 5√2 = AC. When the sum of two distances equals the third, the three points are collinear → A, B, C are in a straight line. Answer: AB = 3√2, BC = 2√2, AC = 5√2. AB + BC = AC ⇒ A, B, C are collinear (seated in a line).
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Check my answer Questions from the NCERT Class 10 Mathematics textbook (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.