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NCERT Solutions · Class 10 · Chapter 7: Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

Exercise 7.1: Distance formula. \(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Use it to test collinearity (the two shorter distances add to the longest), classify triangles and quadrilaterals (compare sides and diagonals) and find equidistant points.

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Exercise 7.1 questions and solutions

Exercise 7.1, Question 1

Find the distance between each pair of points.
(i) \((2, 3),\ (4, 1)\)
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  1. \(\sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{8}\).
Answer: \(2\sqrt2\)
(ii) \((-5, 7),\ (-1, 3)\)
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  1. \(\sqrt{4^2 + (-4)^2} = \sqrt{32}\).
Answer: \(4\sqrt2\)
(iii) \((a, b),\ (-a, -b)\)
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  1. \[\sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2}\]
Answer: \(2\sqrt{a^2 + b^2}\)

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Exercise 7.1, Question 2

Find the distance between \((0, 0)\) and \((36, 15)\). Use it for two towns A and B at these positions (in km).
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  1. \[\begin{aligned}\sqrt{36^2 + 15^2} &= \sqrt{1296 + 225} \\ &= \sqrt{1521}\end{aligned}\]
Answer: \(39\) (so the towns are 39 km apart)

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Exercise 7.1, Question 3

Are \((1, 5),\ (2, 3),\ (-2, -11)\) collinear?
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  1. \(AB = \sqrt{1 + 4} = \sqrt5\), \(BC = \sqrt{16 + 196} = \sqrt{212}\), \(AC = \sqrt{9 + 256} = \sqrt{265}\).
  2. Without a calculator, square: \[(\sqrt5 + \sqrt{212})^2 = 217 + 2\sqrt{1060}\], which equals \(265\) only if \(\sqrt{1060} = 24\); but \(24^2 = 576 \ne 1060\). So \(AB + BC \ne AC\), and no other pair adds up either (\(\sqrt{265}\) is the longest side).
Answer: Not collinear

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Exercise 7.1, Question 4

Are \((5, -2),\ (6, 4),\ (7, -2)\) the vertices of an isosceles triangle?
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  1. \(AB = \sqrt{1 + 36} = \sqrt{37}\), \(BC = \sqrt{1 + 36} = \sqrt{37}\), \(AC = 2\).
Answer: Yes: two sides equal \(\sqrt{37}\)

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Exercise 7.1, Question 5

Four friends stand at \(A(3, 4),\ B(6, 7),\ C(9, 4),\ D(6, 1)\). Is ABCD a square?
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  1. Sides: \(AB = BC = CD = DA = \sqrt{18} = 3\sqrt2\).
  2. Diagonals: \(AC = 6\), \(BD = 6\). Equal sides and equal diagonals make a square.
Answer: Yes, ABCD is a square

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Exercise 7.1, Question 6

Name the quadrilateral formed (if any) by the points, and give a reason.
(i) \((-1, -2),\ (1, 0),\ (-1, 2),\ (-3, 0)\)
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  1. All sides \(\sqrt8 = 2\sqrt2\); both diagonals \(4\).
Answer: Square
(ii) \((-3, 5),\ (3, 1),\ (0, 3),\ (-1, -4)\)
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  1. \((-3, 5), (0, 3), (3, 1)\): distances \(\sqrt{13} + \sqrt{13} = 2\sqrt{13}\), which is the distance from \((-3, 5)\) to \((3, 1)\). Three of the points are collinear.
Answer: No quadrilateral (three points are collinear)
(iii) \((4, 5),\ (7, 6),\ (4, 3),\ (1, 2)\)
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  1. Opposite sides: \(\sqrt{10}, \sqrt{10}\) and \(\sqrt{13}, \sqrt{13}\).
  2. Diagonals: \(2\) and \(\sqrt{52}\), not equal.
Answer: Parallelogram

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Exercise 7.1, Question 7

Find the point on the x-axis equidistant from \((2, -5)\) and \((-2, 9)\).
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  1. Let it be \((x, 0)\): \((x - 2)^2 + 25 = (x + 2)^2 + 81\).
  2. \(-8x = 56 \Rightarrow x = -7\).
Answer: \((-7, 0)\)

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Exercise 7.1, Question 8

Find \(y\) if the distance between \(P(2, -3)\) and \(Q(10, y)\) is 10 units.
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  1. \[\begin{aligned}&64 + (y + 3)^2 = 100 \\ \Rightarrow\ &(y + 3)^2 = 36\end{aligned}\]
  2. \(y + 3 = \pm 6\).
Answer: \(y = 3\) or \(y = -9\)

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Exercise 7.1, Question 9

\(Q(0, 1)\) is equidistant from \(P(5, -3)\) and \(R(x, 6)\). Find \(x\), and the distances QR and PR.
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  1. \(QP^2 = 25 + 16 = 41\); \(QR^2 = x^2 + 25\). So \(x^2 = 16\), \(x = \pm 4\).
  2. \(QR = \sqrt{41}\).
  3. \(x = 4\): \(PR = \sqrt{1 + 81} = \sqrt{82}\). \(x = -4\): \(PR = \sqrt{81 + 81} = 9\sqrt2\).
Answer: \(x = \pm4\); \(QR = \sqrt{41}\); \(PR = \sqrt{82}\) or \(9\sqrt2\)

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Exercise 7.1, Question 10

Find a relation between \(x\) and \(y\) such that \((x, y)\) is equidistant from \((3, 6)\) and \((-3, 4)\).
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  1. \[(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2\]
  2. \[\begin{aligned}&-6x - 12y + 45 = 6x - 8y + 25 \\ \Rightarrow\ &12x + 4y = 20\end{aligned}\]
Answer: \(3x + y - 5 = 0\)

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Done the NCERT exercises? The board paper asks more

Coordinate Geometry has 40 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

Coordinate Geometry in our sample papers: Sample paper 1 (questions 6, 7, 37) · Sample paper 2 (questions 7, 8, 20, 28) · Sample paper 3 (questions 8, 9, 37) · Sample paper 4 (questions 8, 23, 28) · Sample paper 5 (questions 7, 8, 37).

Also useful: free MCQs and case studies for Coordinate Geometry · Class 10 formula sheet · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

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