CBSE • Class 10 • Mathematics • Chapter 7 (Coordinate Geometry) • Exercise 7.1

Exercise 7.1: Coordinate Geometry — NCERT Solutions

Distance formula — computing the length of a segment between two points.

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What this exercise covers

Distance formulaVerifying collinearityGeometric proofs from coordinates

Step-by-step solutions — Exercise 7.1

8 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 7.1 Q1 • 2 marks

Find the distance between the points A(3,4)A(3, 4) and B(7,1)B(7, 1).
Hint (Socratic — try this first)▾
Which formula uses the differences in the xx-coordinates and yy-coordinates?
Step-by-step solution▾

The distance formula is d=(x2−x1)2+(y2−y1)2.d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Here (x1,y1)=(3,4)(x_1, y_1) = (3, 4) and (x2,y2)=(7,1)(x_2, y_2) = (7, 1).

d=(7−3)2+(1−4)2=42+(−3)2d = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{4^2 + (-3)^2} =16+9=25=5.= \sqrt{16 + 9} = \sqrt{25} = 5.

So the distance is 55 units.

Common mistake:
Forgetting to square the negative difference correctly — writing (1−4)2=−9(1-4)^2 = -9 instead of 99.
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Exercise 7.1 Q2 • 2 marks

Find the distance of the point P(−6,8)P(-6, 8) from the origin.
Hint (Socratic — try this first)▾
What are the coordinates of the origin, and how does that simplify the distance formula?
Step-by-step solution▾

The origin is O(0,0)O(0, 0). Using the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} =(−6−0)2+(8−0)2= \sqrt{(-6 - 0)^2 + (8 - 0)^2} =36+64=100=10.= \sqrt{36 + 64} = \sqrt{100} = 10.

The distance from the origin is 1010 units.

Common mistake:
Ignoring the negative sign of −6-6 and computing 62−82\sqrt{6^2 - 8^2} or mixing addition with subtraction under the root.
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Exercise 7.1 Q3 • 3 marks

Show that the points A(1,2)A(1, 2), B(5,2)B(5, 2) and C(3,5)C(3, 5) form an isosceles triangle.
Hint (Socratic — try this first)▾
For an isosceles triangle, how many of the three side lengths must be equal?
Step-by-step solution▾

Compute all three side lengths.

AB=(5−1)2+(2−2)2=16+0=4.AB = \sqrt{(5-1)^2 + (2-2)^2} = \sqrt{16 + 0} = 4. BC=(3−5)2+(5−2)2=4+9=13.BC = \sqrt{(3-5)^2 + (5-2)^2} = \sqrt{4 + 9} = \sqrt{13}. CA=(1−3)2+(2−5)2=4+9=13.CA = \sqrt{(1-3)^2 + (2-5)^2} = \sqrt{4 + 9} = \sqrt{13}.

Since BC=CA=13BC = CA = \sqrt{13}, two sides are equal.

Therefore △ABC\triangle ABC is isosceles.

Common mistake:
Concluding equilateral without checking all three sides, or making an arithmetic error that makes unequal sides appear equal.
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Exercise 7.1 Q4 • 3 marks

Find the value of yy for which the distance between P(2,−3)P(2, -3) and Q(10,y)Q(10, y) is 1010 units.
Hint (Socratic — try this first)▾
Can you set the distance formula equal to 1010 and then square both sides?
Step-by-step solution▾

Using the distance formula and equating to 1010: (10−2)2+(y−(−3))2=10.\sqrt{(10-2)^2 + (y-(-3))^2} = 10. 64+(y+3)2=10.\sqrt{64 + (y+3)^2} = 10.

Squaring both sides: 64+(y+3)2=100.64 + (y+3)^2 = 100. (y+3)2=36.(y+3)^2 = 36. y+3=±6.y + 3 = \pm 6.

So y+3=6⇒y=3y + 3 = 6 \Rightarrow y = 3, or y+3=−6⇒y=−9y + 3 = -6 \Rightarrow y = -9.

Thus y=3y = 3 or y=−9y = -9.

Common mistake:
Taking only the positive square root and reporting just one value of yy instead of both.
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Exercise 7.1 Q5 • 3 marks

Show that the points A(1,1)A(1, 1), B(4,4)B(4, 4) and C(6,6)C(6, 6) are collinear.
Hint (Socratic — try this first)▾
If three points are collinear, how does the sum of two of the distances relate to the third?
Step-by-step solution▾

Find the three distances.

AB=(4−1)2+(4−1)2=9+9=18=32.AB = \sqrt{(4-1)^2 + (4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}. BC=(6−4)2+(6−4)2=4+4=8=22.BC = \sqrt{(6-4)^2 + (6-4)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}. AC=(6−1)2+(6−1)2=25+25=50=52.AC = \sqrt{(6-1)^2 + (6-1)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}.

Since AB+BC=32+22=52=ACAB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC, the points are collinear.

Common mistake:
Assuming collinearity from a rough sketch without verifying AB+BC=ACAB + BC = AC.
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Exercise 7.1 Q6 • 3 marks

Find a point on the xx-axis which is equidistant from A(2,−5)A(2, -5) and B(−2,9)B(-2, 9).
Hint (Socratic — try this first)▾
What is the yy-coordinate of any point on the xx-axis, and what does equidistant tell you about two distances?
Step-by-step solution▾

Let the point be P(x,0)P(x, 0) on the xx-axis.

Equidistant means PA=PBPA = PB, so PA2=PB2PA^2 = PB^2. (x−2)2+(0+5)2=(x+2)2+(0−9)2.(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2. (x−2)2+25=(x+2)2+81.(x-2)^2 + 25 = (x+2)^2 + 81. x2−4x+4+25=x2+4x+4+81.x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81. −4x+25=4x+81.-4x + 25 = 4x + 81. −8x=56⇒x=−7.-8x = 56 \Rightarrow x = -7.

The required point is (−7,0)(-7, 0).

Common mistake:
Taking the point as (0,y)(0, y) (on the yy-axis) instead of (x,0)(x, 0) on the xx-axis.
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Exercise 7.1 Q7 • 4 marks

Prove that the points A(3,0)A(3, 0), B(6,4)B(6, 4) and C(−1,3)C(-1, 3) are the vertices of a right-angled isosceles triangle.
Hint (Socratic — try this first)▾
Which two conditions must hold — one for 'isosceles' and one (via the converse of Pythagoras) for 'right-angled'?
Step-by-step solution▾

Compute the squares of the side lengths.

AB2=(6−3)2+(4−0)2=9+16=25.AB^2 = (6-3)^2 + (4-0)^2 = 9 + 16 = 25. BC2=(−1−6)2+(3−4)2=49+1=50.BC^2 = (-1-6)^2 + (3-4)^2 = 49 + 1 = 50. CA2=(3−(−1))2+(0−3)2=16+9=25.CA^2 = (3-(-1))^2 + (0-3)^2 = 16 + 9 = 25.

Isosceles: AB2=CA2=25AB^2 = CA^2 = 25, so AB=CAAB = CA.

Right-angled: AB2+CA2=25+25=50=BC2AB^2 + CA^2 = 25 + 25 = 50 = BC^2, so by the converse of the Pythagoras theorem the angle at AA is 90∘90^\circ.

Hence △ABC\triangle ABC is a right-angled isosceles triangle.

Common mistake:
Checking the Pythagoras condition with the wrong side as hypotenuse (using AB2+BC2=CA2AB^2 + BC^2 = CA^2 instead of the longest side squared).
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Exercise 7.1 Q8 • 3 marks

Find the value of kk if the point P(k,4)P(k, 4) is equidistant from A(1,2)A(1, 2) and B(3,−2)B(3, -2).
Hint (Socratic — try this first)▾
Equidistant means the two distances are equal — can you square both to avoid the roots?
Step-by-step solution▾

Set PA2=PB2PA^2 = PB^2. (k−1)2+(4−2)2=(k−3)2+(4+2)2.(k-1)^2 + (4-2)^2 = (k-3)^2 + (4+2)^2. (k−1)2+4=(k−3)2+36.(k-1)^2 + 4 = (k-3)^2 + 36. k2−2k+1+4=k2−6k+9+36.k^2 - 2k + 1 + 4 = k^2 - 6k + 9 + 36. −2k+5=−6k+45.-2k + 5 = -6k + 45. 4k=40⇒k=10.4k = 40 \Rightarrow k = 10.

Thus k=10k = 10.

Common mistake:
Expanding (4+2)2(4+2)^2 as 42+22=204^2 + 2^2 = 20 instead of 62=366^2 = 36.
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How to approach Exercise 7.1

  1. Re-read the chapter summary first. Open Coordinate Geometry and refresh the key concepts: Distance formula, Section formula, Midpoint, Area of triangle from coordinates.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Coordinate Geometry

  1. Exercise 7.1 — Distance formula — computing the length of a segment between two points.
  2. Exercise 7.2 — Section formula — finding the point that divides a segment in a given ratio.

Exercise 7.2 →

Section formula — finding the point that divides a segment in a given ratio.

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