Coordinate Geometry: Class 10 Maths · Chapter 7

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

70 min
2 simulations
29 practice questions
19 MCQs

What you will learn in Coordinate Geometry

Recall the abscissa and ordinate of a point and the forms of points on the axes, and plot points in all four quadrants.
Derive the distance formula from the Pythagoras theorem and use it for any two points, including distances from the origin.
Use distances to decide whether points are collinear and to classify triangles and quadrilaterals.
Find unknown coordinates from distance conditions, such as a point equidistant from two given points.
Derive the section formula for internal division using similar triangles, with its k : 1 and midpoint forms.
Find the point dividing a segment in a given ratio, the ratio in which a point or an axis divides a segment, and points of trisection.
Use the midpoint formula with properties of parallelograms and circles to find missing vertices and endpoints.

Original companion text aligned to NCERT Mathematics Class 10, Chapter 7, pp. 99–112. Not affiliated with or endorsed by NCERT.

7.1 Introduction

Open a map app and ask how far the nearest chemist is. The app stores your home and the shop as pairs of numbers, and a short formula turns those numbers into a distance. In this chapter you will learn that formula, and a second one that finds a point part of the way between two places.

First, a quick recap from Class 9. Two perpendicular number lines, the x-axis and the y-axis, cross at the origin $O (0, 0)$ , and every point of the plane gets an ordered pair $(x, y)$ . The first number, the x-coordinate or abscissa (भुज), measures how far the point is from the y-axis: to the right when it is positive and to the left when it is negative. The second number, the y-coordinate or ordinate (कोटि), measures how far the point is above or below the x-axis.

A point on the x-axis is at zero height, so it always looks like $(x, 0)$ . A point on the y-axis has no sideways shift, so it looks like $(0, y)$ . Board questions often hide these forms in words such as "a point on the y-axis", so watch for them.

Try thisNCERT 7.1 Introduction · p. 99

Plot a kite

You’ll need

graph paper or a squared notebook page
a sharp pencil
a ruler

Predict first

Two of these points lie on the y-axis. Which two?

A (0, 7)

and

C (0, - 5)

B (4, 3)

and

D (- 4, 3)

P (- 1, - 6)

and

Q (1, - 7)

Draw both axes with 1 cm for 1 unit: from $- 6$ to $6$ across and from $- 9$ to $8$ up.
Plot $A (0, 7)$ , $B (4, 3)$ , $C (0, - 5)$ , $D (- 4, 3)$ , $P (- 1, - 6)$ , $Q (1, - 7)$ and $R (- 0.5, - 8)$ .
Join A to B to C to D and back to A.
Join A to C and B to D. Then join C to P, P to Q and Q to R.
What picture do you get? In which quadrants are D, P and Q?

You get a kite (patang) with its two sticks and a zigzag tail, ready for Makar Sankranti. D lies in quadrant II, P in quadrant III and Q in quadrant IV. R, at $x = - 0.5$ , sits half a square left of the y-axis, in quadrant III.

3/3Add the sticks AC and BD and the tail C, P, Q, R.

Fig. 7.1Plot, join, and a kite appears.

A coordinate grid from minus 6 to 6 across and minus 9 to 8 up, with a list of the points A(0, 7), B(4, 3), C(0, minus 5), D(minus 4, 3), P(minus 1, minus 6), Q(1, minus 7) and R(minus 0.5, minus 8) on the right. Step 1 shows the seven points plotted as dots with their letters; A and C sit on the y-axis. Step 2 joins A, B, C, D and back to A, making a four-sided kite outline. Step 3 adds the two sticks AC and BD, colours the upper part yellow and the lower part orange, and draws the tail as a zigzag from C through P and Q to R.

You have used the plane for more than single points. A linear equation $a x + b y + c = 0$ , with $a$ and $b$ not both zero, draws a straight line, and in Chapter 2 the graph of $y = a x^{2} + b x + c$ with $a \neq 0$ turned out to be a parabola. Turning shapes into equations, and equations back into shapes, is the business of coordinate geometry (निर्देशांक ज्यामिति). A question about a figure becomes a calculation, and an equation becomes a picture you can look at.

Because it lets algebra do the work of measuring, coordinate geometry sits inside many tools you use: map and GPS apps, computer graphics and animation, engineering drawings, robot arms and the instruments that locate earthquakes.

This chapter adds two tools. The distance formula gives the distance between two points from their coordinates alone. The section formula finds the point that cuts a segment in a given ratio, such as halfway or one-third of the way along.

7.2 Distance Formula

A district hospital in the hills sends medicines to villages by drone. One village, V, is 45 km east and 24 km north of the hospital H. The drone flies in a straight line. How long is its flight?

Put the hospital at the origin, with east along the x-axis and north along the y-axis, and let 1 unit stand for 1 km. Then H is $(0, 0)$ and V is $(45, 24)$ . The east part and the north part of the journey are the two perpendicular sides of a right triangle, and the flight path is its hypotenuse (Fig. 7.2). By the Pythagoras theorem, $H V^{2} = 45^{2} + 24^{2} = 2025 + 576 = 2601$ , so $H V = 51$ km.

3/3The straight path is the hypotenuse: $\sqrt{45^{2} + 24^{2}} = 51$ km.

Fig. 7.2East and north make the legs of a right triangle; the flight is the hypotenuse.NCERT Fig. 7.1 · p. 100

A map with East to the right and North upwards. Step 1 shows the hospital H at the origin, marked with a cross, and the village V at (45, 24). Step 2 adds dashed lines 45 km east and then 24 km north, meeting at a right angle below V. Step 3 draws the straight flight path from H to V, labelled 51 km, with the working: square root of 45 squared plus 24 squared equals 51.

The same idea works for any two points. Let us build it up step by step, starting with points on the axes.

Points on an axis

Take $A (- 4, 0)$ and $B (5, 0)$ on the x-axis (Fig. 7.3). A is 4 units to the left of O and B is 5 units to the right, so $A B = 4 + 5 = 9$ units. Subtracting the x-coordinates gives the same answer at once: $5 - (- 4) = 9$ . On the y-axis, $C (0, 3)$ and $D (0, - 2)$ are $3 - (- 2) = 5$ units apart.

So two points on the x-axis are $∣ x_{2} - x_{1} ∣$ apart, and two points on the y-axis are $∣ y_{2} - y_{1} ∣$ apart. The bars mean "take the size and drop the sign", because a distance is never negative.

Now join A, on the x-axis, to C, on the y-axis. The axes are perpendicular, so triangle AOC has a right angle at O, with $O A = 4$ and $O C = 3$ . The Pythagoras theorem gives $A C = \sqrt{4^{2} + 3^{2}} = 5$ units.

3/3The axes meet at a right angle, so $A C = \sqrt{4^{2} + 3^{2}} = 5$ .

Fig. 7.3On an axis, subtract; across the axes, use the right angle at O.NCERT Fig. 7.2 · p. 100

A grid with the x-axis from minus 5 to 7 and the y-axis from minus 3 to 4. Step 1 highlights the stretch of the x-axis from A(minus 4, 0) to B(5, 0) and shows AB = 5 minus minus 4 = 9. Step 2 highlights the y-axis from C(0, 3) down to D(0, minus 2) and shows CD = 3 minus minus 2 = 5. Step 3 shades the right triangle AOC with its right angle at the origin and shows AC = square root of 4 squared plus 3 squared = 5.

Any two points

For points away from the axes, build the right triangle yourself. Take $K (- 2, 5)$ in quadrant II and $L (4, - 3)$ in quadrant IV (Fig. 7.4). A horizontal line through K and a vertical line through L meet at $C (4, 5)$ , the corner of a right triangle KCL.

The horizontal side KC runs from $x = - 2$ to $x = 4$ : 2 units to reach the y-axis and 4 more beyond it, so $K C = 4 - (- 2) = 6$ . The vertical side CL runs from $y = 5$ down to $y = - 3$ , so $C L = 5 - (- 3) = 8$ . Therefore $K L = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$ units.

4/4By Pythagoras, $K L = \sqrt{6^{2} + 8^{2}} = 10$ .

Fig. 7.4Subtracting coordinates counts the units on both sides of an axis.NCERT Fig. 7.4 · p. 101

A grid from minus 4 to 6 across and minus 5 to 7 up. Step 1 shows K(minus 2, 5) in quadrant II and L(4, minus 3) in quadrant IV. Step 2 adds the corner C(4, 5) where a horizontal line through K meets a vertical line through L, with a right-angle mark. Step 3 colours the legs and states KC = 2 + 4 = 6 and CL = 5 + 3 = 8, since each leg crosses an axis. Step 4 draws KL and shows KL = square root of 6 squared plus 8 squared = 10.

Crossing an axis needed no special care: subtracting the coordinates counted the units on both sides of it. Two points in the same quadrant work the same way. For $(2, 3)$ and $(7, 15)$ the legs are $7 - 2 = 5$ and $15 - 3 = 12$ , so the distance is $\sqrt{25 + 144} = 13$ units.

DerivationThe distance formula

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ be any two points. Draw a horizontal line through P and a vertical line through Q. They meet at the corner $C (x_{2}, y_{1})$ , which is level with P and directly in line with Q (Fig. 7.5).

4/4The right angle at C gives $P Q^{2} = P C^{2} + C Q^{2}$ .

Fig. 7.5The corner

C

takes its x from

Q

and its y from

P

.NCERT Fig. 7.5 · p. 102

Axes with origin O. Step 1 shows P with coordinates x1, y1 and Q with coordinates x2, y2, with dashed guides to both axes. Step 2 adds the corner C(x2, y1) and a right angle at C. Step 3 labels the horizontal leg PC as x2 minus x1 and the vertical leg CQ as y2 minus y1. Step 4 draws PQ and writes PQ squared = PC squared + CQ squared.

Along PC only x changes, so $P C = ∣ x_{2} - x_{1} ∣$ . Along CQ only y changes, so $C Q = ∣ y_{2} - y_{1} ∣$ . The angle at C is a right angle, so the Pythagoras theorem gives $P Q^{2} = P C^{2} + C Q^{2}$ , that is, $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

Squaring removes any minus sign, so the bars are no longer needed.

A length is never negative, so we keep only the positive square root:

Distance formula

P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

(7.1)

where

$(x_{1}, y_{1})$: the coordinates of P
$(x_{2}, y_{2})$: the coordinates of Q

This is the distance formula (दूरी सूत्र). Two consequences are worth remembering. First, putting $(0, 0)$ for one of the points gives the distance of $P (x, y)$ from the origin:

Distance from the origin

O P = \sqrt{x^{2} + y^{2}}

(7.2)

Second, $(x_{1} - x_{2})^{2} = (x_{2} - x_{1})^{2}$ , because a number and its negative have the same square. So it never matters which point you call P and which you call Q.

SimulationDistance explorer

A coordinate grid from minus 8 to 8 on both axes with three draggable points, A at (minus 2, 5), B at (4, minus 3) and C at (4, 5). Points snap to whole numbers. The sides AB, BC and CA are drawn and measured, with the right-triangle legs for each distance, and each length is given exactly, for example 2 root 5, and to two decimal places.

Fig. 7.6Drag A, B and C. Each distance is shown with its horizontal and vertical legs, as an exact surd and as a decimal.

Try this

Drag C until triangle ABC is isosceles. Which two distances are now equal?
Drag C onto segment AB, for example to $(1, 1)$ . How do $A C + C B$ and $A B$ compare?
Move A to $(- 3, - 1)$ and B to $(3, 7)$ . Work out AB first, then check.

Quick check

1 of 3

Find the distance between

(- 1, 3)

and

(5, - 5)

units

Show answer

Answer 10 units

The differences are

5 - (- 1) = 6

and

- 5 - 3 = - 8

, so the distance is

\sqrt{36 + 64} = 10

. A common slip is

- 5 - 3 = - 2

, which gives

\sqrt{40}

Using the distance formula

Distances let you test shapes without a protractor. Three facts do most of the work:

Triangle or not? Three points form a triangle when the sum of any two of the three distances is greater than the third. If the two shorter distances add up exactly to the longest, the points are collinear (संरेखी): they lie on one straight line.
Right angle? If the squares of two sides add up to the square of the third side, the angle opposite that third side is $90^{\circ}$ . This is the converse of the Pythagoras theorem.
Equal sides? Equal distances reveal isosceles and equilateral triangles, and rhombuses and squares among quadrilaterals.

Compare

Telling quadrilaterals apart with distances

Square

All four sides equal
Diagonals equal

Rhombus

All four sides equal
Diagonals unequal

Rectangle

Opposite sides equal
Diagonals equal

Parallelogram

Opposite sides equal
Diagonals unequal

Example 7.1NCERT Example 1 · p. 102

Triangle or not? Which kind?

Do the points $P (2, - 1)$ , $Q (- 4, 2)$ and $R (3, 1)$ form a triangle? If they do, what kind of triangle is it?

Try it yourself first. Then reveal the solution one step at a time.

Example 7.2NCERT Example 2 · p. 103

Proving a square

Show that $A (- 2, 0)$ , $B (1, - 2)$ , $C (3, 1)$ and $D (0, 3)$ are the vertices of a square.

Try it yourself first. Then reveal the solution one step at a time.

3/3Both diagonals are $\sqrt{26}$ , so the rhombus is a square.

Fig. 7.7Four equal sides and equal diagonals: a square, even though it is tilted.

A grid from minus 3 to 4 on both axes, with the coordinates A(minus 2, 0), B(1, minus 2), C(3, 1) and D(0, 3) listed on the right. Step 1 shows the four points. Step 2 draws the four sides in blue and states AB = BC = CD = DA = root 13. Step 3 adds the two diagonals as dashed orange lines, states AC = BD = root 26, and concludes that ABCD is a square.

Example 7.3NCERT Example 3 · p. 103

Can one cable run straight?

On a colony map, an electrician has marked three poles for street lights at $A (- 1, - 3)$ , $B (1, 1)$ and $C (4, 7)$ . She wants one straight cable to pass through all three. Is that possible?

Try it yourself first. Then reveal the solution one step at a time.

3/4 $A B + B C = 5 \sqrt{5} = A C$ : one straight line.

Fig. 7.8When the two shorter distances add up to the longest, the points are collinear.

A grid from minus 2 to 6 across and minus 4 to 8 up with A(minus 1, minus 3), B(1, 1) and C(4, 7). Step 1 shows the three points. Step 2 colours AB and BC and gives AB = root 20 = 2 root 5 and BC = root 45 = 3 root 5. Step 3 draws AC straight through B, gives AC = root 125 = 5 root 5, and notes that AB + BC = AC, so the points are on one straight line. Step 4 adds a point E(5, 4) off the line, with a dotted bent path through B, and notes that for a bent path AB + BE is greater than AE.

Notice why simplifying mattered: $2 \sqrt{5}$ and $3 \sqrt{5}$ add like $2 x + 3 x$ , but $\sqrt{20} + \sqrt{45}$ cannot be added until each is simplified.

Example 7.4NCERT Example 4 · p. 104

All points equally far from A and B

Which condition must $x$ and $y$ satisfy for $P (x, y)$ to be equidistant (समदूरस्थ), that is, equally far, from $A (2, - 1)$ and $B (6, 3)$ ?

Try it yourself first. Then reveal the solution one step at a time.

Test two points that fit $x + y = 5$ , say $(1, 4)$ and $(5, 0)$ : each is exactly as far from A as from B (Fig. 7.9). So the answer is a whole straight line of points. It passes through $(4, 1)$ , halfway between A and B, and crosses AB at a right angle: it is the perpendicular bisector (लंब समद्विभाजक) of AB that you met in Class 9 constructions.

4/4All such points lie on $x + y = 5$ , the perpendicular bisector of AB.

Fig. 7.9Points equidistant from A and B line up on the perpendicular bisector of AB.NCERT Fig. 7.7 · p. 104

A grid from minus 1 to 7 across and minus 2 to 6 up with A(2, minus 1) and B(6, 3) joined. Step 1 shows A and B. Step 2 adds P(1, 4) with dashed blue lines to A and B, both root 26. Step 3 adds Q(5, 0) with dotted orange lines to A and B, both root 10, and marks M(4, 1), the midpoint of AB. Step 4 draws the green line x + y = 5 through P, M and Q, crossing AB at a right angle at M.

Example 7.5NCERT Example 5 · p. 104

A point on the y-axis

Find the point on the y-axis that is equidistant from $A (7, 3)$ and $B (- 1, - 5)$ .

Try it yourself first. Then reveal the solution one step at a time.

Linking this to Example 7.4: P is where the perpendicular bisector of AB meets the y-axis. Here the bisector is $x + y = 2$ , and putting $x = 0$ gives $y = 2$ once more.

Ready for speed practice? The drill below gives you two points at a time. Type the distance as a whole number or a surd such as 2√5.

SimulationDistance drill

A challenge on a grid from minus 10 to 10. Each round marks two points; you type their distance as a whole number, a decimal or a surd such as 2 root 5. After an answer the sim draws the horizontal and vertical legs and shows the exact distance, and it keeps score over five rounds.

Fig. 7.10Five rounds: find the distance between the two marked points, then check the right-triangle legs.

Try this

Before typing, say the two leg lengths aloud.
Leave the answer as a simplified surd when it is not a whole number.

Practice · 10 questions

Practice 7.1: questions with answers

1
Short answerEasyNCERT Exercise 7.1, Q1
Find the distance between each pair of points.
1. $(3, - 2)$ and $(- 5, 4)$
  Show solution
  Answer10 units
  Differences: $- 5 - 3 = - 8$ and $4 - (- 2) = 6$ .
  Distance $= \sqrt{64 + 36} = \sqrt{100} = 10$ .
2. $(- 6, - 1)$ and $(- 2, 7)$
  Show solution
  Answer $4 \sqrt{5}$ units
  Differences: $4$ and $8$ .
  Distance $= \sqrt{16 + 64} = \sqrt{80} = 4 \sqrt{5}$ .
3. $(a + b, a - b)$ and $(a - b, a + b)$ , where $b > 0$
  Show solution
  Answer $2 \sqrt{2} b$
  The x-difference is $(a - b) - (a + b) = - 2 b$ .
  The y-difference is $(a + b) - (a - b) = 2 b$ .
  Distance $= \sqrt{4 b^{2} + 4 b^{2}} = \sqrt{8 b^{2}} = 2 \sqrt{2} b$ .
Short answerEasyNCERT Exercise 7.1, Q2
Find the distance between $(0, 0)$ and $(20, 21)$ . A ferry leaves a jetty and docks 20 km east and 21 km north of it. How far is the straight crossing?
Show solution
Answer29 units; 29 km
By (7.2), the distance is $\sqrt{20^{2} + 21^{2}} = \sqrt{400 + 441} = \sqrt{841} = 29$ .
With the jetty at $(0, 0)$ and the dock at $(20, 21)$ , the straight crossing is 29 km.
3
Short answerMediumNCERT Exercise 7.1, Q3
Use distances to decide whether each set of points is collinear.
1. $A (- 2, 7)$ , $B (1, 1)$ , $C (4, - 5)$
  Show solution
  AnswerCollinear
  $A B = \sqrt{9 + 36} = 3 \sqrt{5}$ and $B C = \sqrt{9 + 36} = 3 \sqrt{5}$ .
  $A C = \sqrt{36 + 144} = \sqrt{180} = 6 \sqrt{5}$ .
  $A B + B C = 6 \sqrt{5} = A C$ , so the points are collinear (B is the middle one).
2. $A (- 1, 2)$ , $B (2, 6)$ , $C (6, 9)$
  Show solution
  AnswerNot collinear
  $A B = \sqrt{9 + 16} = 5$ and $B C = \sqrt{16 + 9} = 5$ .
  $A C = \sqrt{49 + 49} = 7 \sqrt{2} \approx 9.90$ .
  $A B + B C = 10$ , which is more than $A C$ , so the points are not collinear; they form a triangle.
Short answerEasyNCERT Exercise 7.1, Q4
Do $A (- 3, 1)$ , $B (1, 4)$ and $C (4, 0)$ form an isosceles triangle? Is there anything more you can say about it?
Show solution
AnswerYes; it is right-angled at B as well.
$A B^{2} = 16 + 9 = 25$ , $B C^{2} = 9 + 16 = 25$ and $A C^{2} = 49 + 1 = 50$ .
So $A B = B C = 5$ : the triangle is isosceles.
Also $A B^{2} + B C^{2} = 50 = A C^{2}$ , so $∠ B = 90^{\circ}$ . It is a right isosceles triangle.
Short answerMediumNCERT Exercise 7.1, Q5
For Diwali, four diyas are placed on a rangoli grid at $A (2, 5)$ , $B (5, 7)$ , $C (8, 5)$ and $D (5, 3)$ , and joined in order with a line of chalk. Is ABCD a square? If not, what is it?
Show solution
AnswerA rhombus, not a square
Each side has differences of 3 and 2, so $A B = B C = C D = D A = \sqrt{13}$ .
Diagonal $A C = 8 - 2 = 6$ (a horizontal line) and diagonal $B D = 7 - 3 = 4$ (a vertical line).
Four equal sides make a rhombus, but the diagonals are unequal, so it is not a square.
6
Long answerMediumNCERT Exercise 7.1, Q6
Joining each set of four points in the order given, what kind of quadrilateral, if any, do you get? Give reasons.
1. $(0, - 3)$ , $(4, 0)$ , $(1, 4)$ , $(- 3, 1)$
  Show solution
  AnswerSquare
  Each side has differences of 4 and 3 (in some order and sign), so all four sides are 5.
  Diagonals: $\sqrt{1 + 49} = \sqrt{50}$ and $\sqrt{49 + 1} = \sqrt{50}$ .
  Equal sides and equal diagonals: a square.
2. $(- 2, - 1)$ , $(1, 1)$ , $(4, 3)$ , $(3, - 4)$
  Show solution
  AnswerNo quadrilateral
  Call them A, B, C, D. $A B = \sqrt{13}$ , $B C = \sqrt{13}$ and $A C = \sqrt{52} = 2 \sqrt{13}$ .
  $A B + B C = A C$ , so A, B and C are collinear.
  Three of the points lie on one line, so no quadrilateral is formed.
3. $(- 2, 1)$ , $(3, 2)$ , $(5, - 1)$ , $(0, - 2)$
  Show solution
  AnswerParallelogram
  Sides: $\sqrt{26}$ , $\sqrt{13}$ , $\sqrt{26}$ , $\sqrt{13}$ , so opposite sides are equal.
  Diagonals: $\sqrt{49 + 4} = \sqrt{53}$ and $\sqrt{9 + 16} = 5$ , which are unequal.
  Opposite sides equal with unequal diagonals: a parallelogram (not a rectangle).
Short answerMediumNCERT Exercise 7.1, Q7
Find the point on the x-axis that is equidistant from $(7, - 2)$ and $(- 1, 6)$ .
Show solution
Answer $(1, 0)$
Let the point be $P (x, 0)$ . Then $(x - 7)^{2} + 4 = (x + 1)^{2} + 36$ .
Expanding: $- 14 x + 53 = 2 x + 37$ , so $16 x = 16$ and $x = 1$ .
Check: both distances are $\sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}$ .
Short answerMediumNCERT Exercise 7.1, Q8
P is $(1, - 2)$ and Q is $(6, y)$ . If $P Q = 13$ units, what can $y$ be?
Show solution
Answer $y = 10$ or $y = - 14$
$(6 - 1)^{2} + (y + 2)^{2} = 13^{2}$ , so $(y + 2)^{2} = 169 - 25 = 144$ .
$y + 2 = 12$ or $y + 2 = - 12$ .
So $y = 10$ or $y = - 14$ . Both points are 13 units from P, one above and one below.
Long answerMediumNCERT Exercise 7.1, Q9
The point $A (0, 2)$ is equidistant from $B (3, - 2)$ and $C (x, 5)$ . Find the possible values of $x$ , and the lengths AC and BC.
Show solution
Answer $x = \pm 4$ ; $A C = 5$
$A B = \sqrt{9 + 16} = 5$ .
$A C^{2} = x^{2} + 9$ , and this must be 25, so $x^{2} = 16$ and $x = 4$ or $x = - 4$ .
In both cases $A C = A B = 5$ .
For $C (4, 5)$ : $B C = \sqrt{1 + 49} = \sqrt{50} = 5 \sqrt{2}$ .
For $C (- 4, 5)$ : $B C = \sqrt{49 + 49} = \sqrt{98} = 7 \sqrt{2}$ .
Short answerMediumNCERT Exercise 7.1, Q10
Write the equation satisfied by every point $(x, y)$ that is as far from $(4, 1)$ as from $(- 2, 5)$ .
Show solution
Answer $3 x - 2 y + 3 = 0$
$(x - 4)^{2} + (y - 1)^{2} = (x + 2)^{2} + (y - 5)^{2}$ .
Expanding: $- 8 x + 17 - 2 y = 4 x + 29 - 10 y$ .
So $- 12 x + 8 y = 12$ , that is, $3 x - 2 y + 3 = 0$ .
Check with the midpoint $(1, 3)$ : $3 - 6 + 3 = 0$ .

7.3 Section Formula

Back to the medicine drone of Section 7.2. Its battery lasts for most of the 51 km flight, but not all of it. The district plans a battery-swap halt S on the straight route from H to V, placed so that the drone has already flown twice as far as it still has to go: $H S : S V = 2 : 1$ . Where should the halt be?

Think of the route in three equal parts. S is at the end of the second part, so it is $\frac{2}{3}$ of the way from H to V. On a straight route, x and y grow steadily together, so at S the drone has covered $\frac{2}{3}$ of the eastward 45 km and $\frac{2}{3}$ of the northward 24 km. That gives $x = \frac{2}{3} \times 45 = 30$ and $y = \frac{2}{3} \times 24 = 16$ , so the halt belongs at $S (30, 16)$ .

Why may we split the journey like this? Drop perpendiculars SD and VE to the x-axis (Fig. 7.11). The small right triangle HDS and the large right triangle HEV share the angle at H, so they are similar (AA, Chapter 6). Every side of the small one is therefore $\frac{H S}{H V} = \frac{2}{3}$ of the matching side of the large one:

\begin{aligned} \frac{H D}{H E} & = \frac{x}{45} = \frac{2}{3} \\ \frac{D S}{E V} & = \frac{y}{24} = \frac{2}{3} \end{aligned}

Check with the distance formula: $H S = \sqrt{30^{2} + 16^{2}} = 34$ km and $S V = \sqrt{15^{2} + 8^{2}} = 17$ km, and 34 is indeed twice 17.

4/4 $\frac{x}{45} = \frac{2}{3}$ and $\frac{y}{24} = \frac{2}{3}$ give $S (30, 16)$ .

Fig. 7.11The small triangle is the large one scaled by

\frac{2}{3}

.NCERT Fig. 7.9 · p. 106

The drone route from H at the origin to V(45, 24), cut into three equal parts by small dots, with the halt S at the second dot so that HS to SV is 2 to 1. Step 2 drops dashed perpendiculars from S and V to the x-axis at D and E. Step 3 shades the small triangle HDS blue inside the large triangle HEV, notes that they are similar, and labels HD as x, DS as y and EV as 24. Step 4 writes x over 45 equals 2 over 3 and y over 24 equals 2 over 3.

We say that S divides HV internally in the ratio $2 : 1$ : it lies between H and V, with $H S : S V = 2 : 1$ . Now let us do the same for any two points and any ratio.

The general formula

DerivationThe section formula

Let $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ be two points, and let $P (x, y)$ lie on segment AB with $A P : P B = m_{1} : m_{2}$ , where $m_{1}$ and $m_{2}$ are positive numbers. Then $A B = A P + P B$ is $m_{1} + m_{2}$ parts, so P is $\frac{m_{1}}{m_{1} + m_{2}}$ of the way from A to B.

Draw a horizontal line through A and a vertical line through B; they meet at the corner $C (x_{2}, y_{1})$ . The vertical line through P meets AC at $D (x, y_{1})$ (Fig. 7.12).

4/4Each side is a difference of coordinates, so $\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{m_{1}}{m_{1} + m_{2}}$ .

Fig. 7.12A small right triangle inside a large one, sharing the angle at A.NCERT Fig. 7.10 · p. 106

A segment from A(x1, y1) up to B(x2, y2) with P(x, y) on it; the part AP is marked m1 and the part PB is marked m2. Step 2 adds a dashed horizontal line from A to the corner C below B, a dashed vertical line CB, and a dashed vertical line from P down to D on AC. Step 3 shades the small triangle ADP blue inside the large triangle ACB and states that they are similar. Step 4 labels AD = x minus x1, DP = y minus y1, AC = x2 minus x1 and CB = y2 minus y1; the step caption gives the ratio x minus x1 over x2 minus x1 equals m1 over m1 plus m2.

Triangles ADP and ACB have right angles at D and C and share the angle at A, so they are similar. Matching sides are in the same ratio:

\begin{aligned} \frac{A D}{A C} & = \frac{D P}{C B} \\ = \frac{A P}{A B} \\ = \frac{m_{1}}{m_{1} + m_{2}} \end{aligned}

Reading the sides from the coordinates, $A D = x - x_{1}$ , $A C = x_{2} - x_{1}$ , $D P = y - y_{1}$ and $C B = y_{2} - y_{1}$ . The first ratio gives

\begin{array}{c} x - x_{1} \\ = \frac{m_{1} (x_{2} - x_{1})}{m_{1} + m_{2}} \end{array}

Multiply by $m_{1} + m_{2}$ and tidy up: $(m_{1} + m_{2}) x = m_{1} x_{2} - m_{1} x_{1} + (m_{1} + m_{2}) x_{1} = m_{1} x_{2} + m_{2} x_{1}$ . The ratio $\frac{D P}{C B}$ gives the same working with y in place of x. So P is the point

Section formula

\begin{aligned} x & = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \\ y & = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{aligned}

(7.3)

where

$m_{1} : m_{2}$: the ratio $A P : P B$
$(x_{1}, y_{1}), (x_{2}, y_{2})$: the coordinates of A and B

The figure has B above and to the right of A, but the formula holds for any positions: if B is to the left of A, then $x - x_{1}$ and $x_{2} - x_{1}$ are both negative, and their ratio is still $\frac{m_{1}}{m_{1} + m_{2}}$ .

This is the section formula (विभाजन सूत्र). Read it as a recipe: each coordinate of P is a weighted average of the endpoints' coordinates, and each endpoint gets the part of the ratio that sits next to the other endpoint. So $m_{1}$ , the part beside A, multiplies the coordinates of B. That is why P lies closer to B when $m_{1}$ is the larger part.

The derivation also gives a quick mental picture. If $A P : P B = m_{1} : m_{2}$ , cut AB into $m_{1} + m_{2}$ equal steps. Each step changes x by $\frac{x_{2} - x_{1}}{m_{1} + m_{2}}$ and y by $\frac{y_{2} - y_{1}}{m_{1} + m_{2}}$ , and P is reached after $m_{1}$ steps from A. Example 7.6 uses this to check its answer.

Sometimes the ratio is the unknown. Then it is neater to write it as $k : 1$ , where $k = \frac{m_{1}}{m_{2}}$ . Dividing the top and bottom of (7.3) by $m_{2}$ gives the point

Ratio k : 1

\begin{aligned} x & = \frac{k x_{2} + x_{1}}{k + 1} \\ y & = \frac{k y_{2} + y_{1}}{k + 1} \end{aligned}

(7.4)

The midpoint

The midpoint (मध्य-बिंदु) of AB divides it in the ratio $1 : 1$ . Putting $m_{1} = m_{2} = 1$ in the section formula gives

Midpoint formula

\begin{aligned} x & = \frac{x_{1} + x_{2}}{2} \\ y & = \frac{y_{1} + y_{2}}{2} \end{aligned}

(7.5)

Each coordinate of the midpoint is simply the average of the endpoints' coordinates. For example, the midpoint of $(- 3, 8)$ and $(5, 2)$ is $(\frac{2}{2}, \frac{10}{2}) = (1, 5)$ .

InteractiveSection lab

A grid from minus 8 to 8 on both axes with two draggable points A and B that snap to whole numbers; with a keyboard, focus a point and use the arrow keys. Plus and minus buttons set the ratio m1 to m2 from 1 to 9 each. The segment AB is cut into m1 plus m2 equal steps marked by small dots, and P sits m1 steps from A. A panel shows the section formula with the numbers substituted, the exact coordinates of P as fractions, and the lengths AP and PB with their ratio. Buttons set the midpoint (1 to 1) and swap the ratio, and a Hide P switch hides the answer until you press Reveal.

Fig. 7.13Drag A and B, set

m_{1}

and

m_{2}

, and watch P and its coordinates. Turn on Hide P to test yourself first.

Quick check

1 of 3

Which point divides the segment from

A (1, 2)

B (7, 8)

in the ratio

1 : 2

(3, 4)

(5, 6)

(4, 5)

(3, 6)

Show answer

Answer (A) $(3, 4)$

x = \frac{1 (7) + 2 (1)}{3} = 3

and

y = \frac{1 (8) + 2 (2)}{3} = 4

. The point

(5, 6)

comes from multiplying each endpoint by its own part of the ratio; it divides AB in

2 : 1

. The point

(4, 5)

is the midpoint.

Solving problems with the section formula

Example 7.6NCERT Example 6 · p. 108

Dividing in a given ratio

Find the point that divides the segment joining $A (- 2, 7)$ and $B (6, - 5)$ internally in the ratio $3 : 1$ .

Try it yourself first. Then reveal the solution one step at a time.

3/3Three steps from A reach $P (4, - 2)$ .

Fig. 7.14For the ratio

3 : 1

, walk 3 of the 4 equal steps from A.

A grid from minus 3 to 7 across and minus 6 to 8 up with A(minus 2, 7) joined to B(6, minus 5). Step 1 shows the segment and the ratio 3 to 1. Step 2 cuts it into 4 equal steps, drawn as a dashed staircase of 2 right and 3 down each, with dots at the division points. Step 3 highlights the first three steps and rings the point P they reach, (minus 2 + 6, 7 minus 9) = (4, minus 2).

Example 7.7NCERT Example 7 · p. 108

Finding the ratio

In what ratio does the point $P (1, 4)$ divide the segment joining $A (- 3, 8)$ and $B (9, - 4)$ ?

Try it yourself first. Then reveal the solution one step at a time.

You could also start from $m_{1} : m_{2}$ : the x-equation $1 = \frac{9 m_{1} - 3 m_{2}}{m_{1} + m_{2}}$ gives $8 m_{1} = 4 m_{2}$ , so $m_{1} : m_{2} = 1 : 2$ again. And if you are told that the three points lie on one line, comparing the distances AP and PB also gives the ratio.

Example 7.8NCERT Example 8 · p. 109

Points of trisection

Find the points of trisection (समत्रिभाजन बिंदु) of the segment joining $A (- 4, 5)$ and $B (5, - 1)$ , that is, the two points that cut it into three equal parts.

Try it yourself first. Then reveal the solution one step at a time.

Example 7.9NCERT Example 9 · p. 110

Where an axis cuts a segment

In what ratio does the x-axis divide the segment joining $A (5, - 2)$ and $B (- 1, 6)$ ? Find the point where it cuts AB.

Try it yourself first. Then reveal the solution one step at a time.

Fig. 7.15 shows a useful check. The vertical drops from A and B to the x-axis, of lengths 2 and 6, form two similar right triangles, so the x-axis cuts AB in the ratio $2 : 6 = 1 : 3$ . In general, when A and B are on opposite sides of the x-axis, it divides AB in the ratio $∣ y_{1} ∣ : ∣ y_{2} ∣$ . Use this to check your answer, but show the section-formula working in an exam.

3/3The heights 2 and 6 give $A P : P B = 2 : 6 = 1 : 3$ .

Fig. 7.15Heights 2 and 6 on either side of the axis give the ratio

1 : 3

A grid from minus 2 to 6 across and minus 3 to 7 up with A(5, minus 2) below the x-axis and B(minus 1, 6) above it, joined by a segment. Step 2 rings the point P where the segment crosses the x-axis, where y is 0. Step 3 drops a blue vertical of length 2 from A and an orange vertical of length 6 from B to the x-axis, shades the two similar right triangles, and concludes AP to PB equals 2 to 6, which is 1 to 3.

Example 7.10NCERT Example 10 · p. 110

The missing vertex of a parallelogram

Three vertices of a parallelogram ABCD, taken in order, are $A (- 2, - 1)$ , $B (4, 0)$ and $C (5, 4)$ . Find the fourth vertex D.

Try it yourself first. Then reveal the solution one step at a time.

4/4Join the sides: ABCD is a parallelogram.

Fig. 7.16The diagonals share a midpoint, so D is as far beyond M as B is before it.

A grid from minus 3 to 6 across and minus 2 to 5 up with A(minus 2, minus 1), B(4, 0) and C(5, 4) joined by two sides. Step 2 draws the dashed diagonal AC and rings its midpoint M at (1.5, 1.5). Step 3 draws a dotted line from B through M to an equal distance beyond, ending at D(minus 1, 3). Step 4 completes and shades the parallelogram ABCD.

Quick check

1 of 3

In what ratio does the x-axis divide the segment joining

A (1, - 4)

and

B (3, 2)

2 : 1

1 : 2

1 : 1

4 : 3

Show answer

Answer (A) $2 : 1$

With ratio

k : 1

, the y-coordinate

\frac{2 k - 4}{k + 1}

is 0, so

k = 2

: the ratio is

2 : 1

. The heights check agrees:

∣ - 4 ∣ : ∣ 2 ∣ = 2 : 1

. The answer

1 : 2

turns the ratio upside down.

Practice · 10 questions

Practice 7.2: questions with answers

Short answerEasyNCERT Exercise 7.2, Q1
Find the point that divides the segment joining $(- 3, 8)$ and $(7, - 2)$ internally in the ratio $3 : 2$ .
Show solution
Answer $(3, 2)$
$x = \frac{3 (7) + 2 (- 3)}{5} = \frac{15}{5} = 3$ .
$y = \frac{3 (- 2) + 2 (8)}{5} = \frac{10}{5} = 2$ .
Short answerEasyNCERT Exercise 7.2, Q2
Find the points of trisection of the segment joining $(- 5, 2)$ and $(7, - 4)$ .
Show solution
Answer $(- 1, 0)$ and $(3, - 2)$
Ratio $1 : 2$ : $x = \frac{7 + 2 (- 5)}{3} = - 1$ and $y = \frac{- 4 + 2 (2)}{3} = 0$ .
Ratio $2 : 1$ : $x = \frac{2 (7) - 5}{3} = 3$ and $y = \frac{2 (- 4) + 2}{3} = - 2$ .
Check: $(3, - 2)$ is the midpoint of $(- 1, 0)$ and $(7, - 4)$ .
Long answerMediumNCERT Exercise 7.2, Q3
The floor of a school hall is a grid of 1 m tiles. Take the front-left corner as the origin, with x measured across the front wall and y measured down the 60 m side wall. For the annual day, the microphone stand is 3 m across and one-quarter of the way down the hall. The camera stand is 9 m across and 24 m down. How far apart are the two stands? A speaker box must go exactly halfway between them. Where?
Show solution
Answer $3 \sqrt{13}$ m apart; speaker at $(6, 19.5)$
One-quarter of 60 m is 15 m, so the microphone is at $(3, 15)$ ; the camera is at $(9, 24)$ .
Distance $= \sqrt{6^{2} + 9^{2}} = \sqrt{117} = 3 \sqrt{13} \approx 10.8$ m.
Halfway point: $(\frac{3 + 9}{2}, \frac{15 + 24}{2}) = (6, 19.5)$ , that is, 6 m across and 19.5 m down.
Short answerMediumNCERT Exercise 7.2, Q4
In what ratio does the point $(2, - 1)$ divide the segment joining $(- 4, 5)$ and $(6, - 5)$ ?
Show solution
Answer $3 : 2$
$\frac{6 k - 4}{k + 1} = 2$ gives $6 k - 4 = 2 k + 2$ , so $k = \frac{3}{2}$ .
The ratio is $\frac{3}{2} : 1 = 3 : 2$ .
Check with y: $\frac{- 5 k + 5}{k + 1} = \frac{- 2.5}{2.5} = - 1$ .
Short answerMediumNCERT Exercise 7.2, Q5
In what ratio does the y-axis divide the segment joining $A (- 2, 3)$ and $B (6, - 5)$ ? Find the point of division.
Show solution
Answer $1 : 3$ ; $(0, 1)$
With ratio $k : 1$ : $\frac{6 k - 2}{k + 1} = 0$ , so $k = \frac{1}{3}$ and the ratio is $1 : 3$ .
$y = \frac{1 (- 5) + 3 (3)}{4} = \frac{4}{4} = 1$ .
Check with the shortcut: $∣ x_{1} ∣ : ∣ x_{2} ∣ = 2 : 6 = 1 : 3$ .
Short answerMediumNCERT Exercise 7.2, Q6
Joined in this order, $(- 1, 0)$ , $(3, y)$ , $(x, 5)$ and $(1, 4)$ make a parallelogram. Find $x$ and $y$ .
Show solution
Answer $x = 5$ , $y = 1$
Midpoint of the first diagonal: $(\frac{x - 1}{2}, \frac{5}{2})$ .
Midpoint of the second diagonal: $(\frac{4}{2}, \frac{y + 4}{2})$ .
So $x - 1 = 4$ , giving $x = 5$ , and $y + 4 = 5$ , giving $y = 1$ .
Short answerEasyNCERT Exercise 7.2, Q7
AB is a diameter of a circle with centre $(- 1, 2)$ . If B is $(3, - 4)$ , find A.
Show solution
Answer $A (- 5, 8)$
Let $A = (p, q)$ . Then $\frac{p + 3}{2} = - 1$ and $\frac{q - 4}{2} = 2$ .
So $p = - 5$ and $q = 8$ .
Short answerMediumNCERT Exercise 7.2, Q8
A is $(- 3, 4)$ and B is $(7, - 1)$ . Find P on segment AB such that $A P = \frac{2}{5} A B$ .
Show solution
Answer $P (1, 2)$
$A P = \frac{2}{5} A B$ means $P B = \frac{3}{5} A B$ , so $A P : P B = 2 : 3$ .
$x = \frac{2 (7) + 3 (- 3)}{5} = 1$ and $y = \frac{2 (- 1) + 3 (4)}{5} = 2$ .
Short answerMediumNCERT Exercise 7.2, Q9
Find the points that divide the segment joining $A (- 4, 1)$ and $B (4, 7)$ into four equal parts.
Show solution
Answer $(- 2, 2.5)$ , $(0, 4)$ , $(2, 5.5)$
Four equal steps, each of $\frac{8}{4} = 2$ across and $\frac{6}{4} = 1.5$ up.
One step from A: $(- 2, 2.5)$ . Two steps: $(0, 4)$ , the midpoint. Three steps: $(2, 5.5)$ .
Check the last one with the ratio $3 : 1$ : $x = \frac{12 - 4}{4} = 2$ and $y = \frac{21 + 1}{4} = 5.5$ .
Long answerHardNCERT Exercise 7.2, Q10
The vertices of a rhombus, taken in order, are $(5, 3)$ , $(0, 3)$ , $(- 3, - 1)$ and $(2, - 1)$ . Find its area. (For a rhombus, multiply the two diagonals and halve the result.)
Show solution
Answer20 square units
Diagonal from $(5, 3)$ to $(- 3, - 1)$ : $\sqrt{64 + 16} = \sqrt{80} = 4 \sqrt{5}$ .
Diagonal from $(0, 3)$ to $(2, - 1)$ : $\sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$ .
Area $= \frac{1}{2} \times 4 \sqrt{5} \times 2 \sqrt{5} = \frac{1}{2} \times 40 = 20$ square units.

7.4 Summary

One segment can show the whole chapter. In Fig. 7.17, the same two points give a distance, a dividing point and a midpoint.

3/3Two steps of four: the midpoint $M (1, 2)$ .

Fig. 7.17Distance, section point and midpoint from the same two points.

A grid with A(minus 3, minus 1) and B(5, 5) joined by a blue segment, with dashed legs of 8 across and 6 up meeting at a right angle. Step 1 gives AB = square root of 8 squared plus 6 squared = 10. Step 2 marks the three points that cut AB into four equal parts and rings P(minus 1, 0.5), which divides AB in the ratio 1 to 3. Step 3 also rings the midpoint M(1, 2), which divides AB in the ratio 1 to 1.

Recap

Coordinate Geometry: key points and summary

A point on the x-axis has the form $(x, 0)$ ; a point on the y-axis has the form $(0, y)$ (Section 7.1).
Points on one axis are $∣ x_{2} - x_{1} ∣$ or $∣ y_{2} - y_{1} ∣$ apart (Points on an axis).
Distance formula: $P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ . It is the Pythagoras theorem on a right triangle with a horizontal and a vertical leg ((7.1)).
The distance of $(x, y)$ from the origin is $\sqrt{x^{2} + y^{2}}$ ((7.2)).
Three points are collinear when the two shorter distances add up to the longest; otherwise they form a triangle (Using the distance formula).
Sides and diagonals tell squares, rhombuses, rectangles and parallelograms apart (the comparison table).
Points equidistant from A and B satisfy $P A^{2} = P B^{2}$ , a linear equation: the perpendicular bisector of AB (Example 7.4).
Section formula: the point dividing AB internally in the ratio $m_{1} : m_{2}$ has $x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}$ and $y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}$ ((7.3)).
For an unknown ratio, write it as $k : 1$ , find $k$ from one coordinate and check with the other (Example 7.7).
Midpoint: $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ ((7.5)).
Points of trisection divide AB in $1 : 2$ and $2 : 1$ ; an axis cuts AB where that coordinate becomes 0 (Example 7.8, Example 7.9).
The diagonals of a parallelogram share a midpoint, and the centre of a circle is the midpoint of every diameter (Example 7.10).

Recap check

1 of 3

Match each question about segment AB with the tool that answers it.

Tap an item on the left, then its match on the right.

Show answer

Answer

How long is AB? : distance formula
Where is the centre of a circle with diameter AB? : midpoint formula
Where does AB cross the y-axis? : section formula with $k : 1$
Which points are as far from A as from B? : $P A^{2} = P B^{2}$

A length needs the distance formula; a centre is a midpoint; a crossing point needs an unknown ratio; equal distances give the perpendicular bisector.

Flash cards · 7

Key terms

1 / 7

Board exam practice

These questions follow the formats of the CBSE Class 10 board paper: one-mark multiple-choice and assertion–reason items, short and long answers, and a case study. Try each one before you open its solution.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 7.1, Q1	Practice 7.1 · Q1
NCERT Exercise 7.1, Q2	Practice 7.1 · Q2
NCERT Exercise 7.1, Q3	Practice 7.1 · Q3
NCERT Exercise 7.1, Q4	Practice 7.1 · Q4
NCERT Exercise 7.1, Q5	Practice 7.1 · Q5
NCERT Exercise 7.1, Q6	Practice 7.1 · Q6
NCERT Exercise 7.1, Q7	Practice 7.1 · Q7
NCERT Exercise 7.1, Q8	Practice 7.1 · Q8
NCERT Exercise 7.1, Q9	Practice 7.1 · Q9
NCERT Exercise 7.1, Q10	Practice 7.1 · Q10
NCERT Exercise 7.2, Q1	Practice 7.2 · Q1
NCERT Exercise 7.2, Q2	Practice 7.2 · Q2
NCERT Exercise 7.2, Q3	Practice 7.2 · Q3
NCERT Exercise 7.2, Q4	Practice 7.2 · Q4
NCERT Exercise 7.2, Q5	Practice 7.2 · Q5
NCERT Exercise 7.2, Q6	Practice 7.2 · Q6
NCERT Exercise 7.2, Q7	Practice 7.2 · Q7
NCERT Exercise 7.2, Q8	Practice 7.2 · Q8
NCERT Exercise 7.2, Q9	Practice 7.2 · Q9
NCERT Exercise 7.2, Q10	Practice 7.2 · Q10

Quick check: MCQs on Coordinate Geometry

7.1 Introduction3 questions
Any two points3 questions
Using the distance formula4 questions
The midpoint3 questions
Solving problems with the section formula3 questions
Recap check3 questions

Glossary · 11

Key terms (English–Hindi)

abscissaभुजbhuj: The x-coordinate. It tells how far a point is to the right (positive) or left (negative) of the y-axis.First seen in the chapter →

collinearसंरेखीsanrekhi: Lying on one straight line. For three points, the two shorter distances add up to the longest.First seen in the chapter →
coordinate geometryनिर्देशांक ज्यामितिnirdeshank jyamiti: The study of geometry using coordinates, so that points, lines and shapes can be handled with algebra.First seen in the chapter →

distance formulaदूरी सूत्रdoori sutra: How far apart two points are, worked out from their coordinates: $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .First seen in the chapter →

equidistantसमदूरस्थsamdoorasth: At equal distances from two or more given points.First seen in the chapter →

internal divisionअंतः विभाजनantah vibhajan: A point P that lies between A and B, with $A P : P B = m_{1} : m_{2}$ , is said to cut AB internally in that ratio.First seen in the chapter →

midpointमध्य-बिंदुmadhya-bindu: The point halfway along a segment; its coordinates are the averages of the endpoints' coordinates.First seen in the chapter →

ordinateकोटिkoti: The y-coordinate. It tells how far a point is above (positive) or below (negative) the x-axis.First seen in the chapter →

perpendicular bisectorलंब समद्विभाजकlamb samadvibhajak: The line through the midpoint of a segment at right angles to it. Its points are equidistant from the two ends.First seen in the chapter →
points of trisectionसमत्रिभाजन बिंदुsamtribhajan bindu: The two points that cut a segment into three equal parts, dividing it in $1 : 2$ and $2 : 1$ .First seen in the chapter →

section formulaविभाजन सूत्रvibhajan sutra: The coordinates of the point dividing AB internally in the ratio $m_{1} : m_{2}$ : $x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}$ , and similarly for y.First seen in the chapter →

Coordinate Geometry: Class 10 Maths · Chapter 7

What you will learn in Coordinate Geometry

7.1 Introduction

7.2 Distance Formula

Points on an axis

Any two points

Using the distance formula

Practice 7.1: questions with answers

7.3 Section Formula

The general formula

The midpoint

Solving problems with the section formula

Practice 7.2: questions with answers

7.4 Summary

Coordinate Geometry: key points and summary

Board exam practice

Practice questions with answers (NCERT “Board exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Coordinate Geometry

Key terms (English–Hindi)

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