Chapter 7 of 9

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Telangana (TSBIE)Class 10 Mathematics★ The textbook's index allots this chapter 12 periods in November, and it runs from textbook page 163 to page 194. That is four numbered exercises (7.1 with sixteen questions, 7.2 with twelve, 7.3 with five, 7.4 with one eight-part question) plus an Optional Exercise of four, twenty-four worked examples, eight Do This boxes, six Try This boxes, six Think and Discuss boxes and one Activity. No marks distribution is printed anywhere in the volume, so no official weightage can be quoted; the period count is the only number the book itself gives.

Coordinate Geometry

How Pythagoras turns two pairs of coordinates into a distance, how the section formula places a point at a chosen ratio along a segment, why a triangle of zero area means three points are collinear, and what the slope of a line measures. Telangana SCERT Class 10 Mathematics Chapter 7.

⏱ 19 min readMedium difficulty✓ Free · TSBIE syllabus

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By the end of this chapter you'll be able to…

1Find the distance between two points on a line parallel to either axis using the modulus of a coordinate difference
2Derive the distance formula from Pythagoras theorem by dropping perpendiculars to the X-axis
3Apply d = root((x2 - x1)^2 + (y2 - y1)^2) and explain why the order of the two points does not matter
4Test three points for collinearity by checking whether two of the distances sum to the third
5Identify a quadrilateral as a square, rhombus, rectangle or parallelogram from its side and diagonal lengths
6Find a point on an axis, or a relation between x and y, from an equidistance condition
7Apply the section formula to divide a segment internally in a given ratio, crossing m1 with x2 and m2 with x1
8Find the midpoint, the two trisection points, and the points dividing a segment into four equal parts
9Find the centroid of a triangle as the average of its three vertices, and explain the 2 : 1 median property
10Use the section formula backwards to find the ratio in which a given point divides a segment
11Compute the area of a triangle from its vertices, and of a quadrilateral by splitting it into two triangles
12Use zero area as a collinearity test, including solving for an unknown coordinate that makes three points collinear
13Find the slope of a line from two points, and recognise that a horizontal line has slope zero while a vertical line has none

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Why this chapter matters

This is the chapter where geometry stops needing a ruler. Give every point an address and a question about shape becomes a question about arithmetic: whether three friends sit in a line, what kind of quadrilateral four seats make, where to put a relay tower twice as far from one town as the other. The pay-off is not just speed - it is certainty. A drawing can look collinear and not be; an area of exactly zero settles it. The four formulas here also carry forward directly: distance underpins the circle work in chapter 9, the section formula returns whenever something must be divided in a ratio, and slope is the first step towards the idea of a rate of change that calculus is built on.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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The Cartesian plane: axes, quadrants, abscissa, ordinate and plotting points, from class IX

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Pythagoras theorem, which the distance formula is a direct application of

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Properties of quadrilaterals: which have equal sides, which have equal diagonals, which have both

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Area of a triangle as half base times height, and area of a trapezium

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Medians of a triangle and the idea of concurrency, needed for the centroid

Coordinate Geometry

1. What This Chapter Covers

The chapter opens on a chessboard. A knight moves in an L shape, two squares one way and one the other; a bishop moves diagonally as far as the board allows. Put the knight at (0, 0) and its eight possible landing squares acquire coordinates, and the question of how far it has travelled becomes a question about numbers.

That is the move this chapter makes throughout. Geometry is done with arithmetic once every point has an address. Distance, midpoint, area and slope all become formulas in the coordinates.

The chessboard makes a second point without saying so. A square is named by a letter and a number, f6 rather than (6, 6), because the two directions are different kinds of thing. Coordinate geometry keeps that idea and drops the two alphabets, writing both as numbers so that they can be subtracted.

The index allots this chapter 12 periods in November, and it runs from textbook page 163 to page 194, with four numbered exercises and an Optional Exercise.

2. Distance Along an Axis

Start with the easy case. The points (2, 0) and (6, 0) both lie on the X-axis, and the distance between them is plainly 4 units — the difference of the x-coordinates.

But try (−2, 0) and (−6, 0). The difference (−6) − (−2) is −4, and the book is firm about what to do: we never say the distance in negative values, so take the absolute value. The distance is |−4| = 4.

So for two points A(x₁, 0) and B(x₂, 0) on the X-axis, the distance is |x₂ − x₁|, and for (0, y₁) and (0, y₂) on the Y-axis it is |y₂ − y₁|.

Section 7.3 extends this without any new idea. If two points share a y-coordinate they lie on a line parallel to the X-axis, and dropping perpendiculars makes a rectangle whose opposite side lies on the axis. So the distance is still |x₂ − x₁|, and the matching statement holds for lines parallel to the Y-axis.

3. The Distance Formula

Now take A(4, 0) and B(0, 3) with the origin O. Triangle AOB has a right angle at O, its legs are 4 and 3, and Pythagoras gives AB = √(16 + 9) = 5.

Section 7.4 does the same for any two points. Drop perpendiculars from A(x₁, y₁) and B(x₂, y₂) to the X-axis and draw AR across to meet BQ. Then AR = x₂ − x₁ and BR = y₂ − y₁, and triangle ARB is right-angled at R.

So the distance formula is

d = √((x₂ − x₁)² + (y₂ − y₁)²)

and putting A at the origin gives the distance of P(x, y) from O as √(x² + y²).

A Think and Discuss box makes a point worth holding on to: the formula could just as well be written with (x₁ − x₂)² and (y₁ − y₂)². Squaring kills the sign, so the order of the two points does not matter. Another box has Sridhar computing the distance from T(5, 2) to R(−4, −1) as 9.5, then asks for the distance from P(4, 1) to Q(−5, −2) — both are √90, because the differences are the same.

4. What the Formula Is Used For

Once distance is arithmetic, several geometric questions become arithmetic too.

Collinearity. Example-4 takes A(4, 2), B(7, 5) and C(9, 7) and finds AB = 3√2, BC = 2√2 and AC = 5√2. Since AB + BC = AC exactly, the three points lie on one line. Points on the same line are called collinear.

Whether a triangle exists. Example-5 takes (3, 2), (−2, −3) and (2, 3), finds the three lengths as 7.07, 7.21 and 1.41, and observes that the sum of any two exceeds the third, so a triangle is formed. Exercise 7.1 question 13 is the opposite case: (1, 5), (5, 8) and (13, 14) give 5, 10 and 15, and since 5 + 10 = 15 exactly, no triangle can be drawn.

Naming a quadrilateral. Example-6 shows that (1, 7), (4, 2), (−1, −1) and (−4, 4) form a square, by checking that all four sides are √34 and both diagonals are √68. The method generalises: equal sides with equal diagonals give a square, equal sides with unequal diagonals a rhombus, equal opposite sides with equal diagonals a rectangle, and equal opposite sides with unequal diagonals a plain parallelogram.

Finding an unknown point. Example-9 looks for the point on the Y-axis equidistant from A(6, 5) and B(−4, 3). Any such point is (0, y), and setting PA² = PB² gives 61 − 10y = 25 − 6y, so y = 9 and the point is (0, 9). Example-8 does the same in general, and the condition that (x, y) is equidistant from (7, 1) and (3, 5) collapses to the straight line x − y = 2.

That last result is worth pausing on. The set of points equidistant from two fixed points is always a straight line — the perpendicular bisector of the segment joining them — and the algebra shows it without any geometry: squaring both distances cancels the x² and y² terms, and what is left is linear.

The same cancellation is what makes the equidistant questions tractable. Exercise 7.1 asks for a point on the X-axis equidistant from (2, −5) and (−2, 9), and because such a point is (x, 0) there is only one unknown; the answer is (−7, 0). Question 12 uses distance as a radius: the circle centred at (3, 2) through (−5, 6) has radius √80 = 4√5.

5. The Section Formula

Section 7.5 starts with a problem a telephone company might actually have. Town B is 36 km east and 15 km north of town A, and a relay tower must go on the straight road between them so that its distance from B is twice its distance from A.

Being twice as far from B means P divides AB in the ratio 1 : 2. Similar triangles give x/(36 − x) = 1/2 and y/(15 − y) = 1/2, so x = 12 and y = 5.

Running the same similar-triangle argument in general produces the section formula. If P divides the segment from A(x₁, y₁) to B(x₂, y₂) internally in the ratio m₁ : m₂, then

P = ( (m₁x₂ + m₂x₁)/(m₁ + m₂) , (m₁y₂ + m₂y₁)/(m₁ + m₂) )

The pattern to memorise is the cross: m₁ pairs with x₂ and m₂ with x₁, not the other way round. Getting that backwards is the single commonest error in this section, and it produces a point that is a genuine point of the segment — just the wrong one.

Setting m₁ = m₂ = 1 gives the midpoint, ((x₁ + x₂)/2, (y₁ + y₂)/2), which is simply the average of the coordinates.

Section 7.6 applies this to trisection. The two points dividing a segment into three equal parts sit at the ratios 1 : 2 and 2 : 1. For A(2, −2) and B(−7, 4) they work out to P(−1, 0) and Q(−4, 2).

Section 7.7 applies it again. The centroid of a triangle, the point where the three medians meet, divides each median in the ratio 2 : 1 from the vertex. Substituting the midpoint of BC into the section formula collapses everything to

G = ( (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3 )

the plain average of the three vertices.

The formula also runs backwards. Example-14 asks in what ratio (−4, 6) divides the segment from (−6, 10) to (3, −8); setting up the x-equation gives 7m₁ = 2m₂, so the ratio is 2 : 7. Example-15 finds where the Y-axis cuts the segment from (5, −6) to (−1, −4): a point on that axis has abscissa 0, which forces the ratio 5 : 1 and the point (0, −13/3).

Example-16 uses the midpoint as a test. The diagonals of a parallelogram bisect each other, so a quadrilateral is a parallelogram exactly when its two diagonals share a midpoint — no side lengths needed.

6. The Area of a Triangle

Section 7.8 starts with the easy case again. The triangle with vertices O, A(0, 4) and B(6, 0) is right-angled at the origin with base 6 and height 4, so its area is 12.

For a general triangle the book drops perpendiculars AP, BQ and CR to the X-axis, producing three trapezia, and observes that

area ABC = area ABQP + area APRC − area BQRC.

Substituting the side lengths and simplifying leaves a formula that looks nothing like its derivation:

Δ = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

The book adds a Note explaining the modulus signs: as the area cannot be negative, we take the absolute value. The formula's inside can come out negative depending on the order in which the vertices are listed, and that sign is not a mistake — it is discarded.

Example-20 extends the method to a quadrilateral by cutting it into two triangles. For A(−5, 7), B(−4, −5), C(−1, −6) and D(4, 5), the diagonal BD splits it into triangles of area 53 and 19, giving 72 square units in all.

7. Zero Area Means Collinear

Section 7.8.1 draws the consequence that makes the area formula worth having. If three points are collinear they cannot enclose anything, so their triangle has area zero — and conversely, when the area comes out zero the three points must be collinear.

Example-21 verifies that (3, −2), (−2, 8) and (0, 4) are collinear by computing ½|12 − 12| = 0. Example-22 turns it into an equation: for A(1, 2), B(−1, b) and C(−3, −4) to be collinear, the area must vanish, and |4b + 4| = 0 forces b = −1.

This is the cleanest collinearity test in the chapter. The distance method of Example-4 also works, but it needs you to notice which of the three lengths is the longest before you can add the other two — the area method needs no such judgement.

Section 7.8.2 adds Heron's formula for when the side lengths are known but no coordinates are: A = √(s(s − a)(s − b)(s − c)) with s the half-perimeter. For sides 12, 9 and 15 it gives s = 18 and an area of 54.

8. Slope

The last section asks which of two playground slides is faster, and answers with the one that makes the bigger angle with the ground. Slope measures steepness, and it is the ratio of the change in y to the change in x.

An Activity tabulates a line through (0, 0), (1, 2), (2, 4), (3, 6) and (4, 8) and finds that every pair of points gives the same ratio 2. That constancy is what makes slope a property of the line rather than of the pair of points chosen on it.

Section 7.9.2 makes it a formula. For A(x₁, y₁) and B(x₂, y₂),

m = (y₂ − y₁)/(x₂ − x₁) = tan θ

where θ is the angle the line makes with the X-axis. Example-24 runs it backwards: for the line through P(2, 5) and Q(x, 3) to have slope 2, we need −2/(x − 2) = 2, so x = 1.

Two special cases matter. A line parallel to the X-axis has y₂ = y₁, so its slope is zero. A line parallel to the Y-axis has x₂ = x₁, so the denominator vanishes and its slope is undefined — the book asks students to discover this for themselves in a Try This box of three vertical segments.

9. What the Exercises Ask, and Where the Book Goes Wrong

Exercise 7.1 has sixteen questions on distance, Exercise 7.2 twelve on the section formula, Exercise 7.3 five on area, and Exercise 7.4 one eight-part question on slope, with four more in the Optional Exercise.

The spread is worth knowing before revision. Exercise 7.1 runs from plain distances through collinearity, isosceles and equilateral checks, four quadrilateral identifications, two unknown-coordinate questions and a radius; Exercise 7.2 covers ratios in both directions, trisection, division into four equal parts, the endpoints of a diameter, three centroids and two questions where a point is given and an endpoint must be recovered.

Exercise 7.3 is short but does the most per question: three areas, three values of k that force collinearity, a midpoint triangle, a quadrilateral split into triangles, and one area by Heron's formula. Exercise 7.4 is a single eight-part slope drill, including a pair with surd coordinates, a pair with letters, and a horizontal pair whose slope is zero.

The printed answers run from textbook page 381 to page 382. Exercise 7.4 is entirely correct, and so is nearly all of the rest. But four printed answers are wrong, and two of them are wrong in ways worth studying.

Exercise 7.1 question 7 asks for the side of the rhombus on (−4, −7), (−1, 2), (8, 5) and (5, −4). Every side is √90 = 3√10, about 9.49. The key prints it as ∛10, a cube root, which is about 2.15 — the coefficient 3 has been typeset as the radical's index. Its own answer for the area, 72 square units, is right and is impossible with a side of 2.15.

Exercise 7.1 question 8(ii) asks what quadrilateral (−3, 5), (3, 1), (1, −3) and (−5, 1) form, and the key answers Rectangle. Opposite sides are indeed equal, at √52 and √20, so it is a parallelogram. But the diagonals are √80 and 8, which are not equal, and the dot product of two adjacent side vectors is 4, not 0, so there is no right angle. It is a parallelogram, nothing more.

Exercise 7.2 question 12 says P(3, 6) divides AB in the ratio 2 : 3, where A is on the X-axis and B on the Y-axis, and the key answers A(15/2, 0) and B(0, 10). Those points do give P(3, 6) — but at the ratio 3 : 2, not 2 : 3.

With the ratio as the question states it, the answer is A(5, 0) and B(0, 15), which checks: (2×0 + 3×5)/5 = 3 and (2×15 + 3×0)/5 = 6. The key has the two parts of the ratio the wrong way round, which is exactly the slip the cross-pattern in the section formula is there to prevent.

Exercise 7.3 question 3 asks for the area of the midpoint triangle of (0, −1), (2, 1) and (0, 3) and then for the ratio of this area to the area of the given triangle. The key gets the area right at 1 square unit, and then prints the ratio as 4 : 1. The given triangle has area 4, so the ratio asked for is 1 : 4. The key has inverted it.

The key also spells rhombus as "thombus" in question 7.

10. Three Things in the Chapter Text

Two are misprints that make a true statement false as written.

Page 170, in Example-7, concludes that three seated friends are collinear with the line "Since, AB + BC = 3√2 = 2√2 = 5√2 = AC". The plus sign between 3√2 and 2√2 has been set as an equals sign, so the printed line asserts that 3√2 equals 2√2. It should read 3√2 + 2√2 = 5√2, which is what the sentence around it means.

Page 171, at the end of Example-9, states "So (0, 9) is equidistant from (6, 5) and (4, 3)". The point B was given as (−4, 3) and is used as (−4, 3) throughout the working; the minus sign has been dropped in the last line only.

The third is smaller. Page 169 describes a triangle with three unequal sides as a "scelance" triangle, and the Project on page 193 asks for the coordinates of a point dividing a segment "intervally".

One observation rather than an error: the quadrilateral in Example-16, on A(7, 3), B(6, 1), C(8, 2) and D(9, 4), has all four sides equal to √5. It is a rhombus, and the book proves only that it is a parallelogram — true, but less than the figures allow.

11. Summary

Every formula in this chapter comes from one of two ideas. Distance comes from Pythagoras applied to the horizontal and vertical gaps between two points; everything else comes from similar triangles.

The distance between (x₁, y₁) and (x₂, y₂) is √((x₂ − x₁)² + (y₂ − y₁)²), and the order of the points does not matter because the differences are squared.

The section formula places the point dividing a segment in the ratio m₁ : m₂, crossing m₁ with x₂ and m₂ with x₁. Equal ratios give the midpoint as the average of the coordinates; a 2 : 1 division of each median gives the centroid as the average of all three vertices.

The area of a triangle is ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|, taken in absolute value because area cannot be negative. When it comes out zero the three points are collinear, which is the quickest collinearity test the chapter offers.

Slope is the change in y over the change in x, equals tan θ for the angle the line makes with the X-axis, and is the same whichever two points on the line you use. It is zero for a horizontal line and undefined for a vertical one.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Distance on a line parallel to the X-axis

|x2 - x1|

The y-coordinates are equal, so only the x-difference survives. The modulus is there because a distance is never negative.

Distance on a line parallel to the Y-axis

|y2 - y1|

The mirror case. Both follow from the rectangle formed by dropping perpendiculars.

Distance from the origin

OP = root(x^2 + y^2)

The distance formula with one point at (0, 0). A Think and Discuss box asks students to confirm this for themselves.

Distance formula

d = root((x2 - x1)^2 + (y2 - y1)^2)

Order does not matter, because both differences are squared. The book raises this deliberately in a Think and Discuss box.

Section formula (internal division)

P = ((m1 x2 + m2 x1)/(m1 + m2), (m1 y2 + m2 y1)/(m1 + m2))

Note the cross: m1 goes with x2 and m2 with x1. Swapping them gives a real point of the segment, just the wrong one.

Section formula in the form k : 1

P = ((k x2 + x1)/(k + 1), (k y2 + y1)/(k + 1))

Useful when the ratio is unknown, because there is only one variable to solve for.

Midpoint

M = ((x1 + x2)/2, (y1 + y2)/2)

The section formula at the ratio 1 : 1. It is just the average of the coordinates.

Trisection points

the ratios 1 : 2 and 2 : 1

P divides AB as 1 : 2 and Q as 2 : 1, so AP = PQ = QB.

Centroid of a triangle

G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)

The point where the medians meet. It divides each median in the ratio 2 : 1 measured from the vertex.

Area of a triangle from coordinates

Delta = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

The modulus is essential: the bracket can come out negative depending on the order of the vertices, and that sign is discarded.

Collinearity test

three points are collinear exactly when that area is 0

Quicker than comparing distances, because it needs no judgement about which length is the longest.

Heron's formula

A = root(s(s - a)(s - b)(s - c)) with s = (a + b + c)/2

For when the three side lengths are known but the coordinates are not. Sides 12, 9 and 15 give s = 18 and area 54.

Slope of a line

m = (y2 - y1)/(x2 - x1) = tan(theta)

theta is the angle with the X-axis. A horizontal line has slope 0; a vertical line has no slope, because the denominator is zero.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Reporting a negative distance

✓ Take the modulus. The difference (-6) - (-2) = -4, but the distance is |-4| = 4. The book states this rule before it states any formula.

WATCH OUT

✗ Crossing the section formula the wrong way

✓ m1 multiplies x2, and m2 multiplies x1. Writing (m1 x1 + m2 x2)/(m1 + m2) gives the point that divides the segment in the ratio m2 : m1 instead - a real point, so nothing looks wrong until the answer is checked.

WATCH OUT

✗ Forgetting the modulus in the area formula

✓ Listing the vertices in the other order flips the sign of the bracket. Area cannot be negative, so the absolute value is taken, and the book adds a Note saying exactly this.

WATCH OUT

✗ Concluding three points are collinear because two distances nearly add to the third

✓ Use the area formula instead. It gives exactly zero or it does not, with no rounding to judge, and it does not require you to spot which length is the longest first.

WATCH OUT

✗ Calling a quadrilateral a rectangle when only the opposite sides are equal

✓ Equal opposite sides make a parallelogram. A rectangle also needs equal diagonals. The book's own answer key makes this mistake for (-3, 5), (3, 1), (1, -3), (-5, 1), whose diagonals are root80 and 8.

WATCH OUT

✗ Saying a vertical line has slope zero

✓ It is the other way round. A horizontal line has y2 = y1, so the numerator is zero and the slope is 0. A vertical line has x2 = x1, so the denominator is zero and the slope is undefined - not zero.

WATCH OUT

✗ Testing for a parallelogram by measuring all four sides

✓ Checking that the diagonals share a midpoint is faster and uses only the midpoint formula. Example-16 does exactly that.

WATCH OUT

✗ Assuming the centroid is the midpoint of any side or the centre of the triangle's circle

✓ The centroid is the average of the three vertices and lies on every median, two thirds of the way from each vertex. It is not the circumcentre or the incentre.

WATCH OUT

✗ Using the distance formula when the side lengths are already known

✓ Use Heron's formula. Exercise 7.3 question 5 asks for the area of the triangle on (2, 3), (6, 3) and (2, 6) by Heron's, and the sides 3, 4 and 5 give s = 6 and area 6 with no coordinates needed.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Coordinate Geometry?

21 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

21 questions~15 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Distance is never negative, so every coordinate difference is taken in modulus
•On a line parallel to the X-axis the distance is |x2 - x1|; parallel to the Y-axis it is |y2 - y1|
•The distance formula is root((x2 - x1)^2 + (y2 - y1)^2), from Pythagoras on the two coordinate gaps
•The order of the two points does not matter, because both differences are squared
•The distance of P(x, y) from the origin is root(x^2 + y^2)
•Three points are collinear if two of the three distances add exactly to the third
•Equal sides plus equal diagonals give a square; equal sides with unequal diagonals give a rhombus
•Equal opposite sides with equal diagonals give a rectangle; with unequal diagonals, a plain parallelogram
•The set of points equidistant from two fixed points is a straight line - the x^2 and y^2 terms cancel
•The section formula crosses m1 with x2 and m2 with x1; getting it backwards gives the wrong point of the same segment
•The midpoint is the section formula at 1 : 1, which is the average of the coordinates
•The trisection points sit at the ratios 1 : 2 and 2 : 1
•The centroid is the average of all three vertices and divides each median 2 : 1 from the vertex
•Running the section formula backwards finds the ratio: set up the x-equation, solve, then verify with y
•A point on the Y-axis has abscissa 0, which is what fixes the ratio in Example-15
•A quadrilateral is a parallelogram exactly when its two diagonals share a midpoint
•The area of a triangle is 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
•The modulus is essential, because reordering the vertices flips the sign of the bracket
•A quadrilateral's area is found by splitting it along a diagonal into two triangles
•Zero area means collinear, and that is the quickest collinearity test in the chapter
•Setting the area to zero turns a collinearity condition into an equation for an unknown coordinate
•Heron's formula A = root(s(s-a)(s-b)(s-c)) applies when the sides are known but the coordinates are not
•Slope is rise over run, equals (y2 - y1)/(x2 - x1), and equals tan of the angle with the X-axis
•Slope is a property of the line, not of the pair of points chosen on it
•A horizontal line has slope zero; a vertical line has no slope, because the denominator is zero
•The textbook's answer key is wrong in four places for this chapter and must be recomputed

Telangana (TSBIE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: No marks distribution is printed in the textbook for this chapter or anywhere in the volume, so no total is claimed. The index allots 12 periods in November. The categories below are the book's own four numbered exercises plus its Optional Exercise and its in-chapter Do This, Try This, Think and Discuss and Activity boxes; the marks column indicates question size rather than official weightage. Answers for Exercises 7.1 to 7.4 are printed at textbook pages 381 to 382. Exercise 7.4 is entirely correct and most of the rest is too, but FOUR PRINTED ANSWERS ARE WRONG. Exercise 7.1 question 7 gives the rhombus side as a CUBE root of 10, about 2.15, where every side is root90 = 3 root10, about 9.49 - the coefficient 3 has been typeset as the radical's index, and the key's own area of 72 square units is impossible with a side of 2.15. Exercise 7.1 question 8(ii) answers 'Rectangle' for (-3, 5), (3, 1), (1, -3), (-5, 1), whose opposite sides are equal but whose diagonals are root80 and 8 - it is a parallelogram only. Exercise 7.2 question 12 answers A(15/2, 0) and B(0, 10), which correspond to the ratio 3 : 2 while the question states 2 : 3; the correct answer is A(5, 0) and B(0, 15). Exercise 7.3 question 3 prints the ratio as 4 : 1 where the question asks for the ratio of the midpoint triangle (area 1) to the given triangle (area 4), which is 1 : 4. The key also spells rhombus as 'thombus'.

Question type	Marks each	Typical count
Exercise 7.1	44	16
Exercise 7.2	34	12
Exercise 7.3	20	5
Exercise 7.4	16	1
Optional Exercise	16	4
In-chapter boxes	22	20

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Navigation and mapping

Navigation and mapping, where a position is a coordinate pair and a route length is a distance formula

Surveying land

Surveying land, where Heron's formula and the coordinate area formula both give a field's area without measuring a height

Placing infrastructure at a required ratio along a route

Placing infrastructure at a required ratio along a route, exactly as the chapter's relay tower problem does

Computer graphics and game design

Computer graphics and game design, where midpoints, centroids and interpolation along a segment are the section formula

Road and ramp gradients

Road and ramp gradients, where slope is quoted directly as a rise-over-run ratio or a percentage

Structural and mechanical balance

Structural and mechanical balance, where the centroid of a shape is its centre of mass if the density is uniform

Checking whether three measured points lie on a line

Checking whether three measured points lie on a line, which in the laboratory is the question of whether a relationship is linear

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write down which point is (x1, y1) and which is (x2, y2) before substituting; label them on the page

For quadrilateral questions compute all four sides AND both diagonals, then name the figure from both facts together

Use the area formula rather than distances for collinearity - it is shorter and has no judgement step

When a ratio is unknown, take it as k : 1 rather than m1 : m2, so there is only one variable to solve for

After finding a ratio from the x-coordinate, verify it in the y-coordinate; the book does this in Example-14 and it catches sign slips

Remember the modulus in the area formula, and write the modulus bars before you start substituting

For slope questions, state explicitly when a line is horizontal (slope 0) or vertical (no slope) rather than leaving a fraction

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Prove that the midpoint triangle of any triangle has exactly one quarter of its area, whatever the vertices

STRETCH

Derive the section formula for external division, and show how it follows from the internal one with a negative ratio

STRETCH

Show that the centroid, circumcentre and orthocentre of a triangle are always collinear, and find the ratio in which the centroid divides that line

STRETCH

Given three vertices of a parallelogram, show that there are three possible positions for the fourth, and find all of them

STRETCH

Prove the area formula 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| directly by translating one vertex to the origin

STRETCH

Find the condition on a, b, c for the three lines ax + by + c = 0 taken from a family to be concurrent, using the zero-area idea

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

Telangana SSC public examination - Mathematics Paper II, where a distance or section-formula question and an area-or-collinearity question are both standing items

Navodaya and Telangana residential school entrance tests, which favour plotting, midpoints and simple distances

NTSE and state mathematics talent tests, where collinearity conditions and quadrilateral identification appear as quick multiple-choice items

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

No, and the book raises this deliberately. A Think and Discuss box has Ramu writing the formula as root((x1 - x2)^2 + (y1 - y2)^2) and asks why that is the same thing. The reason is that swapping the points changes the sign of each difference, and squaring destroys the sign. So (x2 - x1)^2 and (x1 - x2)^2 are equal for any values at all.

Think of it as a cross. The ratio m1 : m2 is measured from A, so m1 is the part nearest A - and it multiplies the coordinate of the point furthest away, x2. Likewise m2 multiplies x1. A quick check catches it: if m1 is bigger than m2 the point should sit nearer B, so the answer should be closer to B's coordinates. The textbook's own answer key for Exercise 7.2 question 12 has this reversed.

Because the expression inside can be negative. Listing the vertices clockwise rather than anticlockwise flips its sign, and the book adds a Note saying that as the area cannot be negative, the absolute value is taken. The sign is not meaningless - in higher mathematics it records the orientation of the triangle - but at this level it is discarded.

Area, in almost every case. The distance method needs you to work out all three lengths, decide which is the longest, and then check that the other two add to it - and with surds that addition is fiddly. The area formula gives exactly zero or it does not, with no judgement call and no rounding. The distance method has one advantage: it also tells you the order in which the three points lie along the line.

A horizontal line has y2 = y1, so the numerator is 0 and the slope is the number zero. A vertical line has x2 = x1, so the denominator is 0 and the fraction is not a number at all - the line has no slope. The book sets this up carefully, giving three vertical pairs in a Try This box and asking students to justify what happens before telling them. Saying a vertical line has slope zero swaps the two cases exactly.

No. With coordinates, the formula 1/2 |x1(y2 - y3) + ...| is faster and needs no square roots. Heron's formula earns its place when the side lengths are given directly and no coordinates exist - a surveyor's field, say. Exercise 7.3 question 5 asks for a coordinate triangle by Heron's specifically so that you practise the conversion, and the answer, 6 square units, agrees with the coordinate formula.

Mostly, but four of them are wrong. Exercise 7.1 question 7 prints the rhombus side as a cube root of 10 when every side is 3 root10 - and its own area of 72 proves the cube root cannot be right. Exercise 7.1 question 8(ii) calls a parallelogram a rectangle although its diagonals, root80 and 8, are unequal. Exercise 7.2 question 12 answers the ratio 3 : 2 when the question says 2 : 3. And Exercise 7.3 question 3 inverts a ratio, printing 4 : 1 where the question asks for 1 : 4.

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