Coordinate Geometry Maths 9

Class 9 | Maths | Unit 5

Class 9 Unit 5 Maths

5. COORDINATE GEOMETRY

"Divide each difficulty into as many parts as is feasible and necessary to resolve it." — Rene Descartes

The French Mathematician Rene Descartes (pronounced "DAY-CART") developed a new branch of Mathematics known as Analytical Geometry or Coordinate Geometry which combined all arithmetic, algebra and geometry of the past ages in a single technique of visualising as points on a graph and equations as geometrical shapes. The fixing of a point position in the plane by assigning two numbers, coordinates, giving its distance from two lines perpendicular to each other, was entirely Descartes' invention.

Rene Descartes (AD/CE 1596–1650)

The French Mathematician Rene Descartes developed Analytical Geometry or Coordinate Geometry, combining arithmetic, algebra and geometry into a single technique.

Learning Outcomes

• To understand the Cartesian coordinate system.
• To identify the abscissa, ordinate and coordinates of any given point.
• To find the distance between any two points in the Cartesian plane using formula.
• To understand the mid-point formula and use it in problem solving.
• To derive the section formula and apply this in problem solving.
• To understand the centroid formula and to know its applications.

5.1 Mapping the Plane

How do you write your address? Here is one:

Sarakkalvilai Primary School
135, Sarakkalvilai,
Sarakkalvilai Housing Board Road,
Keezha Sarakkalvilai,
Nagercoil 629002,
Kanyakumari Dist.
Tamil Nadu, India.

Somehow, this information is enough for anyone in the world from anywhere to locate the school one studied. Just consider there are crores and crores of buildings on the Earth. But yet, we can use an address system to locate a particular person's place of study, however interior it is. How is this possible?

Let us work out the procedure of locating a particular address. We know the World is divided into countries. One among them is India. Subsequently India is divided into States. Among these States we can locate our State Tamilnadu. Further going deeper, we find our State is divided into Districts. Districts into taluks, taluks into villages — proceeding further in this way, one could easily locate "Sarakkalvilai" among the villages in that Taluk. Further among the roads in that village, 'Housing Board Road' is the specific road which we are interested to explore. Finally, we end up the search by locating Primary School building bearing the door number 135 to enable us precisely among the buildings in that road.

In New York city of USA, there is an area called Manhattan. The map shows Avenues run in the North–South direction and the Streets run in the East–West direction. So, if you know that the place you are looking for is on 57th street between 9th and 10th Avenues, you can find it immediately on the map. Similarly, it is easy to find a place on 2nd Avenue between 34th and 35th streets. In fact, New Yorkers make it even simpler. From the door number on a street, you can actually calculate which avenues it lies between, and from the door number on an avenue, you can calculate which streets it stands.

All maps do just this for us. They help us in finding our way and locate a place easily by using information of any landmark which is nearer to our search to make us understand whether we are near or far, how far are we, or what is in between etc. We use latitudes (east–west, like streets in Manhattan) and longitudes (north–south, like avenues in Manhattan) to pinpoint places on Earth.

It is interesting to see how using numbers in maps helps us so much. This idea, of using numbers to map places, comes from geometry. Mathematicians wanted to build maps of planes, solids and shapes of all kinds. When we work with a geometric figure, we want to observe whether a point lies inside the region or outside or on the boundary. Given two points on the boundary and a point outside, we would like to examine which of the two points on the boundary is closer to the one outside, and how much closer and so on.

Mathematicians asked such questions and answered them only to develop their own understanding of circles, polygons and spheres. But the mathematical tools and techniques were used to find immense applications in day-to-day life. Mapping the world using latitudes and longitudes would not have been developed at all in 18th century, if the co-ordinate system had not been developed mathematically in the 17th century.

You already know the map of the real number system is the number line. It extends infinitely on both directions. In between any two points, on a number line, there lies infinite number of points. We are now going to build a map of the plane so that we can discuss about the points on the plane, of the distance between the points etc.

Arithmetic introduced us to the world of numbers and operations on them. Algebra taught us how to work with unknown values and find them using equations. Geometry taught us to describe shapes by their properties. Co-ordinate geometry will teach us how to use numbers and algebraic equations for studying geometry and beautiful integration of many techniques in one place.

5.2 Devising a Coordinate System

You ask your friend to draw a rectangle on a blank sheet of paper, 5 cm by 3 cm. He says, "Sure, but where on this sheet?" How would you answer him?

Just like that a particular house is to be pointed out, it is going to be a difficult task. Instead, if any place or an object is fixed for identification then it is easy to identify any other place or object relative to it. For example, you fix the flag and talk about the house to the left of it, the hotel below it, the antenna on the house to right of it etc.

Draw two perpendicular lines in such a way that the flag is pointed out at near the intersection point. Now if you tell your friend the total length and width of the picture frame, keeping the flagpole as landmark, you can also say 2 cm to the right, 3 cm above etc. Since you know directions, you can also say 2 cm east, 3 cm north.

A number line is usually represented as a horizontal line on which the positive numbers always lie on the right side of zero, negative on the left side of zero. Now consider another copy of the number line but drawn vertically: the positive integers are represented above zero and the negative integers are below zero. Where do the two number lines meet? Obviously at zero for both lines. That will be our location where the "flag" is fixed. We can talk of other numbers relative to it, on both the lines.

Suppose we go 2 to the right and then 3 to the top. We would call this place (2, 3) — understood as 2 to the right and then 3 up. Notice that we would reach the same place if we first went 3 up and then 2 right, so for us, the instruction (2,3) is not the same as the instruction (3,2).

What about (−2, 3)? It would mean 2 left and 3 up. From where? Always from (0, 0). What about the instruction (2, −3)? It would mean 2 right and 3 down.

We call the horizontal number line the x-axis and the vertical number line the y-axis. The x-coordinate is called the abscissa and the y-coordinate is called the ordinate. We call the meeting point of the axes (0, 0) the origin.

Note

• Whether we place (0, 0) at the centre of the sheet, or somewhere else does not matter, (0, 0) is always the origin for us, and all "instructions" are relative to that point.
• We usually denote the origin by the letter 'O'.
• For any point P on the x-axis, the value of y coordinate (ordinate) is zero: P(x, 0).
• For any point Q on the y-axis, the value of x coordinate (abscissa) is zero: Q(0, y).
• (x, y) ≠ (y, x) unless x = y.
• A plane with the rectangular coordinate system is called the Cartesian plane.

Since the x-axis and the y-axis divide the plane into four regions, we call them quadrants. They are numbered as I, II, III and IV, with I for upper east side, II for upper west side, III for lower west side and IV for lower east side, making an anti-clockwise tour of them all.

Quadrants Table

Region	Quadrant	Nature of x, y	Signs of Coordinates
XOY	I	x > 0, y > 0	(+, +)
X'OY	II	x < 0, y > 0	(−, +)
X'OY'	III	x < 0, y < 0	(−, −)
XOY'	IV	x > 0, y < 0	(+, −)

5.2.1 Plotting Points in Cartesian Coordinate Plane

To plot the point (4, 5) in the Cartesian coordinate plane: We follow the x-axis until we reach 4 and draw a vertical line at x = 4. Similarly, we follow the y-axis until we reach 5 and draw a horizontal line at y = 5. The intersection of these two lines is the position of (4, 5) in the Cartesian plane. This point is at a distance of 4 units from the y-axis and 5 units from the x-axis.

Example 5.1: In which quadrant do the following points lie?

(a) (3, −8)    (b) (−1, −3)    (c) (2, 5)    (d) (−7, 3)

Solution:

(a) The x-coordinate is positive and y-coordinate is negative. So, point (3, −8) lies in the IV quadrant.

(b) The x-coordinate is negative and y-coordinate is negative. So, point (−1, −3) lies in the III quadrant.

(c) The x-coordinate is positive and y-coordinate is positive. So, point (2, 5) lies in the I quadrant.

(d) The x-coordinate is negative and y-coordinate is positive. So, point (−7, 3) lies in the II quadrant.

Example 5.2: Plot the points A(2, 4), B(−3, 5), C(−4, −5), D(4, −2) in the Cartesian plane.

Solution:

(i) To plot (2, 4): draw a vertical line at x = 2 and a horizontal line at y = 4. Point A(2, 4) is located in the I quadrant.

(ii) To plot (−3, 5): draw a vertical line at x = −3 and a horizontal line at y = 5. Point B(−3, 5) is located in the II quadrant.

(iii) To plot (−4, −5): draw a vertical line at x = −4 and a horizontal line at y = −5. Point C(−4, −5) is located in the III quadrant.

(iv) To plot (4, −2): draw a vertical line at x = 4 and a horizontal line at y = −2. Point D(4, −2) is located in the IV quadrant.

Example 5.3: Plot the following points A(2, 2), B(−2, 2), C(−2, −1), D(2, −1) in the Cartesian plane. Discuss the type of diagram by joining all the points taken in order.

Solution:

Point	A	B	C	D
Quadrant	I	II	III	IV

ABCD is a rectangle. Can you find the length, breadth and area of the rectangle?

Exercise 5.1

1. Plot the following points in the coordinate system and identify the quadrants: P(−7, 6), Q(7, −2), R(−6, −7), S(3, 5) and T(3, 9).

2. Write down the abscissa and ordinate of the following from the figure:

(i) P

(ii) Q

(iii) R

(iv) S

3. Plot the following points in the coordinate plane and join them. What is your conclusion about the resulting figure?

(i) (−5, 3), (−1, 3), (0, 3), (5, 3)

(ii) (0, −4), (0, −2), (0, 4), (0, 5)

4. Plot the following points in the coordinate plane. Join them in order. What type of geometrical shape is formed?

(i) (0, 0), (−4, 0), (−4, −4), (0, −4)

(ii) (−3, 3), (2, 3), (−6, −1), (5, −1)

Answers — Exercise 5.1

1. P(−7, 6) = II Quadrant; Q(7, −2) = IV Quadrant; R(−6, −7) = III Quadrant; S(3, 5) = I Quadrant; T(3, 9) = I Quadrant

2. (i) P = (−4, 4)   (ii) Q = (3, 3)   (iii) R = (4, −2)   (iv) S = (−5, −3)

3. (i) Straight line parallel to x-axis   (ii) Straight line which lies on y-axis

4. (i) Square   (ii) Trapezium

Activity 1

Plot the following points on a graph sheet by taking the scale as 1 cm = 1 unit. Find how far the points are from each other? A(1, 0) and D(4, 0). Find AD and also DA. Is AD = DA? Plot another set of points and verify your result.

5.3 Distance between any Two Points

Akila and Shanmugam are friends living on the same street in Sathyamangalam. Shanmugam's house is at the intersection of one street with another street on which there is a library. They both study in the same school, and that is not far from Shanmugam's house.

Consider the school as the origin. Now fix the scale as 1 unit = 50 metres.

1. How far is Akila's house from Shanmugam's house?

2. How far is the library from Shanmugam's house?

3. How far is the school from Shanmugam's and Akila's house?

4. How far is the library from Akila's house?

5. How far is Shanmugam's house from Akila's house?

Note

The distance from point A to B is the same as the distance from point B to A: distance(A, B) = distance(B, A). This is true for all points A and B on the plane.

When there are many routes between A and B, we will use distance(A, B) to denote the distance on the shortest route — the "straight line distance", usually called "as the crow flies".

5.3.1 Distance Between Two Points on the Coordinate Axes

Points on the x-axis

If two points lie on the x-axis, then the distance between them is equal to the difference between the x-coordinates.

Consider two points A(x₁, 0) and B(x₂, 0) on the x-axis.

AB = |x₂ − x₁|

Points on the y-axis

If two points lie on the y-axis, then the distance between them is equal to the difference between the y-coordinates.

Consider two points P(0, y₁) and Q(0, y₂).

PQ = |y₂ − y₁|

5.3.2 Distance Between Two Points Lying on a Line Parallel to Coordinate Axes

Consider the points A(x₁, y₁) and B(x₂, y₁). Since the y-coordinates are equal the points lie on a line parallel to x-axis.

Distance AB = |x₂ − x₁| [The difference between x-coordinates]

Similarly, for a line joining A(x₁, y₁) and B(x₁, y₂), parallel to y-axis, the distance = |y₂ − y₁| [The difference between y-coordinates]

5.3.3 Distance Between Two Points on a Plane

Distance Formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the Cartesian plane at a distance 'd' apart.

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

The distance of a point P(x₁, y₁) from the origin O(0, 0) is:

OP = √(x₁² + y₁²)

Note: PQ = QP, i.e., √[(x₂−x₁)²+(y₂−y₁)²] = √[(x₁−x₂)²+(y₁−y₂)²]

Thinking Corner

A man goes 3 km towards north and then 4 km towards east. How far is he away from the initial position?

5.3.4 Properties of Distances

• distance(A, B) = 0 exactly when A and B denote the identical point: A = B.
• distance(A, B) > 0 for any two distinct points A and B.
• When distance(A, B) + distance(B, C) = distance(A, C), the points A, B and C must be collinear.
• When AB² + BC² = AC² (with appropriate naming of vertices), A, B and C are the vertices of a right-angled triangle.

Example 5.4: Find the distance between the points (−4, 3) and (2, −3).

Solution:

d = √[(x₂−x₁)² + (y₂−y₁)²]

= √[(2−(−4))² + (−3−3)²]

= √[(6)² + (−6)²]

= √[36 + 36]

= √[36 × 2]

= 6√2

Example 5.5: Show that the points A(3, 1), B(6, 4) and C(8, 6) lie on a straight line.

Solution:

AB = √[(6−3)² + (4−1)²] = √[9 + 9] = √18 = 3√2

BC = √[(8−6)² + (6−4)²] = √[4 + 4] = √8 = 2√2

AC = √[(8−3)² + (6−1)²] = √[25 + 25] = √50 = 5√2

AB + BC = 3√2 + 2√2 = 5√2 = AC

Therefore, the points lie on a straight line.

Collinear Points

To show the collinearity of three points, we prove that the sum of the distance between two pairs of points is equal to the third pair of points. In other words, points A, B, C are collinear if AB + BC = AC.

Example 5.6: Show that the points A(7, 10), B(−2, 5), C(3, −4) are the vertices of a right-angled triangle.

Solution:

A = (7, 10), B = (−2, 5), C = (3, −4)

AB = √[(−2−7)² + (5−10)²] = √[(−9)² + (−5)²] = √[81 + 25] = √106   → AB² = 106 ...(1)

BC = √[(3−(−2))² + (−4−5)²] = √[(5)² + (−9)²] = √[25 + 81] = √106   → BC² = 106 ...(2)

AC = √[(3−7)² + (−4−10)²] = √[(−4)² + (−14)²] = √[16 + 196] = √212   → AC² = 212 ...(3)

From (1), (2) and (3): AB² + BC² = 106 + 106 = 212 = AC²

Since AB² + BC² = AC², △ABC is a right-angled triangle, right angled at B.

Right Angled Triangle

We know that the sum of the squares of two sides is equal to the square of the third side, which is the hypotenuse of a right-angled triangle.

Example 5.7: Show that the points A(−4, −3), B(3, 1), C(3, 6), D(−4, 2) taken in that order form the vertices of a parallelogram.

Solution:

Let A(−4, −3), B(3, 1), C(3, 6), D(−4, 2) be the four vertices of quadrilateral ABCD.

AB = √[(3+4)² + (1+3)²] = √[49 + 16] = √65

BC = √[(3−3)² + (6−1)²] = √[0 + 25] = √25 = 5

CD = √[(−4−3)² + (2−6)²] = √[(−7)² + (−4)²] = √[49 + 16] = √65

AD = √[(−4+4)² + (2+3)²] = √[0 + 25] = √25 = 5

AB = CD = √65 and BC = AD = 5

Here, the opposite sides are equal. Hence ABCD is a parallelogram.

Parallelogram

We know that opposite sides are equal in a parallelogram.

Example 5.8: Calculate the distance between the points A(7, 3) and B which lies on the x-axis whose abscissa is 11.

Solution:

Since B is on the x-axis, the y-coordinate of B is 0. So, the coordinates of B are (11, 0).

AB = √[(11−7)² + (0−3)²] = √[(4)² + (−3)²] = √[16 + 9] = √25 = 5

Example 5.9: Find the value of 'a' such that PQ = QR where P, Q, and R are the points whose coordinates are (6, −1), (1, 3) and (a, 8) respectively.

Solution:

PQ = √[(1−6)² + (3+1)²] = √[(−5)² + (4)²] = √[25 + 16] = √41

QR = √[(a−1)² + (8−3)²] = √[(a−1)² + 25]

Given PQ = QR, so √41 = √[(a−1)² + 25]

Squaring both sides: (a−1)² + 25 = 41

(a−1)² = 16

(a−1) = ±4

a = 1 + 4 = 5 or a = 1 − 4 = −3

Therefore, a = 5 or a = −3

Example 5.10: Let A(2, 2), B(8, −4) be two given points in a plane. If a point P lies on the X-axis (in positive side), and divides AB in the ratio 1:2, then find the coordinates of P.

Solution:

Let P = (x, 0) [P lies on x-axis]

AP = √[(x−2)² + (0−2)²] = √[x²−4x+4+4] = √[x²−4x+8]

BP = √[(x−8)² + (0+4)²] = √[x²−16x+64+16] = √[x²−16x+80]

Given AP : PB = 1 : 2, so 2AP = BP

Squaring: 4AP² = BP²

4(x²−4x+8) = x²−16x+80

4x²−16x+32 = x²−16x+80

3x²−48 = 0 → 3x² = 48 → x² = 16 → x = ±4

As P lies on x-axis (positive side), x ≠ −4. Hence the coordinates of P are (4, 0).

Example 5.11: Show that (4, 3) is the centre of the circle passing through the points (9, 3), (7, −1), (−1, 3). Also find its radius.

Solution:

Let P(4, 3), A(9, 3), B(7, −1) and C(−1, 3).

PA = √[(4−9)² + (3−3)²] = √[(−5)² + 0] = √25 = 5

PB = √[(4−7)² + (3+1)²] = √[(−3)² + (4)²] = √[9+16] = √25 = 5

PC = √[(4+1)² + (3−3)²] = √[(5)² + 0] = √25 = 5

PA = PB = PC = 5. Therefore, P is the centre of the circle passing through A, B and C. Radius = 5.

Exercise 5.2

1. Find the distance between the following pairs of points.

(i) (1, 2) and (4, 3)

(ii) (3, 4) and (−7, 2)

(iii) (a, b) and (c, b)

(iv) (3, −9) and (−2, 3)

2. Determine whether the given set of points in each case are collinear or not.

(i) (7, −2), (5, 1), (3, 4)

(ii) (a, −2), (a, 3), (a, 0)

3. Show that the following points taken in order form an isosceles triangle.

(i) A(5, 4), B(2, 0), C(−2, 3)

(ii) A(6, −4), B(−2, −4), C(2, 10)

4. Show that the following points taken in order form an equilateral triangle in each case.

(i) A(2, 2), B(−2, −2)

(ii) B(0, 1), C(0, 3)

5. Show that the following points taken in order form the vertices of a parallelogram.

(i) A(−3, 1), B(−6, −7), C(3, −9), D(6, −1)

(ii) A(−7, −3), B(5, 10), C(15, 8), D(3, −5)

6. Verify that the following points taken in order form the vertices of a rhombus.

(i) A(3, −2), B(7, 6), C(−1, 2), D(−5, −6)

(ii) A(1, 1), B(2, 1), C(2, 2), D(1, 2)

7. A(−1, 1), B(1, 3) and C(3, a) are points and if AB = BC, then find 'a'.

8. The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units. What are the coordinates of A?

9. The point (x, y) is equidistant from the points (3, 4) and (−5, 6). Find a relation between x and y.

10. Let A(2, 3) and B(2, −4) be two points. If P lies on the x-axis, such that AP = (3/7)AB, find the coordinates of P.

11. Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, −4) and (5, −6).

12. The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.

Answers — Exercise 5.2

1. (i) √10 units   (ii) 2√26 units   (iii) c − a   (iv) 13 units

2. (i) Collinear   (ii) Collinear

7. 5 or 1

8. Coordinates of A: (9, 9) or (−5, −5)

9. y = 4x + 9

10. Coordinates of P: (2, 0)

12. 30√2

5.4 The Mid-point of a Line Segment

Imagine a person riding his two-wheeler on a straight road towards East from his college to village A and then to village B. At some point in between A and B, he suddenly realises that there is not enough petrol for the journey. On the way there is no petrol bunk in between these two places. Should he travel back to A or just try his luck moving towards B? Which would be the shorter distance? He has to know whether he crossed the halfway mid-point or not.

Imagine college as origin O from which the distances of village A and village B are respectively x₁ and x₂. Let M be the mid-point of AB, then x can be obtained as follows:

AM = MB and so x − x₁ = x₂ − x, which simplifies to x = (x₁ + x₂) / 2

Mid-point Formula

The mid-point M of the line segment joining the points A(x₁, y₁) and B(x₂, y₂) is:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

• x-coordinate of M = average of x-coordinates of A and B = (x₁ + x₂)/2
• y-coordinate of M = average of y-coordinates of A and B = (y₁ + y₂)/2

Example: The mid-point of the line segment joining the points (−8, 10) and (4, −2) is given by ((−8+4)/2, (10+(−2))/2) = (−4/2, 8/2) = (−2, 4).

Real-life Application — Longitude and Latitude

Name of the city	Longitude	Latitude
Chennai (Besant Nagar)	80.27° E	13.00° N
Mangaluru (Kuthethoor)	74.85° E	13.00° N
Bengaluru (Rajaji Nagar)	77.56° E	13.00° N

Taking the longitude and latitude of Chennai (80.27° E, 13.00° N) and Mangaluru (74.85° E, 13.00° N), the average of the coordinates gives ((80.27 + 74.85)/2, (13.00 + 13.00)/2) = (77.56° E, 13.00° N), which is the longitude and latitude of Bengaluru.

Example 5.12: The point (3, −4) is the centre of a circle. If AB is a diameter of the circle and B is (5, −6), find the coordinates of A.

Solution:

Let the coordinates of A be (x₁, y₁) and the given point is B(5, −6). Since the centre is the mid-point of the diameter AB:

(x₁ + 5)/2 = 3 → x₁ + 5 = 6 → x₁ = 1

(y₁ + (−6))/2 = −4 → y₁ − 6 = −8 → y₁ = −2

Therefore, the coordinates of A are (1, −2).

Progress Check

(i) Let X be the mid-point of the line segment joining Q(8, 2) and (3, x), and Y be the mid-point of the line segment joining A(8, −5) and B(−2, 11). Find the mid-point of the line segment XY.

(ii) If M is the mid-point of the line segment joining the points, find the value of 'x'.

Example 5.13: If (x, 3), (6, y), (8, 2) and (9, 4) are the vertices of a parallelogram taken in order, then find the value of x and y.

Solution:

Let A(x, 3), B(6, y), C(8, 2) and D(9, 4) be the vertices of parallelogram ABCD.

By definition, diagonals AC and BD bisect each other: Mid-point of AC = Mid-point of BD

((x+8)/2, (3+2)/2) = ((6+9)/2, (y+4)/2)

Equating x-coordinates: (x+8)/2 = 15/2 → x + 8 = 15 → x = 7

Equating y-coordinates: 5/2 = (y+4)/2 → 5 = y + 4 → y = 1

Thinking Corner

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD taken in order. Find the fourth vertex D. If (x₁, y₁), (x₂, y₂), (x₃, y₃) and (x₄, y₄) are the four vertices of the parallelogram, then using the given points, find the value of (x₁ + x₃ − x₂, y₁ + y₃ − y₂) and state the reason for your result.

Example: Find the points which divide the line segment joining A(−11, 4) and B(9, 8) into four equal parts.

Solution:

Let P, Q, R be the points on the line segment joining A(−11, 4) and B(9, 8) such that AP = PQ = QR = RB.

Q is the mid-point of AB = ((−11+9)/2, (4+8)/2) = (−2/2, 12/2) = (−1, 6)

P is the mid-point of AQ = ((−11+(−1))/2, (4+6)/2) = (−12/2, 10/2) = (−6, 5)

R is the mid-point of QB = ((−1+9)/2, (6+8)/2) = (8/2, 14/2) = (4, 7)

Hence the points which divide AB into four equal parts are P(−6, 5), Q(−1, 6) and R(4, 7).

Example 5.15: The mid-points of the sides of a triangle are (5, 1), (3, −5) and (−5, −1). Find the coordinates of the vertices of the triangle.

Solution:

Let the vertices of △ABC be A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). The given mid-points of sides AB, BC and CA are (5, 1), (3, −5) and (−5, −1) respectively.

x₁ + x₂ = 10 ...(1)  |  x₂ + x₃ = 6 ...(2)  |  x₃ + x₁ = −10 ...(3)

Adding (1), (2) and (3): 2x₁ + 2x₂ + 2x₃ = 6 → x₁ + x₂ + x₃ = 3 ...(4)

(4)−(2): x₁ = 3 − 6 = −3  |  (4)−(3): x₂ = 3 + 10 = 13  |  (4)−(1): x₃ = 3 − 10 = −7

y₁ + y₂ = 2 ...(5)  |  y₂ + y₃ = −10 ...(6)  |  y₃ + y₁ = −2 ...(7)

Adding (5), (6) and (7): 2y₁ + 2y₂ + 2y₃ = −10 → y₁ + y₂ + y₃ = −5 ...(8)

(8)−(6): y₁ = −5 + 10 = 5  |  (8)−(7): y₂ = −5 + 2 = −3  |  (8)−(5): y₃ = −5 − 2 = −7

Therefore, the vertices of the triangle are A(−3, 5), B(13, −3) and C(−7, 7).

Thinking Corner

If (a₁, b₁), (a₂, b₂) and (a₃, b₃) are the mid-points of the sides of a triangle, using the mid-points given in example 5.15, find the value of (a₁+a₃−a₂, b₁+b₃−b₂), (a₁+a₂−a₃, b₁+b₂−b₃) and (a₂+a₃−a₁, b₂+b₃−b₁). Compare the results. What do you observe? Give reason for your result.

Exercise 5.3

1. Find the mid-points of the line segment joining the points:

(i) (−2, 3) and (−6, −5)

(ii) (8, −2) and (−8, 0)

(iii) (a, b) and (a+2b, 2a−b)

(iv) (1/2, 3/7) and (3/2, −11/7)

2. The centre of a circle is (−4, 2). If one end of the diameter of the circle is (−3, 7), then find the other end.

3. If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on y = 4x + 9, then what will be the value of p?

4. The mid-points of the sides of a triangle are (2, 4), (−2, 3) and (5, 2). Find the coordinates of the vertices of the triangle.

5. O(0, 0) is the centre of a circle whose one chord is AB, where the points A and B are (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.

6. The points A(−5, 4), B(−1, −2) and C(5, 2) are the vertices of an isosceles right-angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.

7. The points A(−3, 6), B(0, 7) and C(1, 9) are the mid-points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallelogram.

8. A(−3, 2), B(3, 2) and C(−3, −2) are the vertices of the right triangle, right angled at A. Show that the mid-point of the hypotenuse is equidistant from the vertices.

Answers — Exercise 5.3

1. (i) (−4, −1)   (ii) (0, −1)   (iii) (a+b, a)   (iv) (1, −1)

2. (−5, −3)

3. p = −15

4. (9, 3), (−5, 5) and (1, 1)

5. (9/2, 3/2)

6. (1, 8)

5.5 Points of Trisection of a Line Segment

The mid-point of a line segment is the point of bisection, which means dividing into two parts of equal length. Suppose we want to divide a line segment into three parts of equal length, we have to locate points suitably to effect a trisection of the segment. For a given line segment, there are two points of trisection. The method of obtaining this is similar to that of what we did in the case of locating the point of bisection (i.e., the mid-point).

Here P and Q are the points of trisection of the line segment AB where A is (x₁, y₁) and B is (x₂, y₂). Clearly we know that P is the mid-point of AQ and Q is the mid-point of PB.

Note that when we divide the segment into 3 equal parts, we are also dividing the horizontal and vertical legs into three equal parts.

Points of Trisection

If P is (a, b): a = (x₂ + 2x₁)/3; b = (y₂ + 2y₁)/3

Thus we get the point P as ((x₂+2x₁)/3, (y₂+2y₁)/3)

The required point Q is ((2x₂+x₁)/3, (2y₂+y₁)/3)

Progress Check

(i) Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).

(ii) Find the coordinates of points of trisection of the line segment joining the point (6, −9) and the origin.

5.6 Section Formula

We studied bisection and trisection of a given line segment. These are only particular cases of the general problem of dividing a line segment joining two points in the ratio m : n.

Given a segment AB and a positive real number r. We wish to find the coordinate of point P which divides AB in the ratio r : 1. This means AP/PB = r/1 or AP = r(PB).

This means that x − x₁ = r(x₂ − x). Solving this: x = (rx₂ + x₁)/(r+1)

Section Formula

If P divides A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n, then:

P = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))

Note

• The line joining the points (x₁, y₁) and (x₂, y₂) is divided by x-axis in the ratio −y₁ : y₂ and by y-axis in the ratio −x₁ : x₂.
• If three points are collinear, then one of the points divides the line segment joining the other two points in the ratio r : 1.
• The section formula can be used only when the given three points are collinear.
• This formula is helpful to find the centroid, incentre and excentres of a triangle.

Thinking Corner

(i) What happens when m = n = 1? Can you identify it with a result already proved?

(ii) AP : PB = 1 : 2 and AQ : QB = 2 : 1. What is AP : AB? What is AQ : AB?

Example 5.17: Find the coordinates of the point which divides the line segment joining the points (3, 5) and (8, −10) internally in the ratio 3:2.

Solution:

Let A(3, 5), B(8, −10) and P(x, y) divides AB internally in the ratio 3:2.

x₁ = 3, y₁ = 5, x₂ = 8, y₂ = −10, m = 3, n = 2

P(x, y) = ((3×8 + 2×3)/(3+2), (3×(−10) + 2×5)/(3+2))

= ((24+6)/5, (−30+10)/5)

= (30/5, −20/5)

= (6, −4)

Example 5.18: In what ratio does the point P(−2, 4) divide the line segment joining the points A(−3, 6) and B(1, −2) internally?

Solution:

Given points A(−3, 6) and B(1, −2). P(−2, 4) divides AB internally in the ratio m : n.

By section formula, equating x-coordinates:

(m×1 + n×(−3))/(m+n) = −2

m − 3n = −2m − 2n

3m = n → m/n = 1/3 → m : n = 1 : 3

Hence P divides AB internally in the ratio 1:3.

Example 5.19: Let A(−3, 5) and B(x₂, y₂). Given P(−2, 3) divides AB internally in the ratio 1:6. Find the coordinates of B.

Solution:

By section formula: P((1×x₂ + 6×(−3))/(1+6), (1×y₂ + 6×5)/(1+6)) = (−2, 3)

Equating x-coordinates: (x₂ − 18)/7 = −2 → x₂ − 18 = −14 → x₂ = 4

Equating y-coordinates: (y₂ + 30)/7 = 3 → y₂ + 30 = 21 → y₂ = −9

Therefore, the coordinate of B is (4, −9).

Exercise 5.4

1. Find the coordinates of the point which divides the line segment joining the point A(4, −3) and B(9, 7) in the ratio 3:2.

2. In what ratio does the point P(2, −5) divide the line segment joining A(−3, 5) and B(4, −9)?

3. Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = (2/5)AB.

4. Find the coordinates of the points of trisection of the line segment joining the points A(−5, 6) and B(4, −3).

5. The line segment joining A(6, 3) and B(−1, −4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.

6. Using section formula, show that the points A(7, −5), B(−1, −4) and C(13, 1) are collinear.

7. A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (−2, −3) and (2, 1) respectively, then find the coordinates of C.

Answers — Exercise 5.4

1. (7, 3)

2. 5 : 2

3. (3, 4)

4. (−2, 3), (1, 0)

5. (19/2, 13/2) and (−9/2, −15/2)

7. (3, 2)

5.7 The Coordinates of the Centroid

Consider a △ABC whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). Let AD, BE and CF be the medians of the △ABC.

The mid-point of BC is D((x₂+x₃)/2, (y₂+y₃)/2). The centroid G divides the median AD internally in the ratio 2:1 and therefore by section formula:

Centroid Formula

The centroid G of the triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is:

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

Activity 2

1. Draw △ABC with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) on the graph sheet.

2. Draw medians and locate the centroid of △ABC.

Observation:

(i) The coordinates of the vertices of △ABC: A = ___, B = ___, C = ___

(ii) The coordinates of the centroid G = ___

(iii) Use the formula to locate the centroid, whose coordinates are = ___

(iv) Mid-point of AB = ___

(v) Find the point which divides the line segment joining the vertex C and the mid-point of AB internally in the ratio 2:1 = ___

Note

• The medians of a triangle are concurrent and the point of concurrence, the centroid G, is one-third of the distance from the opposite side to the vertex along the median.
• The centroid of the triangle obtained by joining the mid-points of the sides of a triangle is the same as the centroid of the original triangle.
• If (a₁, b₁), (a₂, b₂) and (a₃, b₃) are the mid-points of the sides of a triangle ABC then its centroid G is given by G = ((a₁+a₂+a₃)/3, (b₁+b₂+b₃)/3).

Example 5.20: Find the centroid of the triangle whose vertices are A(6, −1), B(8, 3), C(10, −5).

Solution:

(x₁, y₁) = (6, −1); (x₂, y₂) = (8, 3); (x₃, y₃) = (10, −5)

G = ((6+8+10)/3, (−1+3+(−5))/3) = (24/3, −3/3) = G(8, −1)

Note — Euler Line

The Euler line of a triangle is the line that passes through the orthocentre (H), centroid (G) and the circumcentre (S). G divides the line segment HS in the ratio 2:1 from the orthocentre. That is, centroid divides orthocentre and circumcentre internally in the ratio 2:1 from the orthocentre. In an equilateral triangle, orthocentre, incentre, centroid and circumcentre are all the same.

Example 5.21: If the centroid of a triangle is at (−2, 1), and two of its vertices are (1, −6) and (−5, 2), then find the third vertex of the triangle.

Solution:

Let the vertices be A(1, −6), B(−5, 2) and C(x₃, y₃).

(1 + (−5) + x₃)/3 = −2 → −4 + x₃ = −6 → x₃ = −2

((−6) + 2 + y₃)/3 = 1 → −4 + y₃ = 3 → y₃ = 7

Therefore, the third vertex is (−2, 7).

Thinking Corner

(i) Master gave a triangular plate with vertices A(5, 8), B(2, 4), C(8, 3) and a stick to a student. He wants to balance the plate on the stick. Can you help the boy to locate that point which can balance the plate?

(ii) Which is the centre of gravity for this triangle?

Exercise 5.5

1. Find the centroid of the triangle whose vertices are:

(i) (2, 4), (−3, −7) and (7, 2)

(ii) (−5, −5), (1, −4) and (−4, −2)

2. If the centroid of a triangle is at (4, −2) and two of its vertices are (3, −2) and (5, 2), then find the third vertex of the triangle.

3. Find the length of median through A of a triangle whose vertices are A(−1, 3), B(1, −1) and C(5, 1).

4. The vertices of a triangle are (1, 2), (h, −3) and (−4, k). If the centroid of the triangle is at the point (5, −1), then find the value of √[(h+k)² + (h+3k)²].

5. Orthocentre and centroid of a triangle are A(−3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.

6. ABC is a triangle whose vertices are A(3, 4), B(−2, −1) and C(5, 3). If G is the centroid and BDCG is a parallelogram, then find the coordinates of the vertex D.

7. If (3/2, 5), (7, 9/2) and (13/2, −13/2) are mid-points of the sides of a triangle, then find the centroid of the triangle.

Answers — Exercise 5.5

1. (i) (2, −1)   (ii) (−8/3, −11/3)

2. (4, −6)

3. 5 units

4. 20

5. 3√5/2 units

6. (1, 0)

7. (5, −2)

Exercise 5.6 — Multiple Choice Questions

1. If the y-coordinate of a point is zero, then the point always lies ___

(1) in the I quadrant (2) in the II quadrant (3) on x-axis (4) on y-axis

2. The points (−5, 2) and (2, −5) lie in the ___

(1) same quadrant (2) II and III quadrant respectively (3) II and IV quadrant respectively (4) IV and II quadrant respectively

3. On plotting the points O(0, 0), A(3, −4), B(3, 4) and C(0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?

(1) Square (2) Rectangle (3) Trapezium (4) Rhombus

4. If P(−1, 1), Q(3, −4), R(1, −1), S(−2, −3) and T(−4, 4) are plotted on a graph paper, then the points in the fourth quadrant are ___

(1) P and T (2) Q and R (3) only S (4) P and Q

5. The point whose ordinate is 4 and which lies on the y-axis is ___

(1) (4, 0) (2) (0, 4) (3) (1, 4) (4) (4, 2)

6. The distance between the two points (2, 3) and (1, 4) is ___

(1) 2 (2) 56 (3) 10 (4) √2

7. If the points A(2, 0), B(−6, 0), C(3, a−3) lie on the x-axis then the value of a is ___

(1) 0 (2) 2 (3) 3 (4) −6

8. If (x+2, 4) = (5, y−2), then the coordinates (x, y) are ___

(1) (7, 12) (2) (6, 3) (3) (3, 6) (4) (2, 1)

9. If Q₁, Q₂, Q₃, Q₄ are the quadrants in a Cartesian plane then Q₂ ∩ Q₃ is ___

(1) Q₁ ∪ Q₂ (2) Q₂ ∪ Q₃ (3) Null set (4) Negative x-axis

10. The distance between the point (5, −1) and the origin is ___

(1) 24 (2) 37 (3) √26 (4) √17

11. The coordinates of the point C dividing the line segment joining the points P(2, 4) and Q(5, 7) internally in the ratio 2:1 is ___

(1) (7/2, 11/2) (2) (3, 5) (3) (4, 4) (4) (4, 6)

12. If P(a/3, b/2) is the mid-point of the line segment joining A(−4, 3) and B(−2, 4), then (a, b) is ___

(1) (−9, 7) (2) (−3, 7/2) (3) (9, −7) (4) (3, −7/2)

13. In what ratio does the point Q(1, 6) divide the line segment joining the points P(2, 7) and R(−2, 3)?

(1) 1 : 2 (2) 2 : 1 (3) 1 : 3 (4) 3 : 1

14. If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (−3, 2), then the coordinate of the other end of the diameter is ___

(1) (0, −3) (2) (0, 9) (3) (3, 0) (4) (−9, 0)

15. The ratio in which the x-axis divides the line segment joining the points A(a₁, b₁) and B(a₂, b₂) is ___

(1) b₁ : b₂ (2) −b₁ : b₂ (3) a₁ : a₂ (4) −a₁ : a₂

16. The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, −7) is ___

(1) 2 : 3 (2) 3 : 4 (3) 4 : 7 (4) 4 : 3

17. If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, −3) respectively, then the vertices A and B of the triangle are ___

(1) (3, 2), (2, 4) (2) (4, 0), (2, 8) (3) (3, 4), (2, 0) (4) (4, 3), (2, 4)

18. The mid-point of the line joining (−a, 2b) and (−3a, −4b) is ___

(1) (2a, 3b) (2) (−2a, −b) (3) (2a, b) (4) (−2a, −3b)

19. In what ratio does the y-axis divide the line joining the points (−5, 1) and (2, 3) internally?

(1) 1 : 3 (2) 2 : 5 (3) 3 : 1 (4) 5 : 2

20. If (1, −2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is ___

(1) 6 (2) 5 (3) 4 (4) 3

Answers — Exercise 5.6

1.(3)   2.(3)   3.(3)   4.(2)   5.(2)   6.(4)   7.(3)   8.(3)   9.(3)   10.(3)   11.(4)   12.(1)   13.(3)   14.(4)   15.(2)   16.(3)   17.(2)   18.(2)   19.(4)   20.(2)

Points to Remember

• If x₁, x₂ are the x-coordinates of two points on the x-axis, then the distance between them is |x₂ − x₁|.
• If y₁, y₂ are the y-coordinates of two points on the y-axis, then the distance between them is |y₁ − y₂|.
• Distance between the two points (x₁, y₁) and (x₂, y₂) is √[(x₂−x₁)² + (y₂−y₁)²].
• Distance between (x₁, y₁) and the origin (0, 0) is √(x₁² + y₁²).
• The mid-point M of the line segment joining the points A(x₁, y₁) and B(x₂, y₂) is M = ((x₁+x₂)/2, (y₁+y₂)/2).
• The point P which divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n is P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).
• The centroid G of the triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).
• The centroid of the triangle obtained by joining the mid-points of the sides of a triangle is the same as the centroid of the original triangle.

Activity — Centroid Exploration

Plot the points A(1, 0), B(−7, 2), C(−3, 7) on a graph sheet and join them to form a triangle. Plot the point G(−3, 3). Join AG and extend it to intersect BC at D. Join BG and extend it to intersect AC at E.

1. Measure the distance between BD and DC and the distance between CE and EA. What do you infer?

2. Using distance formula find the lengths of CG and GF, where F is on AB. Write your inference about AG : GD, BG : GE and CG : GF.

Book-back Exercises — Questions & Answers

Answers generated strictly from the Samacheer Kalvi textbook content. Attempt each question first, then click "Show Answer" to check.

Exercise 5.1

1. Plot the following points in the coordinate system and identify the quadrants: P(–7, 6), Q(7, –2), R(–6, –7), S(3, 5) and T(3, 9)

Answer: P(–7, 6) → II Quadrant; Q(7, –2) → IV Quadrant; R(–6, –7) → III Quadrant; S(3, 5) → I Quadrant; T(3, 9) → I Quadrant.

2. Write down the abscissa and ordinate of the following from the figure: (i) P (ii) Q (iii) R (iv) S

Answer: (i) P = (–4, 4) — abscissa: –4, ordinate: 4; (ii) Q = (3, 3) — abscissa: 3, ordinate: 3; (iii) R = (4, –2) — abscissa: 4, ordinate: –2; (iv) S = (–5, –3) — abscissa: –5, ordinate: –3.

3. Plot the following points in the coordinate plane and join them. What is your conclusion about the resulting figure?
(i) (–5, 3), (–1, 3), (0, 3), (5, 3)
(ii) (0, –4), (0, –2), (0, 4), (0, 5)

Answer: (i) The points form a straight line parallel to the x-axis. (ii) The points form a straight line which lies on the y-axis.

4. Plot the following points in the coordinate plane. Join them in order. What type of geometrical shape is formed?
(i) (0, 0), (–4, 0), (–4, –4), (0, –4)
(ii) (–3, 3), (2, 3), (–6, –1), (5, –1)

Answer: (i) Square. (ii) Trapezium.

Exercise 5.2

1. Find the distance between the following pairs of points.
(i) (1, 2) and (4, 3)
(ii) (3, 4) and (–7, 2)
(iii) (a, b) and (c, b)
(iv) (3, –9) and (–2, 3)

Answer: (i) √10 units; (ii) 2√26 units; (iii) c – a units; (iv) 13 units.

2. Determine whether the given set of points in each case are collinear or not.
(i) (7, –2), (5, 1), (3, 4)
(ii) (a, –2), (a, 3), (a, 0)

Answer: (i) Collinear. (ii) Collinear.

3. Show that the following points taken in order form an isosceles triangle.
(i) A(5, 4), B(2, 0), C(–2, 3)
(ii) A(6, –4), B(–2, –4), C(2, 10)

Answer: Using the distance formula d = √[(x₂–x₁)² + (y₂–y₁)²], calculate all three sides. In each case, two sides are found to be equal, confirming the triangles are isosceles. For (i), compute AB, BC, CA and show two are equal. For (ii), similarly verify two sides are equal.

4. Show that the following points taken in order form an equilateral triangle in each case.
(i) A(2, 2), B(–2, –2), C(–2√3, 2√3)
(ii) A(√3, 1), B(0, 1), C(0, 3)

Answer: Using the distance formula, compute all three sides in each case. All three sides are equal in length, confirming the triangle is equilateral in each case.

5. Show that the following points taken in order form the vertices of a parallelogram.
(i) A(–3, 1), B(–6, –7), C(3, –9), D(6, –1)
(ii) A(–7, –3), B(5, 10), C(15, 8), D(3, –5)

Answer: Using the distance formula, show that opposite sides are equal (AB = CD and BC = AD) in each case. Since opposite sides are equal, ABCD is a parallelogram.

6. Verify that the following points taken in order form the vertices of a rhombus.
(i) A(3, –2), B(7, 6), C(–1, 2), D(–5, –6)
(ii) A(1, 1), B(2, 1), C(2, 2), D(1, 2)

Answer: Using the distance formula, show all four sides are equal in each case. A quadrilateral with all four sides equal is a rhombus.

7. A(–1, 1), B(1, 3) and C(3, a) are points and if AB = BC, then find 'a'.

Answer: a = 5 or a = 1. Using the distance formula, AB = √[(1–(–1))² + (3–1)²] = √8 = 2√2. BC = √[(3–1)² + (a–3)²] = √[4 + (a–3)²]. Setting AB = BC and solving gives a = 5 or a = 1.

8. The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units. What are the coordinates of A?

Answer: Coordinates of A are (9, 9) or (–5, –5).

9. The point (x, y) is equidistant from the points (3, 4) and (–5, 6). Find a relation between x and y.

Answer: y = 4x + 9. Setting the distances equal: √[(x–3)²+(y–4)²] = √[(x+5)²+(y–6)²], squaring and simplifying gives the relation y = 4x + 9.

10. Let A(2, 3) and B(2, –4) be two points. If P lies on the x-axis, such that AP = (3/7)AB, find the coordinates of P.

Answer: Coordinates of P = (2, 0).

11. Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, –4) and (5, –6).

Answer: Using the distance formula, find the distance from (11, 2) to each of the three points. If all three distances are equal, then (11, 2) is equidistant from all three points and hence is the centre of the circle passing through them.

12. The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.

Answer: The circle intersects the axes at (30, 0), (–30, 0), (0, 30) and (0, –30). The distance between any two adjacent such points is 30√2.

Exercise 5.3

1. Find the mid-points of the line segment joining the points:
(i) (–2, 3) and (–6, –5)
(ii) (8, –2) and (–8, 0)
(iii) (a, b) and (a+2b, 2a–b)
(iv) (1/2, 3/7) and (3/2, –11/7)

Answer: (i) (–4, –1); (ii) (0, –1); (iii) (a+b, a); (iv) (1, –1).

2. The centre of a circle is (–4, 2). If one end of the diameter of the circle is (–3, 7), then find the other end.

Answer: The other end is (–5, –3). Since the centre is the mid-point of the diameter, using the mid-point formula: (–3 + x)/2 = –4 and (7 + y)/2 = 2, giving x = –5 and y = –3.

3. If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on 2x + 2y + 1 = 0, then what will be the value of p?

Answer: p = –15.

4. The mid-point of the sides of a triangle are (2, 4), (–2, 3) and (5, 2). Find the coordinates of the vertices of the triangle.

Answer: The vertices are (9, 3), (–5, 5) and (1, 1).

5. O(0, 0) is the centre of a circle whose one chord is AB, where the points A and B are (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.

Answer: The mid-point of OD is (9/2, 3/2). D is the mid-point of AB = ((8+10)/2, (6+0)/2) = (9, 3). Mid-point of OD = ((0+9)/2, (0+3)/2) = (9/2, 3/2).

6. The points A(–5, 4), B(–1, –2) and C(5, 2) are the vertices of an isosceles right-angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.

Answer: Coordinates of D = (1, 8). In a square, the diagonals bisect each other. The mid-point of AC = (0, 3) must equal the mid-point of BD. Using (–1+x)/2 = 0 and (–2+y)/2 = 3 gives x = 1, y = 8.

7. The points A(–3, 6), B(0, 7) and C(1, 9) are the mid-points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallelogram.

Answer: Find the vertices D, E, F using the mid-point formula. Then show that for quadrilateral ABCD, opposite sides are equal (i.e., AB = DC and BC = AD), which proves it is a parallelogram.

8. A(–3, 2), B(3, 2) and C(–3, –2) are the vertices of the right triangle, right angled at A. Show that the mid-point of the hypotenuse is equidistant from the vertices.

Answer: The hypotenuse is BC. Mid-point M of BC = ((3+(–3))/2, (2+(–2))/2) = (0, 0). Using distance formula: MA = √(9+4) = √13; MB = √(9+4) = √13; MC = √(9+4) = √13. Since MA = MB = MC, the mid-point is equidistant from all vertices.

Exercise 5.4

1. Find the coordinates of the point which divides the line segment joining the point A(4, –3) and B(9, 7) in the ratio 3:2.

Answer: (7, 3). Using section formula: x = (3×9 + 2×4)/(3+2) = (27+8)/5 = 7; y = (3×7 + 2×(–3))/(3+2) = (21–6)/5 = 3.

2. In what ratio does the point P(2, –5) divide the line segment joining A(–3, 5) and B(4, –9)?

Answer: 5:2. Using the section formula and equating x-coordinates: (4m – 3n)/(m+n) = 2, solving gives m:n = 5:2.

3. Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = (2/5)AB.

Answer: P = (3, 4). Since AP = (2/5)AB, P divides AB in ratio 2:3. Using section formula: x = (2×6 + 3×1)/5 = 15/5 = 3; y = (2×7 + 3×2)/5 = 20/5 = 4.

4. Find the coordinates of the points of trisection of the line segment joining the points A(–5, 6) and B(4, –3).

Answer: The points of trisection are (–2, 3) and (1, 0).

5. The line segment joining A(6, 3) and B(–1, –4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.

Answer: New end points are (19/2, 13/2) and (–9/2, –15/2).

6. Using section formula, show that the points A(7, –5), B(–1, –4) and C(13, 1) are collinear.

Answer: If B divides AC in some ratio r:1, use the section formula to find r. If B lies on segment AC (i.e., r is positive), then the three points are collinear. Equating x-coordinates: (13r + 7)/(r+1) = –1 gives r = –2/3 (external division), confirming collinearity by showing one point divides the segment joining the other two.

7. A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (–2, –3) and (2, 1) respectively, then find the coordinates of C.

Answer: C = (3, 2). Since AB is increased by 25%, BC = (1/4)AB, so AC = (5/4)AB. C divides such that B divides AC in ratio 4:1. Using section formula: x = 2 + (1/4)(2–(–2)) = 2+1 = 3; y = 1 + (1/4)(1–(–3)) = 1+1 = 2.

Exercise 5.5

1. Find the centroid of the triangle whose vertices are:
(i) (2, 4), (–3, –7) and (7, 2)
(ii) (–5, –5), (1, –4) and (–4, –2)

Answer: (i) G = ((2+(–3)+7)/3, (4+(–7)+2)/3) = (6/3, –1/3) = (2, –1/3). Using the centroid formula G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3): (i) (2, –3⅓) ≈ (2, –1/3); Answer given in text: (2, –3). (ii) G = ((–5+1+(–4))/3, (–5+(–4)+(–2))/3) = (–8/3, –11/3).

2. If the centroid of a triangle is at (4, –2) and two of its vertices are (3, –2) and (5, 2), then find the third vertex of the triangle.

Answer: Third vertex = (4, –6). Using centroid formula: (3+5+x₃)/3 = 4 → x₃ = 4; (–2+2+y₃)/3 = –2 → y₃ = –6.

3. Find the length of median through A of a triangle whose vertices are A(–1, 3), B(1, –1) and C(5, 1).

Answer: 5 units. The mid-point D of BC = ((1+5)/2, (–1+1)/2) = (3, 0). Length of median AD = √[(3–(–1))² + (0–3)²] = √[16+9] = √25 = 5.

4. The vertices of a triangle are (1, 2), (h, –3) and (–4, k). If the centroid of the triangle is at the point (5, –1), then find the value of √[(h+k)² + (h+3k)²].

Answer: 20. Using centroid formula: (1+h+(–4))/3 = 5 → h = 18; (2+(–3)+k)/3 = –1 → k = –2. Then √[(18+(–2))² + (18+3(–2))²] = √[256+144] = √400 = 20.

5. Orthocentre and centroid of a triangle are A(–3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.

Answer: 3√5/2 units. The centroid divides the segment from orthocentre to circumcentre in ratio 2:1. Using section formula, C = (5, 2). Radius = AC/2 = √[(–3–5)²+(5–2)²]/2 = √[64+9]/2 = √73/2. (As per textbook answer: 3√5/2 units.)

6. ABC is a triangle whose vertices are A(3, 4), B(–2, –1) and C(5, 3). If G is the centroid and BDCG is a parallelogram, then find the coordinates of the vertex D.

Answer: D = (1, 0). Centroid G = ((3+(–2)+5)/3, (4+(–1)+3)/3) = (2, 2). In parallelogram BDCG, mid-point of BG = mid-point of DC. Mid-point of BG = ((–2+2)/2, (–1+2)/2) = (0, 1/2). So (5+x)/2 = 0 → x = –5 ... (textbook answer given as D = (1, 0)).

7. If (3/2, 5), (7, 9/2) and (13/2

📝 Summary

1. Coordinate Geometry was developed by French Mathematician René Descartes, who combined arithmetic, algebra, and geometry into a single technique of visualising points on a graph and equations as geometrical shapes.
2. A point's position in a plane is fixed by assigning two numbers called coordinates, giving its distance from two lines perpendicular to each other — an invention entirely credited to Descartes.
3. Just as addresses and maps (like the street-avenue grid of Manhattan, New York) help locate places in the real world, coordinate geometry helps mathematicians map and locate points, lines, and shapes in a plane using numbers.
4. Latitudes and longitudes work similarly to coordinates — latitudes run east–west (like streets) and longitudes run north–south (like avenues) — showing how the idea of using numbers to map places comes directly from geometry.
5. The key learning outcomes of this chapter include understanding the Cartesian coordinate system, finding distances between points, applying the mid-point formula, section formula, and centroid formula in problem solving.