coordinate geometry - inyatrust€¦ · - danica mckellar coordinate geometry. 324 unit-13 know...

This unit facilitates you in,

explaining the meaning of coordinate axes.

drawing coordinate axes.

locating the given points on the coordinate

plane.

identifying that the graph of the equation

ax + by = c represents a straight line.

defining, slope of a straight line.

calculating the slope of a given line.

explaining the meaning of x-intercept of a

line and y - intercept of a line.

deriving the slope-intercept form of the

equation of a straight line.

calculating the distance between two points

on a plane.

calculating the coordinates of a point which

divides the line in a given ratio.

Basic ideas - coordinate axis,

origin.

Writing the coordinates of a

point.

Graph of ax + by = c

[Linear graph].

Slope of a straight line.

Equation of a straight line in

slope-intercept form.

Distance between two points in

a plane.

Distance of a point in the plane

from origin.

Section Formula.

13

Rene Descartes

[1596 - 1650]

He was a french philosopher,

mathematician whose work

'La geometrie' includes his

application of Algebra to Geo-

metry from which we now have

'cartesian geometry'.

Mathematics is the only place where truth and

beauty mean the samething.

- Danica Mckellar

Coordinate Geometry

324 UNIT-13

Know this

The words ABSCISSA and

ORDINATE are Latin words

ABSCISSA means cut off

ORDINATE means keep it in

order.

We commonly observe that, the roads leading to the top

of a hill are always winding roads. There are no roads

laid straight upto the top of the hill. Why?

To answer the above question, the ideas learnt

about rectangular co-ordinate system would be useful.

This branch of mathematics was mainly developed

by French philosopher and mathematician Rene

Descartes [1596 - 1650]

Hence, Rectangular coordinate system has been named as 'Cartesian system' after

him.

This has lead to the branch of mathematics called COORDINATE GEOMETRY OR

ANALYTICAL GEOMETRY. This branch of mathematics treats geometry algebraically.

Descartes developed the idea of placing two number lines perpendicular to each other on

a plane and locate points on this plane.

Let us recall the terms related to cartesian system.

* The two perpendicular lines are called coordinate

axes.

* Horizontal line (XOX) is called x-axis and vertical

line (YOY) is called y-axis.

* The point of intersection of the two axes is called

the origin 'O'.

* Coordinate axes divides the plane into four

quadrants.

* For a point on the cartesian plane,

(i) its perpendicular distance from y-axis is called

the x-coordinate (or ABSCISSA) of the point.

(ii) its perpendicular distance from x-axis is called

the y-coordinate (or ORDINATE) of the point.

* (i) the coordinates of a point P is represented as

P(x,y).

(ii) coordinates of the origin are (0.0)

(iii) coordinates of any point on x-axis will be of the form (x,o)

(iv) coordinates of any point on y-axis will be of the form (o,y)

EXERCISE 13.1

I. Locate the following points on a graph sheet.

(i) P(4, –3) (ii) R(–1, –1) (iii) I(0, –5) (iv) X(–5, –2)

(v) Y(3, 2) (vi) Z(4, 0) (vii) E(0, 6) (viii) F(–2, 5)

XI X

Y

YI

O

I quadrant(–x, +y)

II quadrant(–x, +y)

IV quadrant(+x, –y)

III quadrant(–x, –y)

Coordinate Geometry 325

II. In which quadrants do these points lie?

(i) (4, –6) (ii) (3, 1) (iii) (–10, –2) (iv) x(–5, –2)

(v) (–5, –1) (vi) (5, –7) (vii) (9, 9) (viii) (–2, 7)

III. Plot the points on a cartesian plane whose

(i) Ordinate is 4, abscissa is 0 (ii) Ordinate is –3, abscissa is –1

(iii) Ordinate is 5, abscissa is 4 (iv) Ordiante is –1, abscissa is 7

Graphical representation of equations of the form ax + by = c :

You are familiar with graphs of linear equations. Observe the graphs the following

equations.

(1) 2x + y = 7

Reduce it to standard form y = mx + c

we get, y = 7 – 2x

0 1 2 3

7 5 3 1

x

y

(2) 3x + 2y = 5

Rewrite this equation as

y = 5 3

2

x

0 1 1 3

2.5 1 4 2

x

y

O

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–1–2–3–4–5

(0, 7)

(1, 5)

(2, 3)

(3, 1)

XI

X

Y

YI

O

4

3

2

1

–1

–2

–3

–1–2–3

(3, –2)

(–1, 4)

(0, 2.5)

(1, 1)

XI X

Y

YI

326 UNIT-13

From the above examples, it is evident that, the graph of equations of the form

ax + by = c is always a straight line.

Note : 2 points are sufficient to draw the line graph. One or two more points are

taken for verification or confirmation

Inclination of a straight line

Observe the graphs obtained for the equations 2x + y = 7 and 3x + 2y = 5

The straight lines form an angle with the x-axis. This angle can be measured both

in clockwise and anti-clockwise directions.

The angle formed by the linear graph with the positive direction of x-axis is called

the INCLINATION of that line.

This angle is generally represented by ‘ ’.

The inclination of a straight line is the angle formed at the point of intersection with

the X-axis which is in positive direction. This angle is measured in anti-clockwise

direction from the X-axis to the line.

B

A

oXI X

Y

YI

P

Q

XI X

Y

YI

o

Q

XI X

Y

YI

o

M

N

Note: It is a convention to consider the inclination of a line to be positive when the

angle is measured in anti-clockwise direction from the axis to the line.

Slope of a line

Observe the following instances:

A ladder is placed against a wall such that its top touches the

wall at a certain height from the floor.

While loading trucks, labourers use a wooden plank placed in

an inclined position which helps them to load the truck easily.


The staircase to a building is constructed in a particular position

so that the strain of climbing the staircase is minimum.

In each of the above instances, the position in which the ladder/wooden plank /

stairs placed reduces the strain of climbing..

At what distance the ladder should be placed from the wall so that the strain is

minimised ?

Study the following three figures. Discuss in groups and try to find the solution

mathematically.

(1) (2) (3)C B

A

A

BC BC

A

This can be done by taking the ratio of two lengths i.e., the vertical height and

horizontal distance. Note that the length of ladder is same in all cases.

In the above figures, observe that ABBC and ABC is a right angled triangle.

Here, AB is called the vertical distance (height) and, CB is called the horizontal

distance.

Observe the following figures

A

BC4.7 cm

1.6 cm

A

BC3.5 cm

3.5

cm

A

BC2.6 cm

4.3

cm

Vertical distance

Horizontal distance

AB 1.6cm0.34

BC 4.7 cm

Vertical distance

Horizontal distance

AB 3.5cm1

BC 3.5 cm

Vertical distance

Horizontal distance

AB 4.3cm1.65

BC 2.6 cm

In each example, the ratio obtained is called the slope of the line AC.

328 UNIT-13

The ratio of the vertical distance to the horizontal distance is called slope.

Slope = Vertical distance

Horizontal distance

In the above examples,

Case (1), AB < BC Case (2), AB = BC Case (3), AB > BC.

We observe that,

* When vertical distance is less than the horizontal distance, slope is less than 1.

* When vertical distance is equal to the horizontal distance, slope is equal to 1.

* When vertical distance is more than the horizontal distance, slope is more than 1.

The strain of climbing will be least when the slope is less than 1.

Slope of a line can also be expressed in trigonometric ratio form. Study the following

example.

A

BCHorizontal distance

Vert

ical

dis

tance

The slope of a line is the tangent of the angle of its inclination. It is generally

denoted by 'm'.

m = tan

Gradient of a straight line

Gradient of a straight line is defined as the

increase in

increase in

y

x moving from one point on the line to another.

Gradient = BC

AC

y

x

Hence, Gradient = Vertical height

Horizontal distance

Gradient of a straight line is nothing but the slope of

a straight line.

With respect to ABC,

Slope of AC = AB

BC

AC forms an angle with BC.

With respect to angle ,

AB is the opposite side and BC is the adjacent side

Slope = Opposite side

Adjacent side

Slope = tan

A

B

C

OX

I

YI

Y

X


Note:

1. The X-axis subtends an angle of 0° with itself.

Hence, slope of X-axis is m = tan 0° = 0 [ = 0°]

2. The y-axis subtends an angle of 90° with the X-axis.

Hence, slope of Y-axis is m = tan 90° = not defined [ = 90°]

3. Value of angle of Value of slope

inclination

= 0° 0

0° < < 90° positive number

= 90° not defined

90° < < 180° negative number

Now you must be able to understand why roads leading up the mountains are not

laid straight upto the top.

Discuss with your friends and teacher.

Slope of a straight line passing through two given points

Let A(xI, yI) and B(x2, y

2) be any two fixed points on a

cartesian plane.

Draw AL X-axis, BM X-axis, AN BM

From the figure, OL = x1, OM = x

2

LM = x2 – x

1 = AN

AL = y1 = NM, BM = y

2

BN= y2 – y

1

In ABN, BAN =

with respect to , tan = BN

AN

slope = tan = 2 1

2 1

y y

x x

m = 2 1

2 1

y - y

x - x

Slope of a straight line passing through two given fixed points

= difference of ordinates of given points

difference of their abscissa

You have learnt to find the slope of a straight line. Let us find the slopes of two lines

in the same cartesian plane.

o

y1

( ,y )x2 2

A(,

)

xy 1

1

( – )x2 x1

x2

L x1 MX

IX

Y

YI

y 1

(–

)y

y2

1y

2

B

N ( – )x2 x1

330 UNIT-13

Parallel lines

Two parallel lines make the same inclination

with X-axis. Hence they have same slopes.

= [ corresponding angles]

tan = tan

slope of AB = slope of CD.

Parallel lines have equal slopes.

Conversely, if the slopes of two lines are equal,

the lines are parallel to each other.

Note: * Slope of X-axis is 0.

slope of any line parallel to X-axis is also 0.

* Slope of Y-axis is not defined.

slope of any line parallel to Y-axis is also not defined.

Perpendicular lines

If AB is perpendicular to CD and their inclinations are and respectively.

= 90 + [ exterior angle = sum of interior opposite angles]

tan = tan (90 + )

tan = – cot [ tan (90 + ) = – cot ]

tan = 1

tan θ

tan . tan = – 1

m1 . m

2= –1

where m1 slope of AB

m2 slope of CD.

m1

= 2

1

m or m

2 =

1

1

m

If two lines are mutually perpendicular,

then, the product of their slopes is –1.

Conversely, if the product of the slopes of two lines is –1,

then the lines are mutually perpendicular.

ILLUSTRATIVE EXAMPLES

1. Find the slope of the line whose inclination is 60°.

Sol. = 60°

Slope = m = tanm = tan 60°

m = 3

A

B

C

D

O

XI

X

Y

YI

A

B

C

DO

XI X

Y

YI


2. Find the angle of inclination of straight line whose slope is 0.

Sol. Slope = tan

0 = tan

we know that, tan 0º = 0

= 0°

3. Determine the slope of the line joining the points (3, –2) and (4, 5)

Sol. Let (x1, y

1) = (3, –2) and (x

2, y

2) = (4, 5)

Slope = 2 1

2 1

y y

x x =

5 ( 2)

4 3 =

5 2

1

Slope = 7

4. Prove that the line joining the points (4, 5) and (0, –2) is perpendicular to the

line passing through (2, –3) and (–5, 1).

Sol. For line 1 : Let (x1, y

1) = (4, 5) and (x

2, y

2)= (0, – 2)

Slope = 2 1

2 1

y y

x x =

2 5

0 4

Slope = 7

4 =

7

4 = m

1

For line 2 : (x1, y

1) = (2, –3), (x

2, y

2) = (–5, 1)

Slope =2 1

2 1

y y

x x =

1 ( 3)

5 2 =

1 3

7

Slope = 4

7 =

4

7 = m

2

m1 × m

2=

7 4

4 7= –1

The product of their slopes is –1

The two lines are perpendicular.

5. Find the slope of the line passing through the points P(7, 3) and Q(0, –4). Also

find slope of a line (i) parallel to PQ (ii) perpendicular to PQ.

Sol. Let (x1, y

1) = (7, 3) and (x

2, y

2) = (0, –4)

Slope = 2 1

2 1

y y

x x =

4 3

0 7 =

7

7

Slope = 1

Slope of PQ = 1

(i) Slope of a line parallel to PQ = 1 [ for parallel lines, m1 = m

2]

(ii) Slope of a line perpendicular to PQ = – 1 [ for perpendicular lines, m1 × m

2 = –1]

332 UNIT-13

EXERCISE 13.2

1. Find the slope of the line whose inclination is

(i) 90° (ii) 45° (iii) 30° (iv) 0°

2. Find the angles of inclination of straight lines whose slopes are

(i) 3 (ii) 1 (iii)1

3

3. Find the slope of the line joining the points

(i) (–4, 1) and (–5, 2) (ii) (4, –8) and (5, –2) (iii) (0, 0) and 3,3

(iv) (–5, 0) and (0, – 7) (v) (2a, 3b) and (a, –b)

4. Find whether the lines drawn through the two pairs of points are parallel or

perpendicular

(i) (5, 2), (0, 5) and (0, 0), (–5, 3) (ii) (3, 3), (4, 6) and (4, 1), (6, 7)

(iii) (4, 7), (3, 5) and (–1, 7), (1, 6) (iv) (–1, –2), (1, 6) and (–1, 1), (–2, –3)

5. Find the slope of the line perpedicular to the line joining the points.

(i) (1, 7) and (–4, 3) (ii) (2, –3) and (1, 4)

6. Find the slope of the line parallel to the line joining the points.

(i) (–4, 3) and (2, 5) (ii) (1, –5) and (7, 1)

7. A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a)

and (11, 3). Find a.

8. A line passing through the points (1, 0) and (4, 3) is perpendicular to the line joining

(–2, –1) and (m, 0). Find the value of m.

Equation of a line in slope intercept form

Observe the linear graph. The straight line PQ is

intersecting the X-axis at P and Y-axis at Q.

The distance OP i.e., the distance from origin to the point

where the line intersects the X-axis is called the

x-intercept of that line.

The distance OQ i.e., the distance from the origin to the

point where the line intersects the Y-axis is called the

y - intercept of that line.

In the above example, x-intercept is 3 and y-intercept is 4.

Identify the x and y intercepts in the following cases.

O

4

3

2

1

–1

–2

–3

–1–2–3P

Q

XI X

Y

YI


Now, study the following graph.

Let line 'l ' meet the Y-axis at A(0,c).

OA is the y-intercept of the line l.

Let P be any point on line l.

Draw PQ OX and AB PQ,

Let 'l ' makes an angle with X-axis

PAB = (corresponding angles)

Slope = m = tan

m = Opposite side

Adjacent side

In PAB, with respect to ,

m = PB

AB [PB = PQ – BQ, AB = OQ]

m = cy

x

mx = y – c

y = mx + c

The equation of a l ine with slope 'm' and whose y-intercept is 'c'

is given by y = mx + c


1. Find the equation of a line having slope 1

2 and y-intercept –3.

Sol. m = 1

2, y intercept = c = – 3

Equation : y = mx + c

y = 1

2x + (–3) =

1

2x – 3 y=

6

2

x

2y = x – 6 or x – 2y – 6 = 0

O

4

3

2

1

–1

–2

–3

–4

–1–2XI X

Y

YI

O

5

4

3

2

1

–1

–2

–3

–1–2–3X

IX

Y

YI

O

y

l

A B

Q x

c x

(–

)

yc

P(,

)x

y

(0, c)

XI

X

Y

YI

334 UNIT-13

2. Find the equation of a line having angle of inclination 45° and y-intercept 2.

Sol. = 45°, y-intercept = c = 2

Slope = m = tan m = tan 45°

m = 1

Equation : y = mx + c

y = 1(x) + 2 = x + 2

or x – y + 2 = 0

3. The equation of a line is 3x + 2y + 1 = 0. Find its slope and y-intercept.

Sol. Equation: 3x + 2y + 1 = 0

Rewrite in the standard form of y = mx + c

2y = –3x – 1

Dividing by 2, we get

y = 3 1

2 2x

compare with y = mx + c

m = 3

2, c =

1

2

Slope = 3

2

y - intercept = 1

2

EXERCISE 13.3

1. Find the equation of the line whose angle of inclination and y-intercept are given.

(i) = 60°, y-intercept is – 2 (ii) = 45°, y-intercept is 3.

2. Find the equation of the line whose slope and y-intercept are given.

(i) Slope = 2, y-intercept = – 4

(ii) Slope = 2

3, y-intercept =

1

2

(iii) Slope = –2, y-intercept = 3

3. Find the slope and y-intercept of the lines

(i) 2x + 3y = 4 (ii) 3x = y

(iii) x – y + 5 = 0 (iv) 3x – 4y = 5

4. Is the line x = 2y parallel to 2x – 4y + 7 = 0. [Hint : Parallel lines have same slopes]

5. Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each

other. [Hint : For perpendicular lines, m1m

2 = –1]


Distance between two points

Observe the points A and B. The distance between them is the length of the line

segment joining them.

A B A B

We use a ruler to measure the length of the line segment or the distance between

two points.

Is it possible to find the distance between any two points without actually

measuring ?

It is interesting to study that we can find the

distance between them, if they are located on a cartesian

plane.

If the points A and B lie on X-axis, we can count the

number of unit segments between A and B

AB= 4 cm or

Coordinates of A are (1, 0) and of B (5, 0)

AB= x2 – x

1 = 5 – 1

AB = 4 cm

In general, if the coordinates A and B respectively are

(x1, 0) and (x

2, 0) then AB = x

2 – x

1.

If two points P and Q lie on y-axis. We can count the number

of unit segments between P and Q.

PQ= 6 cm or

Coordinates of P = (0, 4) and of Q = (0, –2)

PQ = y2 – y

1 or PQ = y

1–y

2

= –2 – 4 = –6 = 4 – (–2) = 6

P Q = 6 cm = 6

In general, if the coordinates of P and Q respectively are

(0, y1) and (0, y

2) then PQ = y

2 – y

1

Length of a line parallel to x-axis

If AB is parallel to X-axis.

Draw, AA' X-axis and BB' X-axis

The figure formed AA'B'B is a rectangle

AB = A'B'

A'B' = x2 – x

1 = 6 – 1

A'B' = 5 AB = 5cm

Note: All points on a line parallel to x-axis have the same

y-coordinate.

Remember: Distance is

always a positive number.

O

3

2

1

–1

–2

–1–2XI

X

Y

YI

A B

O

5

4

3

2

1

–1

–2

–3

–1–2

P

Q

XI

X

Y

YI

XI X

YI

–2

–3

Y

O 1 2 3 4 5 6–1–2

–1

5

4

3

1

2

AI BI

336 UNIT-13

Length of a line parallel to Y-axis

If AB is parallel to Y-axis,

Draw AA' Y-axis

BB' Y-axis

The figure formed by AA'B'B is a rectangle

AB = A'B'

A'B' = y2 – y

1

= –3 – 4 = – 7

A'B' = 7 cm AB = 7 cm

Note : Points on a line parallel to Y-axis all have the

same x coordinate.

In all the above examples, where we found the

distance between two points, two conditions are noted,

* the points either lie on X-axis or Y-axis

* the points form a line which is parallel to either X-axis or

Y-axis

Now observe the points P and Q are on the cartesian plane.

Neither of these points lie either on X-axis or on Y-axis.

Let us join them. Line segment PQ is neither parallel to

X-axis nor to Y-axis.

Now, let us learn how to find the distance between 2 points

which neither lie on X or Y-axis nor form a line parallel to them.

Distance formula

If A(x1, y

1) and B(x

2, y

2) are the two given points, draw

BC Y-axis and AC X-axis the co-ordinates of C are (x2, y

1)

AC= x2 – x

1 and BC = y

2 – y

1

ABC is right angled triangle, By Pythagoras theorem,

AB2 = AC2 + BC2

AB2 = (x2 – x

1)2 + (y

2 – y

1)2

AB = 2 2

2 1 2 1x x y y

distance between two points (x1, y

1) and (x

2, y

2)

d = 2 2

2 1 2 1x - x + y - y

d = 2 2difference of absscissa difference of ordinates

Note : For the sake of convenience, we take points lying in I quadrant to derive formulae.

But, all the formulae remain true for points lying in any quadrant.

O

5

4

3

2

1

–1

–2

–3

–1–2

AI

BI

A(2, 4)

B(2, –3)

–3–4X

IX

Y

YI

P

Q

OX

I X

Y

YI

OX

IX

Y

YI


Distance of a point in a plane from the origin

Given A(x, y) and O(0, 0) are the two points,

By distance formula,

d = 2 2

2 1 2 1x x y y

d = 2 2

0 0x y = 2 2x y


1. Find the distance between the points (2, 3) and (6, 6).

Sol. Let (x1, y

1) = (2, 3) and (x

2, y

2) = (6, 6)

d = 2 2

2 1 2 1x x y y = 2 2

6 2 6 3

d = 2 24 3 = 16 9 = 25

d = 5 units

2. Find the distance between the origin and the point (12, –5).

Sol. Distance between origin and (x, y) = 2 2x y

Here, (x, y) = (12, –5)

d = 2 212 ( 5) 144 25 169

d = 13 units

3. Find the co-ordinates of points on x-axis which are at a distance of 5 units fromthe point (6, –3).

Sol. Any point on X-axis has co-ordinates of form (x, 0)

(x1, y

1) = (6, –3) and (x

2, y

2) = (x, 0) and d = 5

d = 2 2

2 1 2 1x x y y

5 = 22( 6) 0 ( 3)x = 2 2( 6) 3x

Squaring on both sides 52= 2

26 9x

25 = x2 – 12x + 36 + 9

x2 – 12x + 45 – 25 = 0 x2 – 12x + 20 = 0 (x – 10)(x – 2) = 0

x – 10 = 0 or x – 2 = 0 x = 10 or x = 2

Required points on X - axis are (10, 0) and (2, 0)

O(0, 0)

A(x,y)

XI

X

Y

YI

338 UNIT-13

4. Find the value of x such that the distance between the points (2, 5) and

(x, –7) is 13.

Sol. d = 13, (x1, y

1) = (2, 5) and (x

2, x

2) = (x, –7)

d = 2 2

2 1 2 1x x y y

13 = 2 22 ( 7 5)x = 2( 2) 144x

Squaring both sides,132= x2 – 4x + 4 + 144

169 = x2 – 4x +148

x2 – 4x + 148 – 169 = 0

x2 – 4x – 21 = 0

(x – 7) (x + 3) = 0

x = 7 or x = – 3

5. Show that the triangle whose vertices are (8, –4), (9,5) and (0, 4) is an isosceles

triangle.

Sol. To show that the triangle is isosceles, we must show that the length of 2 sides of

the triangle are equal.

Let A(8, –4), B(9, 5), C(0, 4)

d = 2 2

2 1 2 1x x y y

AB = 2 2

9 8 5 ( 4) = 2 21 9 1 81 82

BC = 2 2 2 2(9 0) (4 5) 9 ( 1) 81 1 82

CA = 22 2 2(0 8) 4 ( 4) ( 8) 8 64 64 128

Since AB = BC, the triangle is isosceles.

6. Prove that the points A(–3, –2), B(5, –2), C(9, 3) and D(1, 3) are the vertices of a

parallelogram.

Sol. A quadrilateral is a parallelogram if its opposite sides are equal

d = 2 2

2 1 2 1x x y y

AB = 2 2

5 ( 3) 2 ( 2) = 2 28 0 64 8

CD = 2 2(9 1) (3 3) = 2 28 0 64 8

BC = 2 22 2(5 9) ( 2 3) 4 5 16 25 41

DA = 2 2 2 21 ( 3) 3 ( 2) 4 5 16 25 41

AB= CD and BC = DA

ABCD is a parallelogram.

B(9, 5)

C(0, 4) A(8, –4)

A(–3, –2) B(5, –2)

C(9, 3)


EXERCISE 13.4

1. Find the distance between the following pairs of points

(i) (8, 3) and (8, –7) (ii) (1, –3) and (–4, 7) (iii) (–4, 5) and (–12, 3)

(iv) (6, 5) and (4, 4) (v) (2, 0) and (0, 3) (vi) (2, 8) and (6, 8)

(viii) (a, b) and (c, b) (viii) (cos – sin) and (sin – cos )

2. Find the distance between the origin and the point

(i) (–6, 8) (ii) (5, 12) (iii) (–8, 15)

3. (i) The distance between the points (3, 1) and (0, x) is 5 units. Find x.

(ii) A point P(2, –1) is equidistant from the points (a, 7) and (–3, a). Find 'a'.

(iii) Find a point on y-axis which is equidistant from the points (5, 2) and (–4, 3)

4. Find the perimeter of the triangles whose vertices have the following coordinates

(i) (–2, 1), (4, 6), (6, –3) (ii) (3, 10), (5, 2), (14, 12)

5. Prove that the points A(1, –3), B(–3, 0) and C(4, 1) are the vertices of a right isosceles

triangle.

6. Find the radius of a circle whose centre is (–5, 4) and which passes through the

point (–7, 1).

7. Prove that each of the set of coordinates are the vertices of parallelograms.

(i) (–5, –3), (1, –11), (7, –6), (1, 2) (ii) (4, 0), (–2, –3), (3, 2), (–3, –1)

8. The coordinate of vertices of triangles are given. Identify the types of triangles.

(i) (2, 1) (10, 1) (6, 9) (ii) (1, 6) (3, 2) (10, 8)

(iii) (3, 5) (–1, 1) (6, 2) (iv) (3, –3) (3, 5) (11, –3)

Section Formula

Let AB be a line joining the points A(x1, y

1) and

B(x2, y

2) and point P divides the line segment AB in

the ratio m : n AP

PB=

m

n

Now, let us try to find the coordinates of point P.

Let the coordinates of P be (x, y)

Draw AL, PM and BN, perpendiculars on X-axis.

Also draw, AR PM and PQ BN

From the figure, AR = LM = OM – OL = x – x1

PR = PM – RM = PM – AL = y – y1

PQ = MN = ON – OM = x2 – x

BQ = BN – QN = BN – PM = y2 – y

Observe that, APR ~ PBQ

L M N X

Y

O

R

–

yy

2

Q

B (x ,y )22

(x y)1 1 P y– y1 x– x2

x– x1

Y

X

A

(x, y)

340 UNIT-13

AR PR AP

= =PQ BQ PB

[ corresponding sides of similar s are proportional]

Let us take, AR

PQ =

AP

PB

x - x1

x - x2

= m

n

nx + mx = mx2 + nx

1, x (m + n) = mx

2 + nx

1, x(m + n) = mx

2 + nx

1

x = 2 1mx nx

m n

Consider PR

BQ =

AP

PB

1

2

y y

y y =

m

n

ny – ny1

= my2 – my

ny + my = my2 + ny

1

y(m + n) = my2 + ny

1

y =2 1my ny

m n

coordinates of P = (x, y)

coordinates of P = 2 1 2 1mx + nx my + ny

,m + n m + n

If P is the midpoint of AB, m : n = 1 : 1 then coordinates of P = 2 1 2 1x + x y + y

,2 2

.

This is also called the mid point fomula.


1. Find the coordinates of point P which divides the line joining A(4, –5) and

B(6, 3) in the ratio 2 : 5.

Sol. Let the coordinates of P be (x, y). Given: m : n = 2 : 5

(x1, y

1) = (4, –5) and (x

2, y

2) = (6, 3)

by section formula,

x = 2 1 2(6) 5(4) 12 20 32

2 5 7 7

mx nx

m n

y = 2 1 2(3) 5( 5) 6 25 19

2 (5) 7 7

my ny

m n coordinates of P =

32 19,

7 7

2. Find the ratio in which point (5, 4) divides the line joining points (2, 1) and (7, 6).

Sol. Let the required ratio be m : n

Given: (x1, y

1) = (2, 1), (x

2, y

2) = (7, 6) and (x, y) = (5, 4)


By section formula,

x = 2 1mx nx

m n or y =

2 1my ny

m n

5 = (7) (2)m n

m n4 =

(6) (1)m n

m n

5 = 7 2m n

m n4 =

6m n

m n

5m + 5n = 7m + 2n 4m + 4n = 6m + n

5m – 7m + 5n – 2n = 0 4m – 6m + 4n – n = 0

–2m + 3n = 0 –2m + 3n = 0

–2m = –3n –2m = –3n

m

n =

3

2

m

n =

3

2

m : n = 3 : 2 m : n = 3 : 2

3. Find the coordinates of the midpoint of the line segments joining the points

(2, 3) and (4, 7).

Sol. (x1, y

1) = (2, 3) and (x

2, y

2) = (4, 7)

by midpoint formula, coordinates of mid point= 1 2 1 2,

2 2

x x y y

= 2 4 3 7

,2 2

= 6 10

,2 2

= (3, 5)

EXERCISE 13.5

1) In what ratio does the point (-2, 3) divide the line segment joining the points

(-3, 5) and (4, -9) ?

2) In the point C(1,1) divides the line segment joining A(-2,7) and B in the ratio 3:2,

find the coordinates of B.

3) Find the ratio in which the point (-1, k) divides the line joining the points

(-3, 10) and (6, -8), and find the value of k.

4) Find the coordinates of the midpoint of the line joining the points

(-3, 10) and (6, -8).

5) Three consecutive vertices of a parallelogram are A(1,2), B(2,3) and C(8,5). Find

the fourth vertex. (Hint : diagonals of a parallelogram bisect each other).

342 UNIT-13

ANSWERS

EXERCISE 13.2

1] (i) ND (ii) 1 (iii) 1

3(iv) 0 2] (i) 60º (ii) 45º (iii) 30º

3] (i) –1 (ii) 6 (iii) 3 (iv)7

5(v)

4b

a4] (i) parallel (ii) parallel

(iii) perpendicular (iv) parallel 5] (i)5

4(ii)

1

76] (i)

1

3(ii) 1 7] 5 8] –3

EXERCISE 13.3

1] (i) y = 3 x – 2 (ii) y = x + 3 2] (i) y = 2x – 4 (ii) 6y = –4x + 3 (iii) y = –2x + 3

3] (i) m = 2

3, c =

4

3(ii) m = 3, c = 0 (iii) m = 1, c = 5 (iv) m =

3

4, c =

5

4

EXERCISE 13.4

1] (i) 10 units (ii) 5 5 units (iii) 2 17 units (vi) 5 units (v) 13 units

(vi) 4 units (vii) (c–a) units (viii) 2 (sin cos )units

2] (i) 10 units (ii) 13 units (iii) 17 units 3] (i) 5 or –3 (ii) 7 (iii) (0, –2)

4] (i) 61 80 85 units (ii) 68 125 181 units 6] 13 units

EXERCISE 13.5

1] 1:6 2] (3, –3) 3] 2:7 4] (3

2, 1) 5] (7, 4)

Drawing graph of equation ax+by=c

Coordinate Geometry

Distance between two points

Inclination of a straight line lying on

axisx-

lying onaxisy-

Length of lineparallel to

axisx-

Length of lineparallel to

axisy-

2 1

2 1

y ym =

x x

Distance Formula

2 2

2 1 2 1d ( ) ( y )x x y

Section Formula

p ,2 1 2 1mx +nx my +ny

m+n m+n

Section Formula

p ,2 1 2 1

2 2

x + x y + y

Slope of a straight

line, m = tan

s

coordinate geometry - inyatrust€¦ · - danica mckellar coordinate geometry. 324 unit-13 know...

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