Co-ordinate Geometry

Types of sums

1. Lines

2. Equations

3. Slopes

4. Parallel & Perpendicular lines

5. Circles

6. Areas

Importance

1.CAT

2.CMAT

3.NMAT

4.SNAP

5.XAT

Basics

Scope of the topic

Astronomy: Computing paths of celestial bodies like planets, comets, binary star systems.

Graphs analysis

Company performance Data Interpretation

Geometry (Area, distance, slope, rate of change) Missile Trajectory

Gyaan: Descartes is regarded as the father of analytical geometry. He founded the concept of coordinate plane(Can be asked in GK :D)

What is a Coordinate plane?

2 Dimentsional

Horizontal line:

X-axis

Vertical line: Y-axis

Intersection is origin which is

the reference point.

It is used to locate an object on a plane.

Quadrants

Plotting a point

P(4,3

)Q(-

6,2)

Single variable

x =5

y = -3

x = -2

Two variables X and Y

x – y =0i.e. x = y

Two variables X and Y

2x – y = 4y = 2x - 4

Degree 2: Quadratic Function

Shape:

Parabola

Parabola (Quadratic function)

Degree 2: Two variables

Circle Ellipse

LinesDistance Formula: d (PQ) = √(x2-x1)2 + (y2-y1)2

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

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

Distance between two points:

Distance between (4,1) and (7,5)

d

P(7,5)

Q(4,1)

4

3

Distance between points

Find the distance between (7, 5) & (4, 1)

a) 5

b) 5 √2

c) 7 √3

d) 10

Distance between points

Find the distance between (4, 2) & (6, 6)

a) 5.45

b) 4.45

c) 7.14

d) 5.65

Section of a line segment

Internal Division External division

m n

n

m

Equations• Find the equation of the line

passing through (-1,-2) & (-5,2)

• Point of intersection of 2x + y = 4 and x - y + 1= 0

Equation of line joining two points

Find the equation of the line passing

through (-1,-2) & (-5,2)

(a) 2x + y = 3

(b) 3x - 2y + 7 = 0

(c) x + y + 3 = 0

(d) None of these

(y + 2)/(-2-2) = (x + 1)/(-1+5)4(y + 2) = -4 (x + 1)

y + 2 = -x -1 => x + y = -3

Point of intersection of two lines

Point of intersection of

2x + y = 4 and x - y + 1= 0

a) (3, 1)

b) (1, 2)

c) (3, 2)

d) (1, 4)

Slopesy = mx + c

Slope of a line

x = y y = mx + c

Slope of a line

2x – y = 4

Find slope of line y = 4

a) 0b)2c) 4d)Cannot be det

Find slope of line x = 4

a) 0b)2c) 4d)Cannot be det

Find slope of line y = –2x + 3

a) 0b) -2c) 4d)Cannot be det

Finding the slope of a line given two points

Finding the slope from the equation of a line

Finding the equation of a line given a point and a slope

Finding the equation of a line given two points

Finding the equation of the line that is parallel to a given line and passing through a given point

Finding the equation of the line that is perpendicular to a given line and passing through a given point

ParallelPerpendicular

Parallelism

x – y =0

x –y +3 = 0

m1=m2

Perpendicularity

2x – y = 4

x + 2y = 4

One line passes through the points (–1, -

2) and (1, 2); another line passes through

the points (–2, 0) and (0, 4).

Are these lines..

a) Parallel

b) Perpendicular

c) Neither

One line passes through the points (0, –4)

and (–1, –7); another line passes through

the points (3, 0) and (–3, 2).

Are these lines..

a) Parallel

b) Perpendicular

c) Neither

Find the equation of the line, with the slope

m= -1 and passing through the point (3,0)

a) y = -x + 3

b) 2y = x + 3

c) 4y = 4x + 3

d) y = -2x + 3

Use ( y - y1 ) = m ( x - x1)

Find the equation of the line parallel to the

line 6x + 9y = -5, and passing through

the point ( 7, 4 )

a) y = -x + 11

b) 9x = - 6y + 3

c) 2x + 3y = 26

d) 6x + 9y = 11

m = -6/9 = -2/3

Use ( y - y1 ) = m ( x - x1)

Equation using slope and point

Find the equation of line passing through

point (2,-3) having slope 5/4.

(a) 4x - 5y = 20

(b) 3x – 2y = 5

(c) 5x - 4y = 22

(d) None of these

(y - y1) = m (x-x1)(y + 3) = 5/4 (x - 2)4y + 12 = 5x -10 => 5x - 4y = 22

Note: Sometimes angle will be given. Slope can be found as tanѲ

Find the equation of the line perpendicular

to the line -5x - y = 3, and passing

through the point ( 4, 3 )

a) y = -x + 7

b) 2y = x + 3

c) 5y = x + 3

d) -x + 5y = 11

m1 = 5/-1 = -5

So, use m2 = 1/5

Equation of perpendicular

Find the equation of the line passing through

point (2,7) and perpendicular to 2x + 3y +

8= 0

a) 5x – 2y = 17

b) 3x + 2y = 20

c) 3x – 2y = -4

d) 6x – 4y = -16

Equation of parallel

Find the equation of the line passing through

point (2,7) and parallel to 2x + 3y + 8= 0

a) 5x - 2y = 17

b) 2x + 3y = 25

c) 2x + 3y = -4

d) 4x + 6y = -16

Area

Centroid Incentre

A(x1,y1)

B(x2,y2)C(x3,y3)

c b

a

Area of Triangle

Area of triangle formed by points P1(x1,y1), P2(x2,y2) & P3(x3,y3)

Area of triangle

Find the area of triangle formed by points

(4,2), (-5,7) and (5,-3)

a) 10

b) 14.5

c) 22.5

d) 20

Formula substitution: [4(3-(-7)) – 2(-5 -5)+ 1(15-35)]/2= 40/2 = =20 Option (d)

Circles

Find the centre and radius of the

following circles:

1.

2.

3.

Find the centre and radius of the

following circle

Find the equation of the circle:

1.with centre (0,5) and radius 5

2.with centre (2, 0) and radius 4

3.with centre (5, 7) and radius 18

Find the equation of the circle with

centre (2, 1) which passes through (4,1).

Find the equation of the circle with

centre (-3, -2) which passes through

(1,-4).

Find the equation of a circle (centre O) with

a diameter between two points, P at and

Q at .

a) x2 + y2 = 30

b) x2 + y2 = 50

c) x2 + y2 = 80

Applications

Manufacturing of product includes some

fixed cost and variable cost A firm produces

50 units of a good for Rs. 320 and 80 units for Rs. 380. Estimate the cost for producing

110 units. (Assume cost curve to be linear)

a) Rs. 330

b) Rs. 365

c) Rs. 440

d) Rs. 1665

Manufacturing of product includes some fixed cost and variable cost A firm produces 50 units of a good for Rs. 320 and 80 units for Rs. 380. Estimate the cost for producing 110 units. (Assume cost curve to be linear)

a) Rs. 330b) Rs. 365c) Rs. 440d) Rs. 1665

We have two points (50,320) and (80, 380) lying on the cost line. Obtain the equation of straight line and substitute x = 110

(y-380)/(380-320) = (x-80)/(80-50)6y - 3y= -660y= 2x + 220Ans: 440/-