List of Formulae:
Category: N-Scheme
4.2 INTEGRATION BY SUBSTITUTION
So far we have dealt with functions, either directly integrable using integration formula (or) integrable after decomposing the given functions into sums & differences.
which cannot be decomposed into sums (or) differences of simple functions. In these cases, using proper substitution, we shall reduce the given form into standard form, which can be integrated using basic integration formula.
When the integrand (the function to be integrated) is either in multiplication or in division form and if the derivative of one full or meaningful part of the function is equal to the other function then the integration can be evaluated using substitution method as given in the following examples.
The above integration can be evaluated by taking u = log x.
it can be integrated by taking u =x2 + 3x + 5.
Integrals of the form
and
are all, more or less of the same type and the use of substitution u = f(x) will reduce the given function to simple standard form which can be integrated using integration formulae.
Example:
Soln:
Example:
Soln:
du = 2 dx
= log u + c
Example:
Soln:
Put u = log x
= log u + c
= log ( log x ) + c
Example:
Soln:
du = (2x + 1) dx
= log u + c
Example:
Soln:
Put u = log x
= tan u + c
= tan ( log x ) + c
Example:
Soln:
Put u = sin x
du = cos x dx
Example:
Soln:
du = (2ax + b) dx
4.1 INTEGRATION – DECOMPOSITION METHOD
Sir Sardar Vallabhai Patel, called the Iron Man of India integrated several princely states together while forming our country Indian Nation after independence. Like that in Maths while finding area under a curve through integration, the area under the curve is divided into smaller rectangles and then integrating (i.e) summing of all the area of rectangles together. So, integration means of summation of very minute things of the same kind.
Integration as the reverse of differentiation:
Integration can also be introduced in another way, called integration as the reverse of differentiation
Ex: Suppose we differentiate the function
The symbol for integration is ∫, known as integral sign. Along with the integral sign there is a term dx which must always be written and which indicates the name of the variable involved, in this case ‘x’. Technically integrals of this sort are called indefinite Integrals.
List of Formulae:
Example
Soln:
Example
Soln:
= 100 log x +100 x + c
Example:
Soln:
Example:
Soln:
Example:
Soln:
Trigonometry related formulae:
Example:
Soln:
Example:
Soln:
= tan x – x+ c
Example:
Soln:
= x – cos x + c
Example:
Soln:
Example:
Soln:
Example:
Soln:
APPLICATION OF VECTOR DIFFERENTIATION
is a vector function, defined and differentiable at each point (x, y, z)in a certain region of space [i.e., A defines a vector field], then the divergence of (abbreviated as ‘Div ‘) is defined as,
Basic properties of Divergence:
If A, B are vector functions and ‘f’ is a scalar function, then
Example:
Soln:
= yz + 3x2 ( 1 ) + ( 0 – y3 )
= yz + 3x2 – y3
Example:
Soln:
At ( 1, -1, 2)
= -2 – 4 – 1
= -7
Example:
Soln:
Example:
Soln:
At ( 1, 1, 1)
2 + 3 – 3p = 0
5 – 3p = 0
3p = 5
p = 5/3
Curl of a vector function:
is a vector function,
Irrotational vector :
Example:
Soln:
Example:
Soln:
Example:
Soln:
VECTOR DIFFERENTIATION
Vector point function and Vector field:
Let P be any point in a region ‘D’ of space. Let r be the position vector of P. If there exists a vector function F corresponding to each P, then such a function F is called a vector function and the region D is called a vector field.
Example: consider the vector function
Let P be a point whose position vector is
At P , the value of F is obtained by putting x = 2, Y = I, z = 3 in F
Thus, to each point P of the region D, there corresponds a vector F given by the vector function (I). Hence F is a vector point function (of scalar variables x, y, z) and the region D is a vector field.
Scalar point function and scalar field:
If there exists a scalar ‘f’ given by a scalar function ‘f’ corresponding to each point P (with position vector r) in a region D of space, ‘f’ is called a scalar point function and D is called a scalar field.
Example: let P be a point whose position vector is
Consider f= xyz + xy + z
Then the value of f at P is obtained by putting x = 2, y = I, z = 3
i.e., At P, f= 2.1.3 + 2.1 + 3 = II
Hence the scalar’ II ‘ is attached to the point P.
The function ‘f’ is a scalar point function (of scalar variables x, y, z), and D is a
scalar field.
Note : There can be vector and scalar function of one or more scalar variables.
Vector differential operator
The operator is applied to both vector and scalar functions.
Gradient
defined at each point (x, y, z) in a certain region of space and is differentiable
Basic properties of the Gradient
Example: If f = x2yz, find grad f at the point (1,-2,1)
Soln: f = x2yz
= yz ( 2x )
= x2z ( 1 )
= x2y ( 1 )
Equation (1) becomes
At the point (1,-2, 1),
Example: Find the unit normal to the surface xy +yz + zx= 3 at the point (1, 1, 1).
Soln:
= y ( 1 ) + z ( 1 )
= x ( 1 ) + z ( 1 )
= y ( 1 ) + x ( 1 )
Equation (1) becomes
At the point (1,1, 1),
Example: Find the acute angle between the surface xy2z = 4 and x2 + y2 + z2 =6 at the point (2, 1, 1).
Soln:
Let f = xy2z = 4 be the surface —— (1)
Normal vector to (1) at (2,1,1)
= y2z ( 1)
= x z (2y)
= xy2 (1)
At the point (2,1, 1),
Let g = x2 + y2 + z2 =6 be the surface —— (2)
Normal vector to (2) at (2,1,1)
At the point (2,1, 1),
Angle between the surfaces = Angle between the normal to them
= Angle between a and b
3.1 PRODUCT OF THREE AND FOUR VECTORS
Scalar triple Product
Properties of Scalar triple Product:
Example :
Soln:
= 1 ( 1 – 0 ) – 1 ( 0 – 1 ) + 0 ( 0 – 1 )
= 1 ( 1 ) -1 (- 1 ) + 0
= 1 + 1
= 2
Example :
Soln:
= 2 ( 4 + 2 ) – 1 (3 – 1 ) + 1 ( -6 – 4 )
= 2 ( 6 ) – 1 (2 ) + 1 ( – 10 )
= 12 – 2 – 10 = 10 – 10
The given points are coplanar.
Example :
Soln:
2 ( 10 + 3m ) + 1 ( 5 + 9 ) + 1 ( m – 6 ) = 0
20 + 6m + 14 + m – 6 = 0
7m + 28 = 0
7m = -28
m = – 4
Example :
Soln:
= -4 ( 12 + 3) + 6 ( -3 + 24 ) – 2 ( 1 + 32 )
= -4 (15) + 6 (21) – 2 ( 33 )
= -60 + 126 – 66
= -126 + 126
The given points lie in the same plane.
Vector Triple Product
Note:
Product of Four vectors
Example :
Soln:
Example :
Soln:
= -3 ( 1 ) + 3 ( – 3 )
= -3 -9
= – 12
2.3 APPLICATION OF SCALAR AND VECTOR PRODUCT
Application of Scalar Product
Work done
Example :
Soln:
= 4 ( 2 ) + 9 ( – 7 ) + 4 ( 1 )
= 8 – 63 + 4
Work done = –51 units
Work done = 51 units (by taking positive value)
Example :
is displaced from the point ( 1, 1, 1 ) to the point ( 2, – 3, 5 ). Find the total work done.
Soln:
= 6 ( 1 ) + 10 ( -4 ) – 1 ( 4 )
= 6 – 40 – 4
= – 38
Work done = –38 units
Work done = 38 units (by taking positive value)
Application of Vector Product
Moment (or) Torque of a force about a point

Example :
Soln:
Example :
Soln:
2.2 PRODUCT OF VECTORS

Definition:
Properties of Scalar Product:
Example :
Soln:
= 1(1) + 1(1) + 0
= 1 + 1
= 2
Example :
Soln:
= 2(3) + 3(2) – 2 (6)
= 6 + 6 -12
= 0
Soln:
2(3) + m(1) – 3 (4) = 0
6 + m -12 = 0
m – 6 = 0
m = 6.
Soln:
= 1(0) – 1(4) + 2 (2)
= 0
= 0(-10) + 4(-2) + 2
= 0 – 8 + 8
= 0
= – 10(1) -2 (-1) + 4 (2)
= – 10 + 2 + 8
= 0
Example :
Soln:
Example :
Soln:
= 3 ( 2 ) + 4 ( – 3 ) – 2 ( – 5 )
= 6 – 12 + 10
= 4
Example :
Soln:
=3 ( 7 ) + 1 ( 1) – 2 ( 2 )
= 21 + 1 – 4
= 18.
Vector Product of Two Vectors
Definition

Properties of Vector Product:
4. Vector product in determinant from
Example :
Soln:
Example :
Soln:
Example :
Soln:
Example :
Soln:
Example :
Soln:
2.1 VECTOR – INTRODUCTION
Vectors constitute one of the several Mathematical systems which can be usefully employed to provide mathematical handling for certain types of problems in Geometry, Mechanics and other branches of Applied Mathematics. Vectors facilitate mathematical study of such physical quantities as possess Direction in addition to Magnitude. Velocity of a particle, for example, is one such quantity.
Physical quantities are broadly divided in two categories viz (a) Vector Quantities & (b) Scalar quantities.
( a ) Vector quantities :
Any quantity, such as velocity, momentum, or force, that has both magnitude and direction
( b ) Scalar quantities :
A quantity, such as mass, length, time, density or energy, that has size or magnitude but does not involve the concept of direction is called scalar quantity.
Mathematical Description of Vector: A vector is a directed line segment. The length of the segment is called magnitude of the vector. The direction is indicated by an arrow joining the initial and final points of the line segment. The vector AB i.e, joining the initial point A and the final point B in the direction of AB is denoted as
Triangle Law of Addition of Two vectors:

Position Vector: If P is any point in the space and 0 is the origin, then
is called the position vector of the point P.
Let P be a point in a Three dimensional Space. Let 0 be the origin and i, j and k the unit vectors along the x ,yand z axes . Then if P is (x, y, z) , the position vector of the point P is
Direction Cosines and Direction ratios
When a directed line OP passing through the origin makes α, β and γ angles with the x, y and z axis respectively with O as the reference, these angles are referred as the direction angles of the line and the cosine of these angles give us the direction cosines. These direction cosines are usually represented as l, m and n.

Direction ratio are x, y, z
Example:
Soln:
Distance between two points:
If A and B are two points in the space with co-ordinates A (x1, y1, z1 ) and B (x2, y2, z2), then the position vectors are
Example : If position vectors of the points A and B are
Sol.: Given
If a is a vector, then
Example : Find the unit vector along the vector
Soln:
Condition for three position vectors
Example : Show that the points whose position vectors
Soln: Given
Home work problem
Show that the points whose position vectors
Soln: Given
- Prove that the points
Soln: Given
The given triangle is an equilateral triangle.
2. Prove that the points whose position vectors are
Soln: Given
The given triangle is an right angled triangle.
1.3 CONICS
Conic: A conic is defined as the locus of a point which moves such that its distance from a fixed point is always ‘e’ times its distance from a fixed straight line.
Focus: The fixed point is called the focus of the conic.
Directrix: The fixed straight line is called the directrix of the conic.
Eccentricity: The constant ratio is called the eccentricity of the conic.
General equation of a conic ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents
(i) a circle if a = b and h = 0.
(ii) a parabola if h2 = ab.
(iii) an ellipse if h2 < ab.
(iv) a hyperbola if h2> ab.
PART – A
Example : Prove that the equation x2 + 6xy + 9y2 + 4x + 12y – 5 = 0 is a parabola.
Soln: x2 + 6xy + 9y2 + 4x + 12y – 5 = 0 —————– ( 1 )
Condition for ( 1 ) to represent parabola is h2 = ab
From ( 1 ) a = 1, b = 9
2h = 6 ⇒ h = 3
h2 = ab
32 = 1 ( 9)
9 = 9. ∴ ( 1 ) represents a parabola.
Example : Show that the equation x2 + 2xy + 3y2 + x – y + 1 = 0 represents an
ellipse.
Soln: Given x2 + 2xy + 3y2 + x – y + 1 = 0 –––– (1)
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ––––– (2)
Comparing, we get
a = 1 2h = 2 b = 3
h =1
h2 – ab = (1)2 – 1(3) = 1 – 3 = -2 < 0
Given equation ( 1 ) represents ellipse.
General Equation of a Conic
‘S’ denotes Focus
Line XM denotes Directrix
SP / PM = e
Note:
(i) If e< 1, the conic is called an ellipse.
(ii) If e = 1, the conic is called aparabola.
(iii) If e>1, the conic is called a hyperbola.
Parabola with vertex (0,0), focus- S (a,0), latus rectum and equation of directrix

y2 = 4ax is the equation of the parabola
Types of Parabolas with vertex (h,k), focus, latus rectum and equation of directrix

Part – C
1. Find axis, vertex, focus and equation of directrix for y2 + 8x – 6y + 1 = 0
Soln: y2 – 6y = – 8x – 1
y2 – 6y + 9 = – 8x – 1 + 9 ( Adding 9 on bothsides)
(y – 3)2 = – 8x + 8
= – 8(x – 1)
(y – 3)2 = – 8(x – 1)
This is of the form Y2 = – 8X ( y2 = -4ax) (open leftward)
Where Y = y – 3 and X = x – 1
4a = 8
a = 2
vertex (0 , 0) for X , Y
Y = 0 ⇒ y – 3 = 0
y = 3
X = 0 ⇒ x – 1 = 0
x = 1
The vertex is ( 1, 3)
Focus:
The focus (X = – a,Y = 0)
X = -a
x – 1 = – 2
x = – 1
Y = 0 ⇒ y – 3 = 0
y = 3
Focus is ( -1, 3 )
Equation of directrix is X – a = 0.
i.e., X – 2 = 0
⇒ x – 1 – 2 = 0
⇒ x – 3 = 0
Latus rectum X + a = 0
i.e., x – 1 + 2 = 0 x + 1 = 0
Length of latus rectum = 4a = 4(2) = 8
2) Find vertex, focus , equation of directrix and latus rectum for x2 – 4x – 5y – 1 = 0
Soln: x2 – 4x – 5y – 1 = 0
x2 – 4x = 5y + 1
x2 – 4x + 4 = 5y + 1 + 4 ( Adding 4 on bothsides)
(x – 2)2 = 5y + 5
= 5(y + 1)
(x – 2)2 = 5(y + 1)
This is of the form X2 = 4aY
Where X = x – 2 and Y = y + 1
4a = 5
a = 5/4
Therefore 1. The vertex (X = 0, Y = 0)
X = 0 ⇒ x – 2 = 0 ⇒ x = 2
Y = 0 ⇒ y + 1 = 0 ⇒ y = – 1
The vertex is ( 2, – 1)
2. The focus (X = 0, Y = a)
is (x = 2, y + 1 = 5/4) = (2, 1/4)
3. The directrix equation is Y = -a or y + 1 = -5/4 or 4y + 9 = 0
4. The latus rectum is 4a = 4(5/4) = 5.
1.2 ANALYTICAL GEOMETRY II
EQUATION OF CIRCLE
Definition:
A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle.
Equation of the circle with centre (h, k) and radius ‘r’ units.
CP = r
√(( x – h )2 + (y – k )2 = r (Using distance formula)
(x – h )2 + (y – k )2 = r2
Note:
The equation of the circle with centre (0, 0 ) and radius ‘r’ units is x2 + y2 = r2
Part – A
Example : Find the equation of the circle with centre (- 5, 7 ) and radius 5 units.
Soln: We know that the equation of circle is (x – h )2 + (y – k )2 = r2
Here h = – 5, k = 7 (given) and r = 5
. . (x + 5 )2 + (y + 7 )2 =
x2 + 10x + 25 + y2 + 14y + 49 = 25
x2 + y2 + 10x – 14y + 25 +49-25 = 0
x2 + y2 + 10x – 14y + 49 = 0
Therefore the equation of circle is x2 + y2 + 10x – 14y + 49 = 0
General equation of the circle: x2 + y2 + 2gx + 2fy + c = 0
Centre = (-g , -f ) and radius r = √( g2 + f2 – c)
Part – A
Example : Find the centre and radius of the circle x2 + y2 + 2x + 2y – 7 = 0 .
Soln: Given x2 + y2 + 2x + 2y – 7 = 0 .
We know the equation of circle is x2 + y2 + 2gx + 2fy + c = 0
2g = 2 2f = 2 c = -7
g = 1 f = 1
centre = ( -g , -f ) r = √( g2 + f2 – c)
= ( – 1 , – 1 ) r = √( (1)2 + (1)2 + 7)
r = √9 r = 3
centre = (- 1 , – 1 ) & r = 3 |
Part – B
Example : Find the equation of the circle passing through the point A (2, – 3) and having its centre at
C ( – 5 , 1).
Soln: r = √(( – 5 – 2 )2 + (1 + 3 )2)
= √( (- 7)2 + (4)2 )
= √ ( 49 + 16)
r = √65
We know that the equation of circle is (x – h )2 + (y – k )2 = r2
Here h = -5, k = 1 (given) and r = √65
(x + 5 )2 + (y – 1 )2 = r2
x2 + 10x + 25 + y2 – 2y + 1 = 65
x2 + y2 + 10x – 2y + 26 – 65 = 0
x2 + y2 + 10x – 2y – 39 = 0
Therefore the equation of circle is x2 + y2 + 10x – 2y – 39 = 0
Equation of the Tangent to a circle at the point (x1,y1) on the circle

Equation of the tangent to a circle x2 + y2 + 2gx + 2fy + c = 0 at ( x1, y1) is
x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0
Part – B
Example : Find the equation of the tangent at (4,1) to the circle
x2 + y2 – 8x – 6y + 21 = 0 .
Soln: W.K.T the equation of tangent at ( x1, y1) is
x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0
Given x2 + y2 – 8x – 6y + 21 = 0
We know the equation of circle is x2 + y2 + 2gx + 2fy + c = 0
2g = – 8 2f = – 6 c = 21
g = – 4 f = – 3
x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0
x (4) + y ( 1) – 4 ( x + 4) – 3 ( y + 1) + 21 = 0
4x + y – 4x – 16 – 3y – 3 + 21 = 0
– 2y + 2 = 0
∴ Equation of tangent is y – 1 = 0.
FAMILY OF CIRCLES
Concentric Circles:
Two or more circles having the same centre but differ in radii are called concentric circles.
Note: Equation of concentric circles differ only by the constant term.

Example 1 :Show that the circles x2 + y2 – 4x + 2y +5 = 0 and x2 + y2 – 4x + 2y – 8 = 0
are concentric circles.
Soln: From the two given equations of the circles, we observe that the constant term alone differs
∴The given circles are concentric circles.
Part – A
Example 2 : Find the equation of the circle passing through the point (5 , 4 ) and concentric to the circle
x2 + y2 – 8x + 12y + 15 = 0 .
Soln: Equation of concentric circle be x2 + y2 – 8x + 12y + k = 0 ………….. ( 1 )
Put x = 5, y = 4 in ( 1 )
(5)2 + (4)2 – 8 ( 5 ) + 12 ( 4 ) + k = 0.
25 + 16 – 40 + 48 + k = 0.
89 – 40 + k = 0.
49 + k = 0
K = – 49
The required equation of the circle is x2 + y2 – 8x + 12y – 49 = 0
Orthogonal Circles:
Two circles x2 + y2 + 2g1x +2f1y+c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 are said to be Orthogonal Circles if
2g1g2 + 2 f1 f2 = c1 + c2
Part – B
Example :Prove that the circles x2 + y2 – 4x + 6y + 4 = 0 and
x2 + y2 + 2x + 4y + 4 = 0 cut orthogonally.
Soln: Given x2 + y2 – 4x + 6y + 4 = 0 ––––– (1) and
x2 + y2 + 2x + 4y + 4 = 0 ––––– (2)
From ( 1 )
2 g1 = – 4 2 f1 = 6 c1 = 4
g1= – 2 f1= 3
From ( 2 )
2 g2 =2 2 f2 = 4 c2 = 4
g2= 1 f2= 2
The condition for orthogonally is
2g1g2 + 2 f1 f2 = c1 + c2
2(-2)(1) + 2(3)( 4+ 4
-4 + 12 = 8
8 = 8
∴ The circles cut orthogonally.
Contact of Circles:
Case ( i ) :
Two circles touch externally if the distance between their centres is equal to sum of their radii.
i.e C1C2 = r1 + r2
Co-ordinates of point of contact are
P = ( (r1 x2 + r2 x1 ) / (r1 + r2 ) , (r1 y2 + r2 y1 ) / (r1 + r2 ) )
where C1 = (x1, y1) and C2 = (x2, y2)
Case ( ii ) :
Two circles touch internally if the distance between their centres is equal to difference of their radii.
i.e C1C2 = r1 – r2 or r2 – r1
Co-ordinates of point of contact are
P = ( (r1 x2 – r2 x1 ) / (r1 – r2 ) , (r1 y2 – r2 y1 ) / (r1 – r2 ) )
where C1 = (x1, y1) and C2 = (x2, y2)
Part – C
1. Prove that the circles x2 + y2 + 2x – 4y – 3 = 0 and x2 + y2 – 8x + 6y +7 = 0 touch each other.
Soln: Given x2 + y2 + 2x – 4y – 3 = 0 ––––– (1) and
x2 + y2 – 8x + 6y +7 = 0 ––––– (2)
From ( 1 )
2g1 = 2 2f1 = – 4 c1 = -3
g1 = 1 f1= – 2
centre is C1 = ( – g1 , – f1 ) r1 = √( g12 + f12 – c)
= ( -1 , 2 ) r1 = √( (1)2 + (-2)2 + 3)
r1 = √( 1 + 4 +3 )
r1 = √8 = 2√2
C1= (-1 , 2) & r1 = 2√2 |
From ( 2 )
2g2 = -8 2f2 = 6 c2 = 7
g2 = – 4 f2 = 3
centre is C2 = ( – g2 , – f2 ) r2 = √( g22 + f22 – c)
= ( 4 , -3 ) r2 = √( (-4)2 + (3)2 – 7)
r2 = √( 16 + 9 – 7 )
r2 = √18 = √(9 × 2) = 3√2
C2= ( 4 , -3 ) & r2 = 3√2 |
C1C2 = √(( – 1 – 4 )2 + (2 + 3 )2)
= √( (- 5)2 + (5)2 )
= √( 25 + 25)
= √50 = √(25 × 2) = 5√2
C1C2 = 5√2
r1 + r2 = 2√2 + 3√2
= 5√2 = C1C2
∴The circles touch each other externally.
2. Prove that the circles x2 + y2 – 4x + 6y – 112= 0 and x2 + y2 – 10x – 6y + 14 = 0 touch each other.
Soln: Given x2 + y2 – 4x + 6y – 112 = 0 ––––– (1) and
x2 + y2 – 10x – 6y + 14 = 0 ––––– (2)
From ( 1 )
2g1 = -4 2f1 = 6 c1 = -112
g1= -2 f1= 3
centre is C1 = ( – g1 , – f1 ) r1 = √( g12 + f12 – c)
= ( 2 , -3 ) r1 = √( (-2)2 + (3)2 + 112)
r1 = √( 4 + 9 + 112 )
r1 = √125 = √(125 × 5) = 5√5
C1= ( 2 , -3 ) & r1 = 5√5 |
From ( 2 )
2 g2 = – 10 2 f2 = -6 c2 = 14
g2 = -5 f2 = – 3
centre is C2 = ( – g2 , – f2 ) r2 = √( g22 + f22 – c)
= ( 5 , 3 ) r2 = √( (-5)2 + (-3)2 – 14)
r2 = √( 25 + 9 – 14 )
r2 = √20 = √(4 × 5)= 2√5
C2= ( 5 , 3 ) & r2 = 2√5 |
C1 C2 = √(( 5 – 2 )2 + (3 + 3 )2)
= √( (3)2 + (6)2 )
= √( 9 + 36)
C1 C2 = √45 = √(9× 5) = 3√5
r1 – r2 = 5√5 –
= 3√5 = C1 C2
∴The circles touch each other internally.
Tamil Nadu Diploma Engineering Mathematics – II Unit – I ( 1.1 – Analytical Geometry – I) material 2020-21 ( N-Scheme)
UNIT – I ANALYTICAL GEOMETRY
1.1 ANALYTICAL GEOMETRY I
Straight Line:
When a variable point moves in accordance with a geometrical law, the point will trace some curve. This curve is known as the locus of the variable point.
If a relation in x and y represent a curve then
(i) The co-ordinates of every point on the curve will satisfy the relation.
(ii) Any point whose co-ordinates satisfy the relation will lie on the curve.
Straight line is a locus of a point.

Slope or gradient of a straight line:
The tangent of the angle of inclination of the straight line is called slope or gradient of the line. If θ is the angle of inclination then slope = tan θ and is denoted by m.
i.e m = tan θ
Equation of a straight line:

When ‘c’ is the y intercept and slope is ‘m’, the equation of the straight line is
y = mx+c
Formulae
1) The length of the perpendicular distance from (x1,y1) to the line ax + by + c = 0 is
± (ax1 + by1 + c)/ √(a2 + b2 )
2) The distance between the parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is
± (c1 – c2 )/ √(a2 + b2 )
ANGLE BETWEEN TWO STRAIGHT LINES:
Book Work:
Find the angle between the lines y = m1 x+c1 and y = m2x+c2. Deduce the conditions for the lines to be
(i) parallel (ii) perpendicular.
Proof:

Let θ1 be the angle of inclination of the line y = m1x+c1.
Slope of this line is m1 = tanθ1.
Let ‘θ’2 be the angle of inclination of the line y=m2x+c2 .
Slope of this line is m2 = tanθ2 .
Let ‘θ’ be the angle between the two lines, then θ1 = θ2 + θ
Θ = θ1 – θ2
∴tan θ = tan (θ1-θ2)
= (tan θ1- tan θ2)/(1 + tan θ1 tan θ2 )
tan θ = (m1 – m2)/(1 + m1 m2)
θ = tan^(-1)((m1 – m2)/(1 + m1 m2))
(i) Condition for two lines to be parallel:
If the two lines are parallel then the angle between the two lines is zero
∴ tan θ = tan 0 = 0
(i.e) = 0
m1 – m2 = 0
∴ m1 = m2
(ii)Condition for two lines to be perpendicular:
If the two lines are perpendicular then the angle between them
Θ = 900
∴ tan θ = tan 900 = ∞ = 1/0
(m1 – m2)/(1 + m1 m2) = 1/0
1 + m1 m2 = 0
m1 m2 = – 1
∴ For perpendicular lines, product of the slopes will be -1
Note: 1) Any line parallel to the line ax+by+c = 0
will be of the form ax+by+ k = 0 (differ only constant term)
2) Any line perpendicular to the line ax+by+c =0
will be of the form bx – ay + k = 0
WORKED EXAMPLES
Part – A
. 1. Find the perpendicular distance from the point (2,3) to the straight line 2x+y+3=0.
Soln: W.K.T The length of the perpendicular distance from (x1,y1) to the line ax + by + c = 0 is
± (ax1 + by1 + c)/ √(a2 + b2 )
Given straight line is 2x+y+3=0
Given point (x1,y1) = (2,3)
i.e 2(2 ) + 1(3) + 3/ √(22 + 12 ) = 10 / √5
2. Find the distance between the line 2x+3y+4 = 0 and 2x+3y -1 = 0
Soln: W.K.T the distance between the parallel lines
ax + by + c1 = 0 and ax + by + c2 = 0 is
± (c1 – c2 )/ √(a2 + b2 )
Here a = 2, b = 3 , c1 = 4 and c2 = – 1
i.e 4 + 1/√(22 + 32 ) = 5 /√13
3. Show that the lines 6x+y-11 =0 and 12x+2y+14 =0 are parallel
Soln: 6x+y-11 = 0 (1)
12x+2y+14 = 0 (2)
Slope of the line (1) = m1 = – a/b
= – 6/1
m1 = – 6
Slope of the line (2) = m2 = – a/b
= – 12/2
= – 6
m1 = m2
∴ The lines are parallel.
4. Find ‘p’ such that the lines 7x-4y+13 = 0 and px = 4y+6 are parallel.
Soln: 7x-4y+13 = 0 (1)
px – 4y+ 6 = 0 (2)
Slope of the line (1) = m1 = – 7/-4
m1 = 7/4
Slope of the line (2) = m2 = – p/-4
m2 = p/4
Since (1) and (2) are parallel lines
m1 = m2
7/4 = p/4
4p = 28
∴ p = 7
5. Show that the lines 2x+3y-7 =0 and 3x- 2y+4 =0 are perpendicular.
Soln: 2x+3y-7 =0 (1)
3x- 2y+4 =0 (2)
Slope of the line (1) = m1 = – a/b
m1 = – 2/3
Slope of the line (2) = m2 = – 3/-2
m2 = 3/2
m1 m2 = (- 2/3 ) × (3/2)
= – 1
∴ The lines (1) and (2) are perpendicular.
6. Find the value of m if the lines 2x + my = 4 and x + 5y – 6 = 0 are perpendicular.
Soln: 2x + my – 4 = 0 (1)
x + 5y – 6 = 0 (2)
Slope of the line (1) = m1 = – 2/m
Slope of the line (2) = m2 = – 1/5
Since (1) and (2) are perpendicular
m1 m2 = – 1
(- 2/m ) (- 1/5) = -1
( 2/5m ) = – 1
– 5m = 2
∴ m = – 2/5
Part – B
1. Find the angle between the lines y = √3 x and x- y = 0
Sol: y = √3 x
√3 x – y = 0 (1)
x- y = 0 (2)
Slope of the line (1) = m1 = – coefficient of x / coefficient of y
= – √3 / – 1
m1 = √3
tan θ1 = √3 ⇒ θ1 = 600
Slope of the line (2) = m2 = – coefficient of x / coefficient of y
= – 1 / – 1
m2 = 1
tan θ2 = 1 ⇒ θ2 = 450
∴ Θ = θ1 – θ2
Θ = 600– 450
Θ = 150
2. Find the angle between the lines 3x + 6y = 8 and 2x = – y + 5
Sol: 3x + 6y = 8
3x + 6y – 8 = 0 (1)
2x + y – 5 = 0 (2)
Slope of the line (1) = m1 = – coefficient of x / coefficient of y
= – 3 / 6
m1 = – 1 / 2
Slope of the line (2) = m2 = – coefficient of x / coefficient of y
= – 2 / 1
m2 = – 2
W.K.T tan θ = (m1 – m2)/(1 + m1 m2)
tan θ = (1 / 2 + 2)/(1 + ( – 1/2 ) ( – 2 ))
tan θ = 3/2/2
tan θ = 3/4
3. Find the equation of the straight line passing through (-1,4) and
parallel to x + 2y = 3.
Soln: Let the equation of line parallel to x+2y-3 = 0 (1)
is x+2y+k = 0 (2)
Equation (2) passes through (-1,4)
put x=-1, y =4 in equation (2)
(-1) + 2(4) +k = 0
-1 + 8 + k = 0
k = -7
∴ Required line is x +2y – 7 = 0
4. Find the equation of the line passing through (3,- 3) and
perpendicular to 4x – 3y – 10 = 0.
Soln: Let the equation of line perpendicular to 4x – 3y – 10 = 0 (1)
is – 3x – 4y + k = 0 (2)
Equation (2) passes through (3,-3)
put x = 3, y =- 3 in equation (2)
-3 (3) – 4(-3) +k = 0
-9 + 12+ k = 0
k = – 3
– 3x – 4y – 3 = 0
∴ Required line is 3x +4y + 3 = 0
PAIR OF STRAIGHT LINES
Pair of Straight lines passing through origin
The general equation of pair of straight lines passing through origin is
ax2 + 2hxy + by2 = 0 ———- ( A )
Let y = m1x
y – m1x = 0 ————– ( 1 )
and y = m2x
y – m2x = 0 ————– ( 2 )
(y – m1x) (y – m2x ) = 0
y2 – (m1 + m2 ) xy + m1 m2 x2 = 0
m1 m2 x2 – (m1 + m2 ) xy + y2 = 0 ————– ( B )
Equations (A) and ( B ) represent the same pair of straight lines. Hence the ratios
of the corresponding coefficients of like terms are proportional.
m1 m2 / a = – (m1 + m2 ) / 2h = 1 / b
m1 + m2 = – 2h / b and m1 m2 = a / b
i.e sum of slopes = – 2h / b and product of slopes = a / b
Formulae
1 ) The angle between the pair of straight lines ax2 + 2hxy + by2 = 0 is
tanθ = ± (2√h2 – ab ) / a + b
2) The condition for the pair of lines to be parallel is h2 – ab = 0.
3) The condition for the pair of lines to be perpendicular is a + b = 0.
Part – A
1. Write down the combined equation of the pair of lines
x − 2y = 0 and 3x + 2y = 0
Soln: The two separate lines are x − 2y = 0 and 3x + 2y = 0
The combined equation is
(x − 2y) (3x + 2y) = 0
3x2 + 2xy – 6xy – 4 y2 = 0
3x2 – 4xy – 4 y2 = 0
2. Write down the separate equations of the pair of lines 12x2 + 7xy – 10y2 = 0
Soln: 12x2 + 7xy – 10y2 = 0
12x2 + 15xy – 8xy – 10y2 = 0
3x( 4x + 5 ) – 2y( 4x – 5y) = 0
(3x – 2y) ( 4x – 5y) = 0
∴ The separate equations are3x – 2y = 0 and 4x – 5y = 0
3. Show that the two lines represented by 4x2 + 4xy + y2 = 0 are parallel to each
other.
Soln: 4x2 + 4xy + y2 = 0
This is of the form ax2 + 2hxy + by2 = 0
a = 4, 2h = 4 h = 2, b = 1
If the lines are parallel then h2 – ab = 0
h2 – ab = (2)2 – (4) ( 1 )
= 4 – 4
= 0
∴ pair of lines are parallel
4. Find the value of ‘p’ if the pair of lines 4x2 + pxy + 9y2 = 0 are parallel to each
Other.
Soln: 4x2 + pxy + 9y2 = 0
This is of the form ax2 + 2hxy + by2 = 0
a = 4, 2h = p h = p / 2, b = 9
Given lines are parallel
h2 – ab = 0
( p /2)2 – ( 4 ) ( 9 ) = 0
P2 / 4 – 36 = 0
P2 = 144
∴ p = ± 12
5. Prove that the two lines represented by 7x2 – 48xy – 7y2 = 0 are perpendicular to
each Other.
Soln: 7x2 – 48xy – 7y2 = 0
This is of the form ax2 + 2hxy + by2 = 0
a = 7, 2h = – 48 h = – 24, b = – 7
If the lines are perpendicular then a + b = 0
a + b = 7 – 7
= 0
∴ pair of lines are perpendicular.
6. . Find the value of ‘p’ if the pair of lines px2 – 5xy + 7y2 = 0 are perpendicular
to each Other.
Soln: px2 – 5xy + 7y2 = 0
This is of the form ax2 + 2hxy + by2 = 0
a = p, 2h = – 5 h = – 5 / 2, b = 7
Given lines are perpendicular
a + b = 0
p + 7 = 0
∴ p = – 7
Part – B
1. Find the separate equations of the line 2x2 – 7xy + 3y2 = 0. Also find the angle
between them.
Soln: 2x2 – 7xy + 3y2 = 0
2x2 – 6xy – xy + 3y2 = 0
2x(x – 3y ) – y( x – 3y) = 0
(2x – y) ( x – 3y) = 0
∴ The separate equations are 2x – y = 0 and x – 3y = 0
2x2 – 7xy + 3y2 = 0
This is of the form ax2 + 2hxy + by2 = 0
a = 2, 2h = – 7 h = – 7 / 2, b = 3
Let θ be the angle between the two straight lines
W.K.T tanθ = ± (2√h2 – ab ) / a + b
= ± (2√ (( – 7 / 2)2 – (2) (3) )/ (2 + 3)
= ± (2 √ (49 / 4 – 6 ) / 5
= ± (2 √ (49 – 24) / 4 ) / 5
= ± (2 √ 25 / 4 ) / 5
= ± (2 ( 5 / 2 ) / 5
tanθ = ± 1
tan θ = tan 45, ∴ θ = 45°
2. The slope of one of the lines ax2 + 2hxy + by2 = 0 is thrice that of the other. Show that 3h2= 4ab
Soln: ax2 + 2hxy + by2 = 0 ——————– ( 1 )
Let y = m1x and y = m2x be the separate equations of equation ( 1 )
m1+ m2 = – 2h / b ——————— ( 2 )
m1m2 = a / b ———————- ( 3 )
Slope of one of the line = thrice slope of the other line
i.e m1= 3m2
Equation (2) becomes
3m2 + m2 = – 2h / b
4m2 = – 2h / b
m2 = – 2h / 4b
m2 = – h / 2b
Substitute m1= 3m2 in equation (3 )
3m2m2 = a / b
3 (m2 )2 = a / b
3 (- h / 2b )2 = a / b
3 (h2 / 4b2 ) = a / b
3h2 / 4b2 = a / b
3h2b = 4ab2
(i.e) 3h2 = 4ab
PAIR OF STRAIGHT LINES NOT PASSING THROUGH THE ORIGIN
Formulae
1) The angle between the pair of straight lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
is tanθ = (2√h2 – ab ) / a + b
2) The condition for the pair of lines to be parallel is h2– ab = 0.
3) The condition for the pair of lines to be perpendicular is a + b = 0.
4) Condition for the second degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 to
represent a pair of straight lines is
abc + 2fgh – af2 – bg2 – ch2 = 0
Part – C
1. Show that the equation 3x2 + 7xy + 2y2 + 5x + 5y + 2= 0 represents a pair of
straight lines.
Soln: Given 3x2 + 7xy + 2y2 + 5x + 5y + 2= 0
Comparing with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
a = 3, 2h = 7 , b = 2, 2g = 5 , 2f = 5 , c = 2
h = 7/ 2 g = 5/ 2 f = 5/ 2
To show abc + 2fgh – af2 – bg2 – ch2 = 0
abc + 2fgh – af2 – bg2 – ch2 =
(3) (2) (2) + 2 (5/2) (5/2) (7/2) – 3 (5/ 2)2 – 2 (5/ 2)2 – 2 (7/ 2)2
= 12 + 175/4 – 75/4 – 50/4 – 98/4
= 12 + ( 175 – 75 – 50 – 98)/4
= 12 + ( 175 – 223)/4
= 12 + ( -48 /4 )
= 12 – 12 = 0
Hence, the given equation represents pair of straight lines.
2. Find K if 2x2 – 7xy + 3y2 + 5x – 5y + k= 0 represents a pair of
straight lines.
Soln: Given 2x2 – 7xy + 3y2 + 5x – 5y + k= 0 ————– ( 1 )
Comparing with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
a = 2, 2h = – 7 , b = 3, 2g = 5 , 2f = – 5 , c = k
h = – 7/ 2 g = 5/ 2 f = – 5/ 2
Given equation ( 1 ) represents a pair of straight lines
i.e abc + 2fgh – af2 – bg2 – ch2 = 0
(2) (3) (k) + 2 (- 5/2) (5/2) (-7/2) – 2 (- 5/ 2)2 – 3 (5/ 2)2 – k (- 7/ 2)2 = 0
6k + 175/4 – 50/4 – 75/4 – 49k/4 = 0
(24k – 49k )/4 + ( 175 – 50 – 75)/4 = 0
– 25k/4 + 50/4 = 0
( -25k + 50 ) / 4 = 0
-25 k = – 50
k = 2