# 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.

https://clnk.in/qfwg

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

±  (c–  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:

https://clnk.in/qfwK

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 (θ12)

=       (tan θ1- tan θ2)/(1 + tan θ1 tan θ2 )

tan θ   =  (m1 – m2)/(1 + m1 m2)

θ   =  tan^(-1)⁡((m1 – m2)/(1 + m1 m2)) https://clnk.in/qfwL

(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

∴   m=  m2 https://clnk.in/qfxA

(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  https://clnk.in/qfxB

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

±  (c–  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

m=  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

m=  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 https://clnk.in/qfxF

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 parallel to the line 3x + 2y -7=0 and passing through the point (1, -2).

Soln:   Let the equation of line parallel to   3x + 2y -7=0                    (1)

is 3x + 2y +k = 0                                                                                 (2)

Equation (2) passes through ( 1, -2 )

put x= 1, y =-2 in equation (2)

3(1) + 2(-2) +k = 0

3 – 4 + k = 0

k = 1

∴ Required line is 3x + 2y +1 = 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 https://clnk.in/qfxL

Part – C

$1.\ If\ the\ two\ straight\ lines\ 2x – 3y + 9 = 0\ and\ 6x + ky + 9 = 0\ are\ parallel\ then\ find\ the\ value\ of\ ‘k’$$and\ also\ find\ the\ distance\ between\ them$

Soln:

$Given\ 2x – 3y + 9 = 0\ ———- —————– (1)$
$Given\ 6x + ky + 9 = 0\ ———- —————– (2)$
$From\ (1)\ m_1 = \frac{3}{2}$
$From\ (2)\ m_2 = \frac{-k}{6}$
$Given\ (1)\ and\ (2)\ are\ parallel\ i.e\ m_1=m_2$
$\frac{3}{2} = \frac{-k}{6}$
$18 = -2k$
$k = -9$
$Equation\ (2)\ becomes\ 6x – 9y + 9 = 0$
$3(2x – 3y + 3) = 0$
$To\ find\ distance\ between\ parallel\ lines\ \hspace10cm$
$d = \frac{c_1 – c_2}{\sqrt{a^2 + b^2}}$
$Here\ a = 2,\ b = -3,\ c_1 = 9,\ c_2 = 3$
$d = \frac{9 – 3}{\sqrt{2^2 + (-3)^2}}$
$d = \frac{6}{\sqrt{4 + 9}}$
$d = \frac{6}{\sqrt{13}}$

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.  https://clnk.in/qfxR

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

3x– 4xy  – 4 y2 =  0

2.        Write down the separate equations of the pair of lines 12x+ 7xy  – 10y2 =  0

Soln:   12x+ 7xy  – 10y2 =  0

12x+ 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 4x+ 4xy  + y2 =  0  are parallel to each

other.

Soln:    4x+ 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 4x+ pxy  + 9y2 =  0  are parallel to each

Other.

Soln:    4x+ 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 7x– 48xy  – 7y2 =  0  are perpendicular to

each  Other.

Soln:    7x– 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 px– 5xy  – 7y2 =  0  are perpendicular

to each  Other.

Soln:    px– 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 https://clnk.in/qfx3

Part – B

1.     Find the separate equations of the line 2x– 7xy  + 3y2 =  0.  Also find the angle

between them.

Soln:   2x– 7xy  + 3y2 =  0

2x– 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

2x– 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 https://clnk.in/qfx6

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  https://clnk.in/qfx8