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.

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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

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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 )

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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: 

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                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))

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            (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

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              (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  

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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 

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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. 

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    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.    

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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

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 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     

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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

Ustraa [CPS] IN

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

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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

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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.

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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

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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.

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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 

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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

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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

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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.

Decathlon [CPS] IN

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

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