# 1.1 – N – Analytical Geometry Exercise Problems With Solutions

Part – A

1.  Find the perpendicular distance from the point (3, -5) to the straight line 3x-4y-26=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     3x-4y-26 =0

Given point (x1,y1)  =  (3, -5)

i.e  ±  3(3) – 4(-5) – 26/ √(32 + (-4)2 )   =   3 / 5

2.     Find the distance between the line 3x+4y = 9 and 6x+8y = 15.

Soln: W.K.T  the  distance between the parallel lines

ax + by + c1 = 0  and  ax + by + c2 = 0  is

±  (c–  c2 )/  √(a2 + b2 )

6x+8y = 15

2 ( 3x+4y ) = 15

3x + 4y = 15/2

Here  a = 3,  b = 4  ,   c1 =  9   and   c2 =  15/2

i.e   9 – 15/2/√(32 + 42 )     =   3 /10

3. Show that the lines 3x+2y+9=0 and 12x+8y-15 =0 are parallel.

Soln: 3x+ 2y+9 = 0                                                                       (1)

12x+2y+14 = 0                                                                (2)

Slope of the line (1) = m1 =  – a/b

=  – 3/2

Slope of the line (2) = m2 =  – a/b

=  – 12/8

=  – 3/2

m=  m2

∴  The lines are parallel.

4.     Find ‘p’ such that the lines 3x+4y = 8 and px + 2y = 7 are parallel.

Soln:                3x+4y = 8                                                                       (1)

px + 2y = 7                                                                (2)

Slope of the line (1) = m1 =  – 3/4

Slope of the line (2) = m2 =  – p/2

Since (1) and (2) are parallel lines

m=  m2

-3/4  =  – p/2

∴        p   =  3/2

5.    Show that the lines 27x-18y+25 =0 and 2x + 3y+7 =0 are perpendicular.

Soln:                27x-18y+25 =0                                                                       (1)

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

Slope of the line (1) = m1 =  – a/b

m1  =  – 27/-18

m1  =  3/2

Slope of the line (2) = m2 =  – 2/3

m1  m2 =  ( 3/2 ) ×  ( – 2/3 )

=  – 1

∴  The lines (1) and (2) are perpendicular.

6.    Find the value of k if the lines 2x + ky -11 =0  and  5x – 3y + 4 = 0  are perpendicular.

Soln: 2x + ky -11 = 0                                                                        (1)

5x – 3y + 4 = 0                                                                         (2)

Slope of the line (1) = m1 =  – 2/k

Slope of the line (2) = m2 =  – 5/-3

m2 =  5/ 3

Since (1) and (2) are  perpendicular

m1 m2  =  – 1

(- 2/k ) ( 5/3)  =   -1

( -10/3k )    =  – 1

10     =  3k

∴           k    = 10/3

7. Write down the combined equation of the pair of lines      2x + y  = 0  and  3x – y = 0.

Soln:    The two separate lines are 2x + y   = 0  and  3x – y = 0

The combined equation is

(2x + y) (3x – y)  = 0

6x2 – 2xy + 3xy  – y2 =  0

6x+ xy  – y2 =  0

8. Write down the separate equations of the pair of lines 3y2 + 7xy =  0.

Soln:   3y2 + 7xy =  0

y ( 3y + 7x) = 0

y = 0, 3y + 7x = 0

∴ The separate equations are y = 0   and    3y + 7x = 0

9. Find the value of  ‘k’  if the pair of  lines kx+ 4xy  – 4y2 =  0  are perpendicular to each  Other.

Soln:    kx+ 4xy  – 4y2 =  0

This is of the form   ax2  +  2hxy  + by2 =  0

a = k,   2h = 4     h = 2,   b  = -4

Given lines are perpendicular

a + b = 0

k  –  4  =  0

∴  k  =  4

https://clnk.in/qfv5

Part – B

1.    Find the angle between the lines √3 x + y = 1  and  x + √3y = 1 .

Sol:      √3 x + y = 1                                              (1)

x + √3y = 1                                                (2)

Slope of the line (1) = m1 =    – coefficient of  x /  coefficient of y

=   – √3   /  1

Slope of the line (2) = m2 =    – coefficient of  x /  coefficient of y

=   – 1  /  √3

$W.K.T \ tan\ \theta\ = \displaystyle\left\lvert\frac{m_1-m_2}{1 + m_1m_2} \right\rvert$
$tan\ \theta\ = \displaystyle\left\lvert\frac{\frac{-\sqrt{3}}{1} + \frac{1} {\sqrt{3}}}{1 + \frac{-\sqrt{3}}{1}\frac{1} {\sqrt{3}}} \right\rvert$
$tan\ \theta\ = \displaystyle\left\lvert\frac{-3 + 1}{2\sqrt{3}}\right\rvert$
$tan\ \theta\ = \displaystyle\left\lvert\frac{-2}{2\sqrt{3}}\right\rvert$
$tan\ \theta\ = \displaystyle\left\lvert\frac{-1}{\sqrt{3}}\right\rvert$
$tan\ \theta\ = \frac{1}{\sqrt{3}}$

2.    Find the equation of the straight line passing through (3,5)  and parallel to x – 2y -7 = 0.

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

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

Equation (2) passes through (3 ,5)

put x=3, y = 5 in equation (2)

3 – 2(5) +k = 0

3 – 10 + k = 0

k – 7 = 0

k = 7

∴ Required line is x – 2y + 7 = 0

3.    Find the equation of the  line passing through (2, 3)  and perpendicular  to 4x – 3y = 10 .

Soln:   Let the equation of line perpendicular to  4x – 3y = 10               (1)

is – 3x  – 4y + k = 0                                                                                  (2)

Equation (2) passes through (2, 3)

put x = 2, y = 3  in equation (2)

-3 (2) – 4(3) +k = 0

-6 – 12 + k = 0

k = 18

– 3x  – 4y + 18 = 0

∴ Required line is 3x +4y + 18 = 0

https://clnk.in/qfCP

Part – C

1. Show that the equation  2x2  –  7xy  + 3y2 + 5x – 5y + 2= 0  represents a pair of straight lines.

Soln:  Given  2x2  –  7xy  + 3y2 + 5x – 5y + 2= 0

Comparing with  ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0

a = 2,   2h = – 7   ,  b = 3,    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

(2) (3) (2)  + 2 (-5/2) (5/2) (-7/2) – 2 (-5/ 2)2  – 3 (5/ 2)2  – 2 (-7/ 2)2

= 12  +  175/4   –  50/4  – 75/4  – 98/4

= 12  +  ( 175 – 50 – 75– 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   3x2 + 7xy  + ky2 – 4x – 13y – 7= 0  represents a pair of  straight lines.

Soln: Given  3x2 + 7xy  + ky2 – 4x – 13y – 7= 0 ————– ( 1 )

Comparing with  ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0

a = 3,   2h = 7           ,  b = k,    2g = – 4       ,    2f = – 13                 ,   c = – 7

h = 7/ 2                     g = – 2           f =  – 13/ 2

Given equation ( 1 ) represents a pair of straight lines

i.e     abc  +  2fgh – af2 – bg2 – ch2 =   0

(3) (k) (-7)  + 2 (- 13/2) (-2) (7/2) – 3 (- 13/ 2)2  – k (-2)2  + 7 ( 7/ 2)2 = 0

-21k  + 91   –  507/4  -4 k  + 343/4   =  0

(-84 k  + 364 – 507 – 16k + 343 )/4  =  0

-100k + 707 – 507 = 0

-100k + 200 = 0

k  =  2

https://clnk.in/qfxL