Mathematics_Today_-_October_2016

14. (a) : Any line through the middle point M(1, 5) of
    the intercept AB may be taken as

xy− (^15) = − =r
cosθθsin
...(i)
where ‘r’ is the distance of any point (x, y) on the line
(i) from the point M(1, 5).
Since the points A and B are equidistant from M and
on the opposite sides of it, therefore if the co-ordinates
of A are obtained by putting r = d in (i), then the
co-ordinates of B are given by putting r = –d.
Now the point A(1 + dcosθ, 5 + dsinθ) lies on the line
5 x – y – 4 = 0 and point B(1 – dcosθ, 5 – dsinθ) lies
on the line 3x + 4y – 4 = 0.
Therefore, 5(1 + dcosθ) – (5 + dsinθ) – 4 = 0
and 3(1 – dcosθ) + 4(5 – dsinθ) – 4 = 0
Eliminating ‘d’, we get cosθθsin
35 83
=.
Hence the required equation of line is
xy−^1 = −
35
5
83
or 83x – 35y + 92 = 0

15. (c) : Point of intersection is (2, 3). Therefore, the
    equation of line passing through (2, 3) is
    y – 3 = m(x – 2) or mx – y – (2m – 3) = 0 ...(i)
    According to the condition,
    3223
    1

7

5

3

4

4

(^23)
mm
m
−− − m

• ()=± ⇒ = ,
  Hence the equations are
  3 x – 4y + 6 = 0 and 4x – 3y + 1 = 0.

16. (b) : The equation of any line parallel to
    2 x + 6y + 7 = 0 is 2x + 6y + k = 0.
    This meets the axes at A⎛⎝⎜−k ⎠⎞⎟ B⎛⎝⎜ −k⎞⎠⎟
    2

00

6

,, and.

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AB = 10

⇒ kk+=⇒ k=

22 2

436

10 10

36

10

⇒ 10 kk^2 = 3600 ⇒ =±6 10

Hence there are two lines given by
266100 xy+± =

17. (b) : Point P(a, b) is on 3x + 2y = 13
    ⇒ 3 a + 2b = 13 ...(i)
    Point Q(b, a) is on 4x – y = 5
    ⇒ 4 b – a = 5 ...(ii)
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a = 3, b = 2
P(a, b) ≡ (3, 2) and Q(b, a) ≡ (2, 3)

Now, equation of PQ is

yx− =

−

−

2 32 −

23

() 3

⇒ y – 2 = –(x – 3) ⇒ x + y = 5

18. (b) : The point of intersection of lines 2x – 3y + 4 = 0
    and 3x + 4y – 5 = 0 is ⎜⎛⎝−^2 ⎞⎠⎟
    34

22

17

,

The slope of required line =−^7

6

∴ Equation of required line is

yx−^22 =− ⎝⎜⎛ + ⎞⎠⎟

17

7

6

2

34

⇒ 119 x + 102y = 125

19. (a) : Equation of the line passing through (–4, 6)
    and (8, 8) is
    yxyx− = −
    
    
    • 
      
    
    
    
    

⎛

⎝⎜

⎞

(^6) ⎠⎟ + ⇒−=+
86
84
462
12
() () 4
⇒ 6 y – 36 = x + 4 ⇒ 6 y – x – 40 = 0 ... (i)
Now equation of any line perpendicular to (i) is
6 x + y + λ = 0 ... (ii)
This line passes through the mid point of (–4, 6) and
(8, 8) i.e., (2, 7)
∴ 6 × 2 + 7 + λ = 0
⇒ 19 + λ = 0 ⇒ λ = –
From (ii), the required equation of line is
6 x + y – 19 = 0

20. (c) : Obviously, slope of AC = 5/2.
    Let m be the slope of a line  (1,–1)

(3, 4 ) 

45 °

45 °

45 °

45 °

inclined at an angle of 45°

to AC, then

tan 45 ,

5

2

(^152)
7
3
3
7
°=±
−

• ⇒ =−
  m
  m
  m
  Let the slope of AB or DC be 3/7and that of AD or BC
  be –7/3. Then equation of AB is 3x – 7y + 19 = 0.
  Also the equation of BC is 7x + 3y – 4 = 0.
  On solving these equations, we get B≡−⎛
  ⎝⎜
  ⎞
  ⎠⎟
  1
  2
  5
  2
  ,.
  Now let the coordinates of the vertex D be (h, k). Since
  the middle points of AC and BD are same, therefore
  1
  2
  1
  2
  1
  2
  31 9
  2
  ⎛⎜⎝hh− ⎞⎠⎟=+()⇒ =
  And,^1
  2
  5
  2
  1
  2
  ⎛⎜⎝k+ ⎟⎞⎠= () 41 − ⇒ k= 1
  2
  Hence, D≡⎛
  ⎝⎜
  ⎞
  ⎠⎟
  9
  2
  1
  2
  ,.