Mathematics_Today_-_October_2016

SOLUTIONS

1. (b) : A line perpendicular to the line 5x – y = 1 is
   given by x + 5y – λ = 0 = L, (given)

In intercept form : xy
λλ

+=

/

1

So, area of triangle =×⎛⎝⎜ ⎞⎠⎟= ⇒ =±

1

25

()λ λ 552 λ

Hence, the equation of required straight lines is
xy+=± 552

2. (b) : Mid point of

AB≡E (^) ⎝⎜⎛aabb+ ′ + ′⎞⎠⎟
22
,
and mid point of
(, )   (, )
CD≡F⎛aabb′−−′  (,–)^ (– , )
⎝⎜
⎞
22 , ⎠⎟
.
Hence equation of line EF is
y bb bb bb
a aaa
−⎛⎝⎜ + ′⎞⎟⎠= − ′−−′ x aa
′−−−′
⎛⎝⎜ − + ′⎞⎠⎟
(^22)
On simplification, we get 2ay – 2b′x = ab – a′b′

3. (b) : Slope of DE= −
   −

(^73) =
51
1
⇒ Slope of AB = 1

(–5, 7) (5, 7)
  
(1, 3)
Hence equation of AB is
y – 7 = (x + 5)
⇒ x – y + 12 = 0

4. (d) : The given line is bx – ay = ab ...(i)
   Obviously it cuts x-axis at (a, 0). The equation of line
   perpendicular to (i) is ax + by = k, but it passes through
   (a, 0) ⇒ k = a^2.
   Hence required equation of line is ax + by = a^2

ie x

b

y

a

a

b

.. +=

5. (a) : Slope ()m bb

aa

aa

bb

= ′−−

′−

= ′−

− ′

(^1).
Mid point of the given points is ⎜⎝⎛aabb+ ′ + ′⎞⎠⎟
22
,
Therefore required equation of line is
y bb a a
bb
−⎜⎛⎝ + ′⎞⎠⎟= ′− x aa
− ′
⎛⎝⎜ − + ′⎞⎠⎟
22
⇒ 2(b – b′)y + 2(a – a′)x – b^2 + b′^2 – a^2 + a′^2 = 0

6. (a) : The equation of any straight line passing through
   (3, –2) is y + 2 = m(x – 3) ...(i)
   The slope of the given line is − 3.

So, tan ()

()

60 3

13

o m

m

=± −−

+ −

On solving, we get m = 0 or 3

Putting the values of m in (i), the required equation of

lines are y + 2 = 0 and 3233 xy− =+.

7. (a) : The gradient of line 2x + 3y + 4 = 0 is –2/3.
   Now the equation of line passing through (–1, 1) and
   having slope m=−
   −

(^1) =
23
3
/ 2
is 2(y – 1) = 3(x + 1).

8. (a) : Let the intercepts be a and 2a, then the equation
   of line is x
   a

y

a

+=

2

1 , but it also passes through (1, 2),

therefore a = 2.

Hence the required equation is 2x + y = 4.

9. (a) : Intersection point on x-axis is (2x 1 , 0) and on
   y-axis is (0, 2y 1 ). Thus equation of line passes through
   these points is x
   x

y

11 y

+= 2

10. (b) : The point of intersection of the lines
    4 x – 3y – 1 = 0 and 5x – 2y – 3 = 0 is (1, 1). The
    equation of line parallel to 2y – 3x + 2 = 0 is
    2 y – 3x + k = 0. It also passes through (1, 1), therefore
    k = 1. Hence the required equation is 2y – 3x + 1 = 0
    or 3x – 2y = 1


11. (c) : Since the hour, minute
    and second hands always pass
    through origin because one
    end of these hands is always at
    origin. Now at 4 O’ clock, the
    hour hand makes 30° angle in
    fourth quadrant.



 











So the equation of hour hand is

ymx y= ⇒ =−^1 x

3

⇒ xy+= 30

12. (a) : Slope =− 3
    ∴ Equation of line is yxc xyc=−^33 + ⇒ +=
    Now

c cxy

2

=|| 4838 ⇒ =± ⇒ +=±

13. (c) : Equation of any line through (0, a) is
    y – a = m(x – 0) or mx – y + a = 0 ... (i)
    If the length of perpendicular from (2a, 2a) to the line
    (i) is ‘a’, then a ma a a
    m

=± − + m

+

() (^22) ⇒ = ,
1
04
(^23)
.
Hence the required equation of lines are
y – a = 0, 4x – 3y + 3a = 0