Mathematics_Today_-_October_2016

Line (i) is parallel to x-axis,

∴ab+=⇒ =−a

b

λλ 0

Put the value of λ in (i), we get

yba

b

b a

b

2 2 3 3 0

22

+

⎛

⎝⎜

⎞

⎠⎟

++ =

⇒ y

ba

b

ba

b

⎛ 2222 + 3322

⎝⎜

⎞

⎠⎟

=− +

⎛

⎝⎜

⎞

⎠⎟

⇒ (^) y=−^3
2
So, it is 3/2 units below x-axis.

30. (c) : Given equation of line having its intercepts on
    the x-axis and y-axis in the ratio 2 : 1 i.e., 2a and a

∴ x+=⇒ +=

a

y

a

xya

2

122

....(i)

According to question,
Line (i) also passes through midpoint of (3, –4) and
(5, 2) i.e., (4, –1).
∴ 4 + 2(–1) = 2a ⇒ a = 1
Hence the required equation of line is, x + 2y = 2

31. (d) : Parallel to x-axis ⇒ l must be zero.


32. (a) : Let L 1 ≡ 2 x + 3y – 7 = 0 and
    L 2 ≡ 2 x + 3y – 5 = 0
    Here slope of L 1 = slope of L 2 = –2/
    Hence the lines are parallel.


33. (c) : Gradient of the line which passes through
    (1, 0) and (, )− 23 is

m= −
−−

(^30) =−
21
1
3
⇒ θ=tan−^1 ⎝⎛⎜−^1 ⎟⎞⎠=°
3
150

34. (b) : θ=°^90 − −⎜⎛⎝^1 ⎟⎠⎞
    3

tan^1

⇒ (^) tanθ=cot tan ⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
− (^11) =
3
3 ⇒^ θ = tan–1(3)

35. (a) : Mid point of (acosα, asinα) and (acosβ, asinβ)

is ⎛aa(cosαβ αβ++cos ), (sin sin )
⎝⎜

⎞

22 ⎠⎟

 





     

         

Slope of line AB is

aa

aa

sin sin m

cos cos

sin sin

cos cos

βα

βα

βα

βα

−

−

= −

−

= 1

And slope of line OP is sin sin

cos cos

αβ

αβ

+

+

=m 2

Now, mm 12

22

×= 221

−

−

sin sin =−

cos cos

βα

βα

Hence the lines are perpendicular.

36. (c) : P is centroid of ΔABC
    ∴ Area of ΔABC = 6 × 5 = 30 sq. units


37. (a, d) : Note that lines u = 0, v = 0 are perpendicular.
    Make the co-ordinate axes coincide with u = 0, v = 0.
    Now the lines L 1 ≡ 0, L 2 ≡ 0 are equally inclined with
    uv axes.
    ∴ u = 0, v = 0 are bisectors.


38. (a, b, c) : Inclinations of two lines are θ and φ
    ∴ Inclination of angle bisector is θφ+
    2
    ⇒ α

θφ α γ

β

= + ×=−

2

and tan 1 ⇒ tanα=−β

γ^

∴ β = –sinα, γ = cosα ⇒ β^2 + γ^2 = 1

39. (b, c) : (3, 2), (–4, 1), (–5, 8) form a right angled
    triangle at (–4, 1).
    Orthocentre is (–4, 1), circumcentre is mid point of
    (3, 2) and (–5, 8) is (–1, 5)


40. (a, d) : Dividing point of P(–5, 1), Q(3, 5) in the
    ratio k : 1 is
    A k
    k

k

k

35

1

51

1

−

+

+

+

⎛

⎝⎜

⎞

, ⎠⎟, B (1, 5), C (7, –2)

Area of triangle ABC = 2

On solving, k= 7 31

9

,

41. (a, b, c, d) : Equation of the curve passing through
    all four points A, B, C, D can be written as
    (3x + 4y – 24)(4x + 3y – 24) + λxy = 0
    Now for different values of λ we will get different curves.


42. (a, b) : (2a + 3b – c)(3a – b + c) = 0
    ⇒ –2a – 3b + c = 0 or 3a – b + c = 0


43. (b) : The quadrilateral formed by angular bisectors
    is a rectangle whose sides are

|(ab−−)|sin , |(ab)|cos

αα

22

S = absinα

Qab=^1 −

2

()sin^2 α