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Straight Lines and Pair of Straight Lines

JEE Mathematics · 144 questions · Page 8 of 15 · Click an option or "Show Solution" to reveal answer

Q71
The number of integral values of m so that the abscissa of point of intersection of lines 3x + 4y = 9 and y = mx + 1 is also an integer, is :
A 1
B 2
C 3
D 0
Correct Answer
Option B
Solution

3x + 4(mx + 1) = 9 ⇒\Rightarrow x(3 + 4m) = 5 ⇒\Rightarrow

x=5(3+4m)x = {5 \over {(3 + 4m)}}

⇒\Rightarrow (3 + 4m) = ±\pm1, ±\pm5 ⇒\Rightarrow 4m = −-3 ±\pm 1, −-3 ±\pm 5 ⇒\Rightarrow 4m = −-4, −-2, −-8, 2 ⇒\Rightarrow m = −-1, −-

12{1 \over 2}

, −-2,

12{1 \over 2}

∴\therefore Two integral value of m.

Q72
The equation of one of the straight lines which passes through the point (1, 3) and makes an angles tan⁡−1(2){\tan ^{ - 1}}\left( {\sqrt 2 } \right) with the straight line, y + 1 = 32{\sqrt 2 } x is :
A 42x+5y−(15+42)=04\sqrt 2 x + 5y - \left( {15 + 4\sqrt 2 } \right) = 0
B 52x+4y−(15+42)=05\sqrt 2 x + 4y - \left( {15 + 4\sqrt 2 } \right) = 0
C 42x+5y−42=04\sqrt 2 x + 5y - 4\sqrt 2 = 0
D 42x−5y−(5+42)=04\sqrt 2 x - 5y - \left( {5 + 4\sqrt 2 } \right) = 0
Correct Answer
Option A
Solution

Let slope of line be m ∴\therefore

∣m−321+32m∣=2\left| {{{m - 3\sqrt 2 } \over {1 + 3\sqrt 2 m}}} \right| = \sqrt 2
⇒m−32=± 2±6m\Rightarrow m - 3\sqrt 2 = \pm \,\sqrt 2 \pm 6m
⇒m∓6m=±2+32\Rightarrow m \mp 6m = \pm \sqrt 2 + 3\sqrt 2
⇒m=−425\Rightarrow m = - {{4\sqrt 2 } \over 5}

or

227{{2\sqrt 2 } \over 7}

Hence line can be

y−3=−425(x−1)y - 3 = {{ - 4\sqrt 2 } \over 5}(x - 1)
⇒5y−15=−42x+42\Rightarrow 5y - 15 = - 4\sqrt 2 x + 4\sqrt 2
⇒42x+5y−(15+42)=0\Rightarrow 4\sqrt 2 x + 5y - (15 + 4\sqrt 2 ) = 0
Q73
Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of Δ\DeltaABC, then (R + r) is equal to :
A 727\sqrt 2
B 92{9 \over {\sqrt 2 }}
C 222\sqrt 2
D 323\sqrt 2
Correct Answer
Option B
Solution
r=∣0+0−32∣=32r = \left| {{{0 + 0 - 3} \over {\sqrt 2 }}} \right| = {3 \over {\sqrt 2 }}
sin⁡30∘=rR=12\sin 30^\circ = {r \over R} = {1 \over 2}

R = 2r So,

r+R=3r=3×(32)=92r + R = 3r = 3 \times \left( {{3 \over {\sqrt 2 }}} \right) = {9 \over {\sqrt 2 }}
Q74
The point P (a, b) undergoes the following three transformations successively : (a) reflection about the line y = x. (b) translation through 2 units along the positive direction of x-axis. (c) rotation through angle π4{\pi \over 4} about the origin in the anti-clockwise direction. If the co-ordinates of the final position of the point P are (−12,72)\left( { - {1 \over {\sqrt 2 }},{7 \over {\sqrt 2 }}} \right), then the value of 2a + b is equal to :
A 13
B 9
C 5
D 7
Correct Answer
Option B
Solution

Image of A(a, b) along y = x is B(b, a). Translating it 2 units it becomes C(b + 2, a). Now, applying rotation theorem

−12+72i=((b+2)+ai)(cos⁡π4+isin⁡π4)- {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {(b + 2) + ai} \right)\left( {\cos {\pi \over 4} + i\sin {\pi \over 4}} \right)
−12+72i=(b+22−a2)+i(b+22+a2)- {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {{{b + 2} \over {\sqrt 2 }} - {a \over {\sqrt 2 }}} \right) + i\left( {{{b + 2} \over {\sqrt 2 }} + {a \over {\sqrt 2 }}} \right)

⇒\Rightarrow b −- a + 2 = −-1 ......(i) and b + 2 + a = 7 ...... (ii) ⇒\Rightarrow a = 4; b = 1 ⇒\Rightarrow 2a + b = 9

Q75
Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x + 2y = 0. If the equation of one of the diagonals of the parallelogram is 11x + 7y = 9, then other diagonal passes through the point :
A (1, 2)
B (2, 2)
C (2, 1)
D (1, 3)
Correct Answer
Option B
Solution

Both the lines pass through origin. point D is equal to intersection of 4x + 5y = 0 & 11x + 7y = 9 So, coordinates of point

D=(53,−43)D = \left( {{5 \over 3}, - {4 \over 3}} \right)

Also, point B is point of intersection of 7x + 2y = 0 and 11x + 7y = 9 So, coordinates of point

B=(−23,73)B = \left( { - {2 \over 3},{7 \over 3}} \right)

diagonals of parallelogram intersect at middle let middle point of B, D

⇒(53−232,−43+732)=(12,12)\Rightarrow \left( {{{{5 \over 3} - {2 \over 3}} \over 2},{{{{ - 4} \over 3} + {7 \over 3}} \over 2}} \right) = \left( {{1 \over 2},{1 \over 2}} \right)

equation of diagonal AC

⇒(y−0)=1α−01α−0(x−0)\Rightarrow (y - 0) = {{{1 \over \alpha } - 0} \over {{1 \over \alpha } - 0}}(x - 0)
y=xy = x

diagonal AC passes through (2, 2)

Q76
Let the equation of the pair of lines, y = px and y = qx, can be written as (y −- px) (y −- qx) = 0. Then the equation of the pair of the angle bisectors of the lines x2 −- 4xy −- 5y2 = 0 is :
A x2 −- 3xy + y2 = 0
B x2 + 4xy −- y2 = 0
C x2 + 3xy −- y2 = 0
D x2 −- 3xy −- y2 = 0
Correct Answer
Option C
Solution

Equation of angle bisector of homogeneous equation of pair of straight line ax2 + 2hxy + by2 is

x2−y2a−b=xyh{{{x^2} - {y^2}} \over {a - b}} = {{xy} \over h}

for x2 – 4xy – 5y2 = 0 a = 1, h = – 2, b = – 5 So, equation of angle bisector is

x2−y21−(−5)=xy−2{{{x^2} - {y^2}} \over {1 - ( - 5)}} = {{xy} \over { - 2}}
x2−y26=xy−2{{{x^2} - {y^2}} \over 6} = {{xy} \over { - 2}}
⇒x2−y2=−3xy\Rightarrow {x^2} - {y^2} = - 3xy

So, combined equation of angle bisector is

x2+3xy−y2=0{x^2} + 3xy - {y^2} = 0
Q77
Let ABC be a triangle with A(−-3, 1) and ∠\angleACB = θ\theta, 0 < θ\theta < π2{\pi \over 2}. If the equation of the median through B is 2x + y −- 3 = 0 and the equation of angle bisector of C is 7x −- 4y −- 1 = 0, then tanθ\theta is equal to :
A 12{1 \over 2}
B 34{3 \over 4}
C 43{4 \over 3}
D 2
Correct Answer
Option C
Solution

∴\therefore

M(a−32,b+12)M\left( {{{a - 3} \over 2},{{b + 1} \over 2}} \right)

lies on 2x + y −- 3 = 0 ⇒\Rightarrow 2a + b = 11 ...........(i) ∵\because C lies on 7x −- 4y = 1 ⇒\Rightarrow 7a −- 4b = 1 ......... (ii) ∴\therefore by (i) and (ii) : a = 3, b = 5 ⇒\Rightarrow C(3, 5) ∴\therefore mAC = 2/3 Also, mCD = 7/4

⇒tan⁡θ2=∣23−441+1412∣⇒tan⁡θ2=12\Rightarrow \tan {\theta \over 2} = \left| {{{{2 \over 3} - {4 \over 4}} \over {1 + {{14} \over {12}}}}} \right| \Rightarrow \tan {\theta \over 2} = {1 \over 2}
⇒tan⁡θ=2.121−14=43\Rightarrow \tan \theta = {{2.{1 \over 2}} \over {1 - {1 \over 4}}} = {4 \over 3}
Q78
Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
A 3x2 −- 2y −- 6 = 0
B 3x2 + 2y −- 6 = 0
C 2x2 + 3y −- 9 = 0
D 2x2 −- 3y + 9 = 0
Correct Answer
Option C
Solution

A(0, 6) and B(2t, 0) Perpendicular bisector of AB is

(y−3)=t3(x−t)(y - 3) = {t \over 3}(x - t)

So,

C=(0,3−t23)C = \left( {0,3 - {{{t^2}} \over 3}} \right)

Let P be (h, k)

h=t2;k=(3−t26)h = {t \over 2};k = \left( {3 - {{{t^2}} \over 6}} \right)
⇒k=3−4h26⇒2x2+3y−9=0\Rightarrow k = 3 - {{4{h^2}} \over 6} \Rightarrow 2{x^2} + 3y - 9 = 0
Q79
If p and q are the lengths of the perpendiculars from the origin on the lines, x cosec α\alpha −- y sec α\alpha = k cot 2α\alpha and x sinα\alpha + y cosα\alpha = k sin2α\alpha respectively, then k2 is equal to :
A 4p2 + q2
B 2p2 + q2
C p2 + 2q2
D p2 + 4q2
Correct Answer
Option A
Solution

First line is

xsin⁡α−ycos⁡α=kcos⁡2αsin⁡2α{x \over {\sin \alpha }} - {y \over {\cos \alpha }} = {{k\cos 2\alpha } \over {\sin 2\alpha }}
⇒xcos⁡α−ysin⁡α=k2cos⁡2α\Rightarrow x\cos \alpha - y\sin \alpha = {k \over 2}\cos 2\alpha
⇒p=∣k2cos⁡α∣⇒2p=∣kcos⁡2α∣\Rightarrow p = \left| {{k \over 2}\cos \alpha } \right| \Rightarrow 2p = \left| {k\cos 2\alpha } \right|

.... (i) second line is

xsin⁡α+ycos⁡α=ksin⁡2αx\sin \alpha + y\cos \alpha = k\sin 2\alpha
⇒q=∣ksin⁡2α∣\Rightarrow q = \left| {k\sin 2\alpha } \right|

.... (ii) Hence,

4p2+q2=k24{p^2} + {q^2} = {k^2}

(From (i) & (ii))

Q80
Let A be the set of all points (α\alpha, β\beta) such that the area of triangle formed by the points (5, 6), (3, 2) and (α\alpha, β\beta) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :
A 45{4 \over {\sqrt 5 }}
B 165{16 \over {\sqrt 5 }}
C 85{8 \over {\sqrt 5 }}
D 125{12 \over {\sqrt 5 }}
Correct Answer
Option C
Solution
∣12∣561321αβ1∣∣=12\left| {{1 \over 2}\left| \begin{array}{lll}5 & 6 & 1 \\ 3 & 2 & 1 \\ \alpha & \beta & 1 \end{array} \right|} \right| = 12

4α\alpha −- 2β\beta = ±\pm 24 + 8 ⇒\Rightarrow 4α\alpha −- 2β\beta = + 24 + 8 ⇒\Rightarrow 2α\alpha −- β\beta = 16 2x −- y −- 16 = 0 ..... (1) ⇒\Rightarrow 4α\alpha −- 2β\beta = −- 24 + 8 ⇒\Rightarrow 2α\alpha −- β\beta = −-8 2x −- y + 8 = 0 ......

(2) Perpendicular distance of (1) from (0, 0)

∣0−0−165∣=165\left| {{{0 - 0 - 16} \over {\sqrt 5 }}} \right| = {{16} \over {\sqrt 5 }}

Perpendicular distance of (2) from (0, 0)

∣0−0+85∣=85\left| {{{0 - 0 + 8} \over {\sqrt 5 }}} \right| = {8 \over {\sqrt 5 }}
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