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Straight Lines and Pair of Straight Lines

JEE Mathematics · 144 questions · Page 3 of 15 · Click an option or "Show Solution" to reveal answer

Q21
Locus of mid point of the portion between the axes of xx coscos α+y sin⁡α=p\alpha + y\,\sin \alpha = p where pp is constant is :
A x2+y2=4p2{x^2} + {y^2} = {4 \over {{p^2}}}
B x2+y2=4p2{x^2} + {y^2} = 4{p^2}
C 1x2+1y2=2p2{1 \over {{x^2}}} + {1 \over {{y^2}}} = {2 \over {{p^2}}}
D 1x2+1y2=4p2{1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}
Correct Answer
Option D
Solution

Equation of

ABAB

is

xcos⁡α+ysin⁡α=p;x\cos \alpha + y\sin \alpha = p;
⇒xcos⁡αp+ysin⁡αp=1;\Rightarrow {{x\cos \alpha } \over p} + {{y\sin \alpha } \over p} = 1;
⇒xp/cos⁡α+yp/sin⁡α=1\Rightarrow {x \over {p/\cos \alpha }} + {y \over {p/\sin \alpha }} = 1

So co-ordinates of

AA

and

BB

are

(pcos⁡α,0)\left( {{p \over {\cos \alpha }},0} \right)

and

(0,psin⁡α);\left( {0,{p \over {\sin \alpha }}} \right);

So coordinates of midpoint of

ABAB

are

(p2cos⁡ α,p2sin⁡α)=(x1,y1)(let);\left( {{p \over {2\cos \,\alpha }},{p \over {2\sin \alpha }}} \right) = \left( {{x_1},{y_1}} \right)\left( {let} \right);
x1=p2 cos⁡ α  &  y1=p2sin⁡α;{x_1} = {p \over {2\,\cos \,\alpha }}\,\,\& \,\,{y_1} = {p \over {2\sin \alpha }};
⇒cos⁡α=p/2x1\Rightarrow \cos \alpha = p/2{x_1}

and

sin⁡α=p/2y1;\sin \alpha = p/2{y_1};
cos⁡2α+sin⁡2α=1⇒p24(1x12+1y12)=1{\cos ^2}\alpha + {\sin ^2}\alpha = 1 \Rightarrow {{{p^2}} \over 4}\left( {{1 \over {{x_1}^2}} + {1 \over {{y_1}^2}}} \right) = 1

Locus of

(x1,y1)\left( {{x_1},{y_1}} \right)

is

1x2+1y2=4p2.{1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}.
Q22
A triangle with vertices (4,0),(−1,−1),(3,5)\left( {4,0} \right),\left( { - 1, - 1} \right),\left( {3,5} \right) is :
A isosceles and right angled
B isosceles but not right angled
C right angled but not isosceles
D neither right angled nor isosceles
Correct Answer
Option A
Solution
AB=(4+1)2+(0+1)2=26;AB = \sqrt {{{\left( {4 + 1} \right)}^2} + {{\left( {0 + 1} \right)}^2}} = \sqrt {26} ;
BC=(3+1)2+(5+1)2=52BC = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( {5 + 1} \right)}^2}} = \sqrt {52}
CA=(4−3)2+(0−5)2=26;CA = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - 5} \right)}^2}} = \sqrt {26} ;

In isosceles triangle side

AB=CAAB=CA

For right angled triangle,

BC2=AB2+AC2B{C^2} = A{B^2} + A{C^2}

So, here

BC=52BC = \sqrt {52}

or

BC2=52B{C^2} = 52

or

         (26)2+(26)2=52\,\,\,\,\,\,\,\,\,{\left( {\sqrt {26} } \right)^2} + {\left( {\sqrt {26} } \right)^2} = 52

So, the given triangle is right angled and also isosceles.

Q23
If the pair of straight lines x2−2pxy−y2=0{x^2} - 2pxy - {y^2} = 0 and x2−2qxy−y2=0{x^2} - 2qxy - {y^2} = 0 be such that each pair bisects the angle between the other pair, then :
A pq=−1pq = -1
B p=qp = q
C p=−qp = -q
D pq=1pq = 1.
Correct Answer
Option A
Solution

Equation of bisectors of second pair of straight lines is,

qx2+2xy−qy2=0             ...(1)q{x^2} + 2xy - q{y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

It must be identical to the first pair

x2−2 pxy−y2=0               (...2){x^2} - 2\,pxy - {y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(... 2 \right)

from

(1)(1)

and

(2)(2)
q1=2−2p=−q−1{q \over 1} = {2 \over { - 2p}} = {{ - q} \over { - 1}}
⇒pq=−1.\Rightarrow pq = - 1.
Q24
A square of side a lies above the xx-axis and has one vertex at the origin. The side passing through the origin makes an angle α(0<α<π4)\alpha \left( {0 < \alpha < {\pi \over 4}} \right) with the positive direction of x-axis. The equation of its diagonal not passing through the origin is :
A y(cos⁡α+sin⁡α)+x(cos⁡α−sin⁡α)=ay\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a
B y(cos⁡α−sin⁡α)−x(sin⁡α−cos⁡α)=ay\left( {\cos \alpha - \sin \alpha } \right) - x\left( {\sin \alpha - \cos \alpha } \right) = a
C y(cos⁡α+sin⁡α)+x(sin⁡α−cos⁡α)=ay\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha - \cos \alpha } \right) = a
D y(cos⁡α+sin⁡α)+x(sin⁡α+cos⁡α)=ay\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha + \cos \alpha } \right) = a
Correct Answer
Option A
Solution

Co-ordinate of

A=(a cos⁡ α,  a sin⁡ α)A = \left( {a\,\cos \,\alpha ,\,\,a\,\sin \,\alpha } \right)

Equation of

OB,OB,
y=tan⁡(π4+α)xy = \tan \left( {{\pi \over 4} + \alpha } \right)x
CA⊥rCA{ \bot ^r}

to

OBOB

∴\therefore slope of

CA=−CA=-
cot⁡(π4+α)\cot \left( {{\pi \over 4} + \alpha } \right)

Equation of

CACA
y−asin⁡α=−cot(π4+α)(x−a cos⁡ α)y - a\sin \alpha = - cot\left( {{\pi \over 4} + \alpha } \right)\left( {x - a\,\cos \,\alpha } \right)
⇒(y−asin⁡α)(tan⁡(π4+α))=(a cos⁡ α−x)\Rightarrow \left( {y - a\sin \alpha } \right)\left( {\tan \left( {{\pi \over 4} + \alpha } \right)} \right) = \left( {a\,\cos \,\alpha - x} \right)
⇒(y−asin⁡α)(tan⁡π4+tan⁡α1−tan⁡π4tan⁡α)(a cos⁡ α−x)\Rightarrow \left( {y - a\sin \alpha } \right)\left( {{{\tan {\pi \over 4} + \tan \alpha } \over {1 - \tan {\pi \over 4}\tan \alpha }}} \right)\left( {a\,\cos \,\alpha - x} \right)
⇒(y−asin⁡α)(1+tan⁡α)=(acos⁡α−x)(1−tan⁡α)\Rightarrow \left( {y - a\sin \alpha } \right)\left( {1 + \tan \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {1 - \tan \alpha } \right)
⇒(y−asin⁡α)(cos⁡α+sin⁡α)=(acos⁡α−x)(cos⁡α−sin⁡α)\Rightarrow \left( {y - a\sin \alpha } \right)\left( {\cos \alpha + \sin \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {\cos \alpha - \sin \alpha } \right)
⇒y(cos⁡+sin⁡α)−asin⁡αcos⁡α−asin⁡2α\Rightarrow y\left( {\cos + \sin \alpha } \right) - a\sin \alpha \cos \alpha - a{\sin ^2}\alpha
                     =acos⁡2α−acos⁡αsin⁡α−x(cos⁡α−sin⁡α)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a{\cos ^2}\alpha - a\cos \alpha \sin \alpha - x\left( {\cos \alpha - \sin \alpha } \right)
⇒y(cos⁡α+sinα)+x(cos⁡α−sin⁡α)=a\Rightarrow y\left( {\cos \alpha + sin\alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a
y(sin⁡α+cos⁡α)+x(cos⁡α−sin⁡α)=a.y\left( {\sin \alpha + \cos \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a.
Q25
If the equation of the locus of a point equidistant from the point (a1,b1)\left( {{a_{1,}}{b_1}} \right) and (a2,b2)\left( {{a_{2,}}{b_2}} \right) is (a1−a2)x+(b1−b2)y+c=0\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0 , then the value of ′c′'c' is :
A a12+b12−a22−b22\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2}
B 12(a22+b22−a12−b12){1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)
C a12−a22+b12−b22{{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}
D 12(a12+a22+b12+b22){1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right).
Correct Answer
Option B
Solution

Since, the points (a1,b1)\left(a_1, b_1\right) and (a2,b2)\left(a_2, b_2\right) satisfy the equation. So, that

a1(a1−a2)+b1(b1−b2)+c=0  ........(1) and a2(a1−a2)+b2(b1−b2)+c=0  .........(2) On adding Eqs. (i) and (ii), we get (a1+a2)(a1−a2)+(b1+b2)(b1−b2)+2c=0⇒2c=−(a12−a22+b12−b22)⇒c=12(a22+b22−a12−b12)\begin{aligned} & a_1\left(a_1-a_2\right)+b_1\left(b_1-b_2\right)+c=0 ~~........(1) \\\\ & \text{ and } a_2\left(a_1-a_2\right)+b_2\left(b_1-b_2\right)+c=0 ~~.........(2) \\\\ & \text{ On adding Eqs. (i) and (ii), we get } \\\\ & \left(a_1+a_2\right)\left(a_1-a_2\right)+\left(b_1+b_2\right)\left(b_1-b_2\right)+2 c=0 \\\\ & \Rightarrow 2 c=-\left(a_1^2-a_2^2+b_1^2-b_2^2\right) \\\\ & \Rightarrow c=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right) \end{aligned}
Q26
If x1,x2,x3{x_1},{x_2},{x_3} and y1,y2,y3{y_1},{y_2},{y_3} are both in G.P. with the same common ratio, then the points (x1,y1),(x2,y2)\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right) :
A are vertices of a triangle
B lie on a straight line
C lie on an ellipse
D lie on a circle
Correct Answer
Option B
Solution

Taking co-ordinates as

(xr,yr);(x,y)  &  (xr,yr)\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)

Then slope of line joining

(xr,yr),(x,y)=y(1−1r)x(1−1r)=yx\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}

and slope of line joining

(x,y)(x,y)

and

(xr,yr)(xr, yr)
=y(r−1)x(r−1)=yx= {{y\left( {r - 1} \right)} \over {x\left( {r - 1} \right)}} = {y \over x}

∴\therefore

m1=m2{m_1} = {m_2}

⇒\Rightarrow Points lie on the straight line.

Q27
Locus of centroid of the triangle whose vertices are (acos⁡t,asin⁡t),(bsin⁡t,−bcos⁡t)\left( {a\cos t,a\sin t} \right),\left( {b\sin t, - b\cos t} \right) and (1,0),\left( {1,0} \right), where tt is a parameter, is :
A (3x+1)2+(3y)2=a2−b2{\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}
B (3x−1)2+(3y)2=a2−b2{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}
C (3x−1)2+(3y)2=a2+b2{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}
D (3x+1)2+(3y)2=a2+b2{\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}
Correct Answer
Option C
Solution
x=acos⁡t+bsin⁡t+13x = {{a\cos t + b\sin t + 1} \over 3}
⇒acos⁡t+bsin⁡t=3x−1\Rightarrow a\cos t + b\sin t = 3x - 1
y=asin⁡t−bcos⁡t3y = {{a\sin t - b\cos t} \over 3}
⇒asin⁡t−bcos⁡t=3y\Rightarrow a\sin t - b\cos t = 3y
Q28
If the sum of the slopes of the lines given by x2−2cxy−7y2=0{x^2} - 2cxy - 7{y^2} = 0 is four times their product cc has the value :
A −2-2
B −1-1
C 22
D 11
Correct Answer
Option C
Solution

Let the lines be

y=m1xy = {m_1}x

and

y=m2xy = {m_2}x

then

m1+m2=−2c7{m_1} + {m_2} = - {{2c} \over 7}

and

m1m2=−17{m_1}{m_2} = - {1 \over 7}

Given

m1+m2=4m1m2{m_1} + {m_2} = 4m{}_1{m_2}
⇒2c7=−47⇒c=2\Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2
Q29
If a vertex of a triangle is (1,1)(1, 1) and the mid points of two sides through this vertex are (−1,2)(-1, 2) and (3,2)(3, 2) then the centroid of the triangle is :
A (−1,73)\left( { - 1,{7 \over 3}} \right)
B (−13,73)\left( {{{ - 1} \over 3},{7 \over 3}} \right)
C (1,73)\left( { 1,{7 \over 3}} \right)
D (13,73)\left( {{{ 1} \over 3},{7 \over 3}} \right)
Correct Answer
Option C
Solution

Vertex of triangle is

(1, 1)\left( {1,\,1} \right)

and midpoint of sides through - this vertex is

(−1, 2)\left( { - 1,\,2} \right)

and

(3,2)\left( {3,2} \right)

⇒\Rightarrow vertex

BB

and

CC

come out to be

(−3,3)\left( { - 3,3} \right)

and

(5,3)\left( {5,3} \right)

∴\therefore centroid is

1−3+53,1+3+53⇒(1,73){{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)
Q30
The line parallel to the xx - axis and passing through the intersection of the lines ax+2by+3b=0ax + 2by + 3b = 0 and bx−2ay−3a=0,bx - 2ay - 3a = 0, where (a,b)(a, b) ≠ \ne (0,0)(0, 0) is :
A below the xx - axis at a distance of 32{3 \over 2} from it
B below the xx - axis at a distance of 23{2 \over 3} from it
C above the xx - axis at a distance of 32{3 \over 2} from it
D above the xx - axis at a distance of 23{2 \over 3} from it
Correct Answer
Option A
Solution

The line passing through the intersection of lines

ax+2by=3b=0ax + 2by = 3b = 0

and

bx−2ay−3a=0bx - 2ay - 3a = 0

is

ax+2by+3b+λ(bx−2ay−3a)=0ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0
⇒(a+bλ)x+(2b−2aλ)y+3b−3λa=0\Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0

As this line is parallel to

xx

-axis. ∴\therefore

a+bλ=0⇒λ=−a/ba + b\lambda = 0 \Rightarrow \lambda = - a/b
⇒ax+2by+3b−ab(bx−2ay−3a)=0\Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0
⇒ax+2by+3b−ax+2a2by+3a2b=0\Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0
y(2b+2a2b)+3b+3a2b=0y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0
y(2b2+2a2b)=−(3b2+3a2b)y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)
y=−3(a2+b2)2(b2+a2)=−32y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}

So it is

3/23/2

units below

xx

-axis.

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