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Straight Lines and Pair of Straight Lines

JEE Mathematics · 144 questions · Page 13 of 15 · Click an option or "Show Solution" to reveal answer

Q121
Let a variable line of slope m>0m>0 passing through the point (4,−9)(4,-9) intersect the coordinate axes at the points AA and BB. The minimum value of the sum of the distances of AA and BB from the origin is
A 30
B 15
C 10
D 25
Correct Answer
Option D
Solution
L:(y+9)=m(x−4)A:(4+9m,0)B:(0,−9−4m)OA+OB]min⁡⇒Emin⁡=4+9m+9+4mE=13+9m+4mdEdM=0⇒−9m2+4=0⇒m=±32\begin{aligned} & L:(y+9)=m(x-4) \\ & A:\left(4+\frac{9}{m}, 0\right) \\ & B:(0,-9-4 m) \\ & O A+O B]_{\min } \\ & \Rightarrow E_{\min }=4+\frac{9}{m}+9+4 m \\ & E=13+\frac{9}{m}+4 m \\ & \frac{d E}{d M}=0 \Rightarrow-\frac{9}{m^2}+4=0 \Rightarrow m= \pm \frac{3}{2} \end{aligned}
d2EdM2=18m3>0 for m=32∴Emin⁡=4+6+9+6=25\begin{aligned} & \frac{d^2 E}{d M^2}=\frac{18}{m^3}>0 \text{ for } \quad m=\frac{3}{2} \\ & \therefore E_{\min }=4+6+9+6=25 \end{aligned}
Q122
Let ΔABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y – 31 = 0 and 9x – 2y – 19 = 0. Let the point (h, k) be the image of the centroid of ΔABC in the line 3x + 6y – 53 = 0. Then h2 + k2 + hk is equal to :
A 47
B 37
C 40
D 36
Correct Answer
Option B
Solution
∴ centroid of △ABC=(9+3+53,11+4+133)=(173,283)\begin{aligned} \therefore \text{ centroid of } \triangle \mathrm{ABC} & =\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right) \\ & =\left(\frac{17}{3}, \frac{28}{3}\right) \end{aligned}

Let image of centroid with respect to line mirror is (h, k)

∴(k−283 h−173)(−12)=−1&3( h+1732)+6(k+2832)=53 Solving (1) & (2) we get h=3,k=4∴h2+k2+hk=37\begin{aligned} &\begin{aligned} & \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{~h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \\ & \& 3\left(\frac{\mathrm{~h}+\frac{17}{3}}{2}\right)+6\left(\frac{\mathrm{k}+\frac{28}{3}}{2}\right)=53 \end{aligned}\\ &\text{ Solving (1) \& (2) we get } \mathrm{h}=3, \mathrm{k}=4\\ &\therefore \mathrm{h}^2+\mathrm{k}^2+\mathrm{hk}=37 \end{aligned}
Q123
Let the line x + y = 1 meet the axes of x and y at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is 49 \dfrac{4}{9} of the area of the triangle OAB and AN : NB = λ:1 \lambda : 1 , then the sum of all possible value(s) of λ \lambda is:
A 12\dfrac{1}{2}
B 52\dfrac{5}{2}
C 2
D 136\dfrac{13}{6}
Correct Answer
Option C
Solution

Area of △AOB=12\triangle \mathrm{AOB}=\dfrac{1}{2} Area of △AMN=49×12=29\triangle \mathrm{AMN}=\dfrac{4}{9} \times \dfrac{1}{2}=\dfrac{2}{9} Equation of AB is x+y=1\mathrm{x}+\mathrm{y}=1

OA=1,AM=sec⁡(45∘−θ)AN=sec⁡(45∘−θ)cos⁡θMN=sec⁡(45∘−θ)sin⁡θ\begin{aligned} & \mathrm{OA}=1, \mathrm{AM}=\sec \left(45^{\circ}-\theta\right) \\ & \mathrm{AN}=\sec \left(45^{\circ}-\theta\right) \cos \theta \\ & \mathrm{MN}=\sec \left(45^{\circ}-\theta\right) \sin \theta \end{aligned}
Ar⁡(△AMN)=12×sec⁡2(45∘−θ)sin⁡θ⋅cos⁡θ=29⇒tan⁡θ=2,12tan⁡θ=2 is rejected ANNB=λ1=cot⁡θ=2\begin{aligned} & \operatorname{Ar}(\triangle \mathrm{AMN})=\frac{1}{2} \times \sec ^2\left(45^{\circ}-\theta\right) \sin \theta \cdot \cos \theta=\frac{2}{9} \\ & \Rightarrow \tan \theta=2, \frac{1}{2} \\ & \tan \theta=2 \text{ is rejected } \\ & \frac{\mathrm{AN}}{\mathrm{NB}}=\frac{\lambda}{1}=\cot \theta=2 \end{aligned}
Q124
Let the triangle PQR be the image of the triangle with vertices (1,3),(3,1)(1,3),(3,1) and (2,4)(2,4) in the line x+2y=2x+2 y=2. If the centroid of △PQR\triangle \mathrm{PQR} is the point (α,β)(\alpha, \beta), then 15(α−β)15(\alpha-\beta) is equal to :
A 21
B 19
C 22
D 24
Correct Answer
Option C
Solution

Let ' GG ' be the centroid of Δ\Delta formed by (1,3)(3,1)(1,3)(3,1) \& (2,4)(2,4)

G≅(2,83)\mathrm{G} \cong\left(2, \frac{8}{3}\right)

Image of G w.r.t. x+2y−2=0x+2 y-2=0

α−21=β−832=−2(2+163−2)1+4=−25(163)⇒α=−3215+2=−215,β=−32×215+83=−241515(α−β)=−2+24=22\begin{aligned} & \frac{\alpha-2}{1}=\frac{\beta-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4} \\ & =\frac{-2}{5}\left(\frac{16}{3}\right) \\ & \Rightarrow \alpha=\frac{-32}{15}+2=\frac{-2}{15}, \beta=\frac{-32 \times 2}{15}+\frac{8}{3}=\frac{-24}{15} \\ & 15(\alpha-\beta)=-2+24=22 \end{aligned}
Q125
Two equal sides of an isosceles triangle are along −x+2y=4 -x + 2y = 4 and x+y=4 x + y = 4 . If m m is the slope of its third side, then the sum, of all possible distinct values of m m , is:
A −210-2\sqrt{10}
B 12
C -6
D 6
Correct Answer
Option D
Solution
tan⁡θ=m−121+12⋅m=−1−m1−m=m+1m−12m−12+m=m+1m−12m2−3m+1=m2+3m+2\begin{aligned} & \tan \theta=\frac{m-\frac{1}{2}}{1+\frac{1}{2} \cdot m}=\frac{-1-m}{1-m}=\frac{m+1}{m-1} \\ & \frac{2 m-1}{2+m}=\frac{m+1}{m-1} \\ & 2 m^2-3 m+1=m^2+3 m+2 \end{aligned}
m2−6m−1=0 sum of root =6 sum is 6\begin{aligned} & m^2-6 m-1=0 \\ & \text{ sum of root }=6 \\ & \text{ sum is } 6 \end{aligned}
Q126
If A and B are the points of intersection of the circle x2+y2−8x=0x^2 + y^2 - 8x = 0 and the hyperbola x29−y24=1\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1 and a point P moves on the line 2x−3y+4=02x - 3y + 4 = 0, then the centroid of ΔPAB\Delta PAB lies on the line :
A x+9y=36x + 9y = 36
B 9x−9y=329x - 9y = 32
C 4x−9y=124x - 9y = 12
D 6x−9y=206x - 9y = 20
Correct Answer
Option D
Solution
x2+y2−8x=0,x29−y24=1.... (1)4x2−9y2=36.... (2) Solve (1)&(2)4x2−9(8x−x2)=3613x2−72x−36=0(13x+6)(x=6)=0x=−613,x=6x=−613( rejected )y→ Imaginary n=6,369−y24=1y2=12,y=I12A(6,12),B(6,−12)p(α,2α+43)P lies on \begin{aligned} & x^2+y^2-8 x=0, \frac{x^2}{9}-\frac{y^2}{4}=1 \quad\text{.... (1)}\\ & 4 x^2-9 y^2=36 \quad\text{.... (2)}\\ & \text{ Solve }(1) \&(2) \\ & 4 x^2-9\left(8 x-x^2\right)=36 \\ & 13 x^2-72 x-36=0 \\ & (13 x+6)(x=6)=0 \\ & x=\frac{-6}{13}, x=6 \\ & x=\frac{-6}{13}(\text{ rejected }) \\ & y \rightarrow \text{ Imaginary } \\ & n=6, \frac{36}{9}-\frac{y^2}{4}=1 \\ & y^2=12, y=I \sqrt{12} \\ & A(6, \sqrt{12}), B(6,-\sqrt{12}) \\ & p\left(\alpha, \frac{2 \alpha+4}{3}\right) P \text{ lies on } \end{aligned}
 centroid (h,k)2x−3y+y=0h=12+α3,α=3 h−12k=2α−3y33⇒2α+4=9yα=9k−426 h−2y=9k−46x−9y=20\begin{aligned} & \text{ centroid }(\mathrm{h}, \mathrm{k}) \quad 2x-3y+y=0\\ & \mathrm{h}=\frac{12+\alpha}{3}, \alpha=3 \mathrm{~h}-12 \\ & \mathrm{k}=\frac{\frac{2 \alpha-3 y}{3}}{3} \Rightarrow 2 \alpha+4=9 \mathrm{y} \\ & \alpha=\frac{9 \mathrm{k}-4}{2} \\ & 6 \mathrm{~h}-2 \mathrm{y}=9 \mathrm{k}-4 \\ & 6 \mathrm{x}-9 \mathrm{y}=20 \end{aligned}
Q127
A rod of length eight units moves such that its ends AA and BB always lie on the lines x−y+2=0x-y+2=0 and y+2=0y+2=0, respectively. If the locus of the point PP, that divides the rod ABA B internally in the ratio 2:12: 1 is 9(x2+αy2+βxy+γx+28y)−76=09\left(x^2+\alpha y^2+\beta x y+\gamma x+28 y\right)-76=0, then α−β−γ\alpha-\beta-\gamma is equal to :
A 24
B 22
C 21
D 23
Correct Answer
Option D
Solution
h=3β+α3k=−4+α+23α=3k+22β=3 h−a=3 h−3k−2 so AB=8(α−β)2+(α+4)2=64(3k+2−(3 h−3k−22))2+(3k+2+4)2=64(9k−3 h+6)24+(3k+6)2=649[(3k−h+2)2+4(k+2)2]=64×49(x2+13y2−6xy−4x+28y)=76α−β−γ=13+6+4=23\begin{aligned} & \mathrm{h}=\frac{3 \beta+\alpha}{3} \\ & \mathrm{k}=\frac{-4+\alpha+2}{3} \\ & \alpha=3 \mathrm{k}+2 \\ & 2 \beta=3 \mathrm{~h}-\mathrm{a}=3 \mathrm{~h}-3 \mathrm{k}-2 \\ & \text{ so } \mathrm{AB}=8 \\ & (\alpha-\beta)^2+(\alpha+4)^2=64 \\ & \left(3 \mathrm{k}+2-\left(\frac{3 \mathrm{~h}-3 \mathrm{k}-2}{2}\right)\right)^2+(3 \mathrm{k}+2+4)^2=64 \\ & \frac{(9 \mathrm{k}-3 \mathrm{~h}+6)^2}{4}+(3 \mathrm{k}+6)^2=64 \\ & 9\left[(3 \mathrm{k}-\mathrm{h}+2)^2+4(\mathrm{k}+2)^2\right]=64 \times 4 \\ & 9\left(\mathrm{x}^2+13 \mathrm{y}^2-6 \mathrm{xy}-4 \mathrm{x}+28 \mathrm{y}\right)=76 \\ & \alpha-\beta-\gamma=13+6+4=23 \end{aligned}
Q128
Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant :
A third
B fourth
C second
D first
Correct Answer
Option C
Solution

mBD ×\times mAD = −- 1 ⇒\Rightarrow

(3−24−0)×(b−0a−0)=−1\left( {{{3 - 2} \over {4 - 0}}} \right) \times \left( {{{b - 0} \over {a - 0}}} \right) = - 1

⇒\Rightarrow b + 4a = 0 . . . . (i) mAB ×\times mCF = −- 1 ⇒\Rightarrow

((b−2)a−0)×(34)=−1\left( {{{\left( {b - 2} \right)} \over {a - 0}}} \right) \times \left( {{3 \over 4}} \right) = - 1

⇒\Rightarrow 3b −- 6 = −- 4a ⇒\Rightarrow 4a + 3b = 6 . . . . .(ii) From (i) and (ii) a =

−34{{ - 3} \over 4}

, b = 3 ∴\therefore IInd quadrant.

Q129
Let the lines 3x−4y−α=0,8x−11y−33=03 x-4 y-\alpha=0,8 x-11 y-33=0, and 2x−3y+λ=02 x-3 y+\lambda=0 be concurrent. If the image of the point (1,2)(1,2) in the line 2x−3y+λ=02 x-3 y+\lambda=0 is (5713,−4013)\left(\dfrac{57}{13}, \dfrac{-40}{13}\right), then ∣αλ∣|\alpha \lambda| is equal to
A 91
B 113
C 101
D 84
Correct Answer
Option A
Solution
∵PM=QM\because \mathrm{PM}=\mathrm{QM}

So, M(5713+12,−4013+22)M\left(\dfrac{\dfrac{57}{13}+1}{2}, \dfrac{\dfrac{-40}{13}+2}{2}\right)

=(3513,−713)=\left(\frac{35}{13}, \frac{-7}{13}\right)

∵M\because \mathrm{M} lies on the time

2x−3y+λ=02(3513)−3(−713)+λ=0λ=−7013+2113=−9113=−7\begin{aligned} & 2 x-3 y+\lambda=0 \\ & 2\left(\frac{35}{13}\right)-3\left(\frac{-7}{13}\right)+\lambda=0 \\ & \lambda=-\frac{70}{13}+\frac{21}{13} \\ & =\frac{-91}{13}=-7 \end{aligned}
∣3−4−α8−11−3323λ∣=0⇒3(−11λ−99)+4(8λ+66)−α(−24+22)=0⇒33λ−297+32λ+264+24α−22α=0⇒−λ+2α−33=0.... (1)∴λ=−7−(−7)+2α−33=02α=26α=13∴∣αλ∣=∣13×(−7)∣=91\begin{aligned} & \left|\begin{array}{ccc} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & 3 & \lambda \end{array}\right|=0 \\ & \Rightarrow 3(-11 \lambda-99)+4(8 \lambda+66)-\alpha(-24+22)=0 \\ & \Rightarrow 33 \lambda-297+32 \lambda+264+24 \alpha-22 \alpha=0 \\ & \Rightarrow-\lambda+2 \alpha-33=0 \quad\text{.... (1)}\\ & \therefore \lambda=-7 \\ & -(-7)+2 \alpha-33=0 \\ & 2 \alpha=26 \\ & \alpha=13 \\ & \therefore|\alpha \lambda|=|13 \times(-7)| \\ & =91 \end{aligned}
Q130
Let the points (112,α)\left(\dfrac{11}{2}, \alpha\right) lie on or inside the triangle with sides x+y=11,x+2y=16x+y=11, x+2 y=16 and 2x+3y=292 x+3 y=29. Then the product of the smallest and the largest values of α\alpha is equal to :
A 22
B 33
C 55
D 44
Correct Answer
Option B
Solution

Point of intersection of x=112x=\dfrac{11}{2} with L1&L3L_1 \& L_3 gives, αmin⁡=112\alpha_{\min }=\dfrac{11}{2} and αmax⁡=6\alpha_{\max }=6

∴αmin⁡⋅αmax⁡=112×6=33\therefore \alpha_{\min } \cdot \alpha_{\max }=\frac{11}{2} \times 6=33
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