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Straight Lines and Pair of Straight Lines

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

Q141
The sides of a rhombus ABCD are parallel to the lines, x −- y + 2 = 0 and 7x −- y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :
A 52{5 \over 2}
B 74{7 \over 4}
C 2
D 72{7 \over 2}
Correct Answer
Option A
Solution

Let the coordinate A be (0, c) Equations of the given lines are x −- y + 2 = 0 and 7x −- y + 3 = 0 We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3

∴   \therefore\,\,\,

equation of angle bisectors is given as :

x−y+22=±7x−y+352{{x - y + 2} \over {\sqrt 2 }} = \pm {{7x - y + 3} \over {5\sqrt 2 }}

5x −- 5y + 10 = ±\pm (7x −- y + 3)

∴   \therefore\,\,\,

Parallel equations of the diagonals are 2x + 4y −- 7 = 0 and 12x −- 6y + 13 = 0

∴   \therefore\,\,\,

slopes of diagonals are

−12{{ - 1} \over 2}

and 2. Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 −- c)

∴   \therefore\,\,\,

2 −- c = 2 ⇒\Rightarrow c = 0 (not possible) ∴\therefore

   \,\,\,

2 −- c =

−12{{ - 1} \over 2}

⇒\Rightarrow c =

52{5 \over 2}
∴   \therefore\,\,\,

Coordinate of A is

52{5 \over 2}

.

Q142
The foot of the perpendicular drawn from the origin, on the line, 3x + y = λ\lambda (λ\lambda ≠ \ne 0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is :
A 1 : 3
B 3 : 1
C 1 : 9
D 9 : 1
Correct Answer
Option D
Solution

Equation of the line, which is perpendicular to the line, 3x + y = λ\lambda(λ\lambda ≠\ne0) and passing through origin , is given by

x−03=y−01=r{{x - 0} \over 3} = {{y - 0} \over 1} = r

For foot of perpendicular r =

−((3×0)+(1×0)−λ)32+12{{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)} \over {{3^2} + {1^2}}}

=

λ10{\lambda \over {10}}

So, foot of perpendicular P =

(3λ10,λ10)\left( {{{3\lambda } \over {10}},{\lambda \over {10}}} \right)

Given the line meets X-axis where y = 0, so 3x + 0 = λ\lambda ⇒\Rightarrow x =

λ3{\lambda \over 3}

Hence, coordinates of A =

(λ3,0)\left( {{\lambda \over 3},0} \right)

and meets Y-axis at B = (0, λ\lambda) So, BP =

(3λ10)2+(λ10−λ)2\sqrt {{{\left( {{{3\lambda } \over {10}}} \right)}^2} + {{\left( {{\lambda \over {10}} - \lambda } \right)}^2}}

⇒\Rightarrow BP =

9λ2100+81λ2100\sqrt {{{9{\lambda ^2}} \over {100}} + {{81{\lambda ^2}} \over {100}}}

= BP =

90λ2100\sqrt {{{90{\lambda ^2}} \over {100}}}

Now, PA =

(λ3−3λ10)2+(0−λ10)2\sqrt {{{\left( {{\lambda \over 3} - {{3\lambda } \over {10}}} \right)}^2} + {{\left( {0 - {\lambda \over {10}}} \right)}^2}}

⇒\Rightarrow

   \,\,\,

PA =

λ2900+λ2100⇒PA\sqrt {{{{\lambda ^2}} \over {900}} + {{{\lambda ^2}} \over {100}}} \Rightarrow PA

=

10λ2900\sqrt {{{10{\lambda ^2}} \over {900}}}

Therefore BP : PA = 9 : 1

Q143
Let the equations of two sides of a triangle be 3x −- 2y + 6 = 0 and 4x + 5y −- 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is :
A 122y −- 26x −- 1675 = 0
B 122y + 26x + 1675 = 0
C 26x + 61y + 1675 = 0
D 26x −- 122y −- 1675 = 0
Correct Answer
Option D
Solution

4x + 5y −- 20 = 0 . . .(1) 3x −- 2y + 6 = 0 . . . (2) orthocentre is (1, 1) line perpendicular to 4x + 5y −- 20 = 0 and passes through (1, 1) is (y −- 1) =

54{5 \over 4}

(x −- 1) ⇒\Rightarrow 5x −- 4y = 1 . . .(3) and line ⊥\bot to 3x −- 2y + 6 = 0 and passes through (1, 1) y −- 1 = −-

23{2 \over 3}

(x −- 1) ⇒\Rightarrow 2x + 3y = 5 . . .(4) Solving (1) and (4) we get C

(352,−10)\left( {{{35} \over 2}, - 10} \right)

Solving (2) and (3) we get A

(−13,−332)\left( { - 13,{{ - 33} \over 2}} \right)

Side BC is y + 10 =

−332+10−13−352(x−352){{{{ - 33} \over 2} + 10} \over { - 13 - {{35} \over 2}}}\left( {x - {{35} \over 2}} \right)

⇒\Rightarrow y + 10 =

1361(x−352){{13} \over {61}}\left( {x - {{35} \over 2}} \right)

⇒\Rightarrow 26x −- 122y −- 1675 = 0

Q144
Let R\mathrm{R} be the interior region between the lines 3x−y+1=03 x-y+1=0 and x+2y−5=0x+2 y-5=0 containing the origin. The set of all values of aa, for which the points (a2,a+1)\left(a^2, a+1\right) lie in RR, is :
A (−3,0)∪(23,1)(-3,0) \cup\left(\dfrac{2}{3}, 1\right)
B (−3,0)∪(13,1)(-3,0) \cup\left(\dfrac{1}{3}, 1\right)
C (−3,−1)∪(13,1)(-3,-1) \cup\left(\dfrac{1}{3}, 1\right)
D (−3,−1)∪(−13,1)(-3,-1) \cup\left(-\dfrac{1}{3}, 1\right)
Correct Answer
Option B
Solution
P(a2,a+1)L1=3x−y+1=0\begin{aligned} & \mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right) \\ & \mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0 \end{aligned}

Origin and

P\mathrm{P}

lies same side w.r.t.

L1\mathrm{L}_1
⇒L1(0)⋅L1(P)>0∴3(a2)−(a+1)+1>0\begin{aligned} & \Rightarrow \mathrm{L}_1(0) \cdot \mathrm{L}_1(\mathrm{P})>0 \\ & \therefore 3\left(\mathrm{a}^2\right)-(\mathrm{a}+1)+1>0 \end{aligned}
⇒3a2−a>0a∈(−∞,0)∪(13,∞)..... (1)\begin{aligned} & \Rightarrow 3 \mathrm{a}^2-\mathrm{a}>0 \\ & \mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \quad \text{..... (1)} \end{aligned}

Let

L2:x+2y−5=0L_2: x+2 y-5=0

Origin and

P\mathrm{P}

lies same side w.r.t.

L2\mathrm{L}_2
⇒L2(0)⋅L2(P)>0⇒a2+2(a+1)−5<0⇒a2+2a−3<0⇒(a+3)(a−1)<0\begin{aligned} & \Rightarrow \mathrm{L}_2(0) \cdot \mathrm{L}_2(\mathrm{P})>0 \\ & \Rightarrow \mathrm{a}^2+2(\mathrm{a}+1)-5<0 \\ & \Rightarrow \mathrm{a}^2+2 \mathrm{a}-3<0 \\ & \Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0 \end{aligned}
∴a∈(−3,1)..... (2)\therefore \mathrm{a} \in(-3,1)\quad \text{..... (2)}

Intersection of (1) and (2)

a∈(−3,0)∪(13,1)\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)
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