JEE Advanced Level Test: Straight Lines- 2 - EmSAT Achieve MCQ

Let P≡(1,2) 
Using the parametric form ,  the coordinates of the point at a distance of (2/3)^1/2 is given by, 
(1+(2/3)^1/2cosθ, 2+(2/3)^1/2 sinθ)
This point lies on the line  x+y=4
Hence, 
3+(2/3)^1/2(sinθ+cosθ)=4
On squaring and simplifying, we get
sin^2θ+cos^2θ+2sinθcosθ=3/2
1+sin2θ=3/2
sin2θ=1/2
The possible values of θ are π/12 and 5π/12

Correct Answer :- A

Explanation : Consider a family of straight lines.   

(x+y)+λ(2x−y+1)=0

This family consists of two lines:  

x+y=0   and  2x−y+1=0

Add the two equations to get,    

3x+1 = 0  

3x = −1         

x= −1/3

Put this value in x+y=0    

−1/3 + y = 0         

y = 1/3

So we have (x,y) = A(−1/3, 1/3)

Let the given point be B(1,−3)

Note that the required line is perpendicular to line AB.

The slope of the line AB is,

m1 = (−3−1/3)/(1+1/3)

= -10/4

m1 = - 5/2

So the slope of the required line is,   m2 = −1/m1 

= −1/(−5/2)

= 2/5

Use point slope form to find the required equation of line.    

y − y1 = m2(x − x1)    

(y−1/3) = 2/5(x+1/3)    

(3y−1) = 2/5(3x+1)   

5(3y−1) = 2(3x+1)   

15y−5=6x+2  

15y−6x−7=0

So the required equation of line is,  

15y − 6x − 7 = 0

Length of the perpendicular from
(x1,y1) to the line ax+by+c=0 is 
p= (ax1+by1+c)/[(a)² + (b)²]^1/2| 
Let a be the length of each side of the equilateral triangle. 
The length of the perpendicular p from (1,2) on 3x+4y−9=0
⇒p= |(3+8−7)/[(3)² + (4)²]^1/2|
= 4/5
sin60^∘=p/a
⇒ [(3)^1/2]/2 = [(4/5)]/a 
⇒  a = (2)/(3)^1/2 = 4/5
a = [8/5(3)^1/2]

Point of intersection (-2,1) and verification

a₁a₂ +b₁b₂  = -32 < 0, c₁c₂ = 130 > 0

Using formula given in synopsis

Equation of the straight line having equal intercepts is x + y = k and proceed.

y₀ = 2x + 3 + √2
x₀ + y₀ = √2.............(1)
x₀ + 2x₀ + 3 + √2 = √2
x₀ = -1
Put the value of x in eq (1)
y₀ = √2 - 1
2x₀ + y₀ = √2 - 1

Given P (1 + t/√2, 2 + t/√2) lies on x + 2y = 1 

Then 1 + t/√2 + 2 (2 + t/√2) = 1

⇒  1 + t/√2 + 4 + 2t/√2 = 1

⇒ 5 + 3t/√2 = 1

⇒ 3t/√2 = -4

⇒ 3t = -4√2

⇒ t = -4√2/3

Also P lies on 2x + 4y = 15

Then, 2 (1 + t/√2) + 4 (2 + t/√2) = 15

⇒ 2 + 2t/√2 + 8 + 4t/√2 = 15

⇒ 10 + 6t/√2 = 15

⇒ 6t/√2 = 5

⇒ 6t = 5√2

⇒ t = 5√2/6

Hence, the range will be -4√2/3 < t < 5√2/6

By observation, both of them are non parallel lines. Locus of non parallel lines is a circle. 

Use sine rule that is a = 2R sinA,b = 2R sinB, c = 2R sinC.

Since, equation has non-zero solution.

Hence, number of values of k is two.

A (2, 3) , B (1, 2) , P (x, y)
AB = BP

y - 2 = -m (x - 8) , where m >0

(x, y) = (2, 0), (X, Y) = (0, 4)
(x – X, y – Y) = (2, -4)
New origin, (x - X, y - Y) 

After shift of origin, New coordinates: X = x - 1, Y = y - 1
The equation X² + Y² + 2X – Y + 2 = 0, w.r.t to original origin:
(x-1)² + (y-1)² + 2(x-1) – (y-1) + 2 = 0
x² + 1 - 2x + y² + 1 - 2y + 2x - 2 – y + 1 + 2 = 0
x² + y² - 3y + 3 = 0

a = 2, b = 7, g = 4, f = -7, c = 15 

Foxed point P ≡ (-2,1)
Distance of P from origin = 2.

P lies in the acute angle
⇒ α² - 3α > 0 and α² - 5α < 0 
⇒ α ∈ (-∞, 0) ∪ (3,∞) and α ∈ (0,5)
∴ α ∈ (3,5)

∴ Fixed point ≡ (2, -3) which lies on x² + y² = 13

The point of intersection of the two lines are (-1,- 2).
Distance PM = √10
Hence the required line is one which passes through (-1, -2) and is perpendicular to P.M.

 (x² + 2xy secα + y²) = 0
(a=1) (h = secα) (b=1)
tanθ = 2[(h²- ab)]^½/(a+b)
tanθ = 2[((secα)^2 - 1)]^½/(1+1)
tanθ  = [(tan²)α]^½
tanθ = tanα
⇒ θ = α

x² + 3xy - 4y² = 0
a = 1, b=-4
h = 3/2
tanθ = 2[(h² - ab)^½]/a+b
= 2[(9/4 - 1(-4))^½]/1+(-4)
= 2[(9/4 + 4)^½]/(-3)
= [2(5/2)]/(-3)
tanθ = 5/3 which is (perpendicular/base)
sinθ = 5/[(5)² + (3)²]
sinθ = 5/(34)^½

Correct Answer :- a

Explanation : x² + 4xy + 3y² = 0

x² + xy + 3xy + 3y² = 0

x(x+y) + 3y(x+y) = 0

(x+3y) (x+y) = 0

Points (k,k)

d1 = |(k+k)/(2)^1/2|

2k/(2)^1/2

d₂ = |(k+3k)/(1+9)^1/2|

d₂ = |4k/(10)1/2|

d₁ * d₂ = 4/(5)^1/2

|(k+k)/(2)^1/2| * |4k/(10)^1/2| = 4/(5)^1/2

|8k²/(20)^1/2| = 4/(5)^1/2

8k²/2(5)^1/2 = 4/(5)^1/2

k² = +-1

Δ = 0

Here, a=2,b=3 and h=−72
Let θ be the angle between the lines represented by 2x²−7xy+3y²=0. Then ,
tanθ=(2√49/4−6)/5=1
⇒ θ = 45^∘.
⇒ π/4

2x² + λxy + 3y² + 8x + 14y + 8 = 0
ax² + 2hxy + by² + 2gx + 2fy + c = 0
Compare eq (1) and (2), we get
a =2, b = 3, 2h = λ
tanθ=(2√h² - ab)/(a+b)
=> tan(π/4) = [(2√(λ/2)² - (2)(3))/(2+3)]
=> 1 = [(2√((λ)²)/4 - 6]/5
=> 5 = [(2√((λ)²)/4 - 6]
=> 25 = [4((λ)²)/4 - 6]
=> 25 = (λ)² - 24
=> (λ)² = 49
=> λ = +-7

Here x is a complex number. So circle and line does not intersect. So there is no solution

(P + q + r) (p² + q² + r² - pq - qr - rp) = 0

We know that concurrent lines have same intersection points.Then
∴ 3x+2y−5=0 --(1)
2x−5y+3=0 ---(2)
5x+by+c=0 ---(3)
From 1 & 2,
6x+4y−10=0  
6x−15y+9=0 
⇒19y−19=0
y=1 & x=1
So, substituting  x=1 & y=1, in equation (3),
5x+by+c=0
5+b+c = 0
∴ b+c = −5