JEE Advanced Maths straight lines previous year questions with solutions are available on this page. Students can expect questions from this topic for the JEE Advanced exam. The important topics in straight lines are distance formula, section formula, the slope of a straight line, the angle between two lines, and the bisector of the angle between two lines, etc. Those who are preparing for JEE Advanced should revise a lot of previous years’ question papers so that they become familiar with the type of questions asked for the exam. These solutions can also be downloaded in pdf format for offline use.
Download Straight Lines Previous Year Solved Questions PDF
Question 1: If the perpendicular bisector of the line segment joining the points P(1, 4) and Q(k, 3) has y-intercept equal to -4, then a value of k is
(a) 2
(b) -4
(c) 1
(d) -2
Solution:
Given points are P(1, 4) and Q (k, 3).
Slope of PQ, m1 = (y2 – y1)/(x2 – x1)
= (3 – 4)/(k – 1)
= -1/(k -1)
Slope of the perpendicular line, m2 = (k – 1) (because m1m2 = -1)
Midpoint of PQ = ((k+1)/2, (3+4)/2)
= ((k+1)/2, 7/2)
The perpendicular line passes through the midpoint ((k+1)/2, 7/2).
The required equation of line is y – y1 = m(x- x1)
y – 7/2 = (k -1)(x – (k+1)/2)
Given that perpendicular bisector has y intercept = -4
At x = 0,y = -4
So -4 – 7/2 = (k -1)(0 – (k+1)/2)
-15/2 = -(k2 – 1)/2
k2 – 1 = 15
k2 = 16
=> k = ±4
Hence option (b) is the answer.
Question 2: A straight line L through the point (3, -2) is inclined at an angle 600 to the line √3x + y = 1. If L also intersects the x-axis, then the equation of L is
(a) y + √3x + 2 – 3√3 = 0
(b) y – √3x + 2 + 3√3 = 0
(c) √y – x + 3 + 2√3 = 0
(d) √3y + x – 3 + 2√3 = 0
Solution:
Given line is √3x + y = 1
The slope of this line = -√3
We know tan θ = |(m2-m1)/(1+m1m2)|
tan 60 = |(m2 + √3)/(1 – √3m2)|
=> m2 = 0 or √3
Equation of line L passing through (3, -2) is y – y1 = m(x – x1)
=> y + 2 = √3(x – 3)
=> y + 2 – √3x + 3√3 = 0
Hence option b is the answer.
Question 3: If λx2 – 5xy + 6y2 + x – 3y = 0 represents a pair of straight lines, then their point of intersection is
(a) (1, 3)
(b) (-1, -3)
(c) (3, 1)
(d) (-3, -1)
Solution:
Given λx2 – 5xy + 6y2 + x – 3y = 0 ..(i)
Comparing (i) with general equation, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
a = λ, h = -5/2, b = 6, g = 1/2, f = -3/2, c = 0
The condition for straight lines is abc + 2fgh – af2 – bg2 – ch2 = 0
=> λ×6×0 + 2×-3/2×1/2×-5/2 – λ×9/4 – 6×1/4 – 0 = 0
=> 0 + 15/4 – 9λ/4 – 3/2 = 0
=> 9/4 = 9λ/4
=> λ = 1
So the equation becomes x2 – 5xy + 6y2 + x – 3y = 0
=> x2 – 3xy – 2xy + 6y2 + x – 3y = 0
=> x(x – 3y) – 2y(x – 3y) + x – 3y =0
=> (x – 3y)(x – 2y + 1) = 0
So x – 3y = 0 …(ii)
x – 2y + 1 = 0 ..(iii)
Solving (ii) and (iii)
=> x = -3, y = -1
The point of intersection of the pair of straight lines x2 – 5xy + 6y2 + x – 3y = 0 is (-3, -1)
Hence option (d) is the answer.
Question 4: If a triangle ABC has vertices A(-1, 7), B(-7, 1), and C(5, -5), then its orthocentre has coordinates
(a) (-3,3)
(b) (-3/5, 3/5)
(c) (3/5, -3/5)
(d) (3,-3)
Solution:
Let H(x, y) be the orthocentre.
AH perpendicular to BC.
So product of slopes = -1
mAH. mBC = -1
((y-7)/(x+1))(1+5)/(-7-5) = -1
=> 2x – y + 9 = 0 …..(i)
Also BH perpendicular to AC.
So product of slopes = -1
mBH. mAC = -1
((y-1)/(x+7)).(7+5)/(-1-5) = -1
x – 2y + 9 = 0 …..(ii)
Solving equation (i) and (ii)
(x, y) = (-3,3)
Hence option (a) is the answer.
Question 5: For a > b> c> 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than 2√2. Then
(a) a + b – c > 0
(b) a – b + c < 0
(c) a + b – c < 0
(d) none of these
Solution:
The point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is (-c/(a+b), -c/(a+b))
The distance between (1, 1) is (-c/(a+b), -c/(a+b)) is less than 2√2
=> ((a+b+c)/(a+b))√2 < 2√2
=> a + b+ c < 2a + 2b
=> a+b-c>0
Hence option (a) is the answer.
Question 6: If the acute angles between the pairs of lines 3x2 – 7xy + 4y2 = 0 and 6x2 – 5xy + y2 = 0 be θ1 and θ2 respectively, then
(a) θ1 = θ2
(b) θ1 = 2θ2
(c) 2θ1 = θ2
(d) none of these
Solution:
We know that the angle between the two lines given by ax2 + 2hxy + by2 = 0 is
tan θ = |(2√(h2 – ab)/(a + b)|
Given 3x2 – 7xy + 4y2 = 0
Comparing with general equation, we get a = 3, h = -7/2, b = 4
tan θ1= |(2√(49/4 – 12)/(3 + 4)|
= |(2√(1/4)/7|
= 1/7
Second equation is 6x2 – 5xy + y2 = 0
Comparing with general equation, we get a = 6, h = -5/2, b = 1
tan θ2 = |(2√(25/4 – 6)/(6 + 1)|
= 1/7
So we get θ1 = θ2
Hence option (a) is the answer.
Question 7: If the angle between the lines represented by the equation y2 + kxy – x2 tan2 A = 0 be 2A, then k =
(a) 0
(b) 1
(c) 2
(d) tan A
Solution:
Given y2 + kxy – x2 tan2 A = 0 ..(i)
We know that the angle between the two lines given by ax2 + 2hxy + by2 = 0 is
tan θ = |(2√(h2 – ab)/(a + b)|
Comparing (i) with general equation, we get a = -tan2A, b = 1, h = k/2
The given angle is 2A.
So tan 2A = |(2√(k2/4 + tan2A)/(1 – tan2A)|
2 tan A/(1 – tan2A) = |(2√(k2/4 + tan2A)/(1 – tan2A)|
=> tan A = √(k2/4 + tan2A)
Squaring both sides
tan2 A = k2/4 + tan2A
=> k2/4 = 0
=> k = 0
Hence option (a) is the answer.
Question 8: If the equation 12x2 + 7xy – py2 – 18x + qy + 6 = 0 represents a pair of perpendicular straight lines, then
(a) p = 12, q = 1
(b) p = 1, q = 12
(c) p = -1, q = 12
(d) p = 1, q = -12
Solution:
Given 12x2 + 7xy – py2 – 18x + qy + 6 = 0 …(i)
Comparing with general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
We get a = 12, b = -p, h = 7/2, g = -9, f = q/2, c = 6
The condition for the pair of straight lines to be perpendicular is a + b = 0
=> 12 – p = 0
=> p = 12
The condition that given lines represents pair of straight lines is
abc + 2fgh – af2 – bg2 – ch2 = 0
=> 12×-p×6 + 2×(q/2)×-9×(7/2) – 12(q/2)2 + p(-9)2 – 6(7/2)2 = 0
=> -72p – 63q/2 – 3q2 + 81p – 147/2 = 0
=> 9p – 63q/2 – 3q2 – 147/2 = 0
Put p = 12
=> 108 – 63q/2 – 3q2 – 147/2 = 0
=> 3q2 + 63q/2 – 69/2 = 0
=> 2q2 + 21q – 23 = 0
Solving using the quadratic formula, we get q = 1, or -23/2
Hence option (a) is the answer.
Question 9: If the coordinates of points A and B be (3,3) and (7,6) then the length of the portion of the line AB intercepted between the axes is
(a) 5/4
(b) √10/4
(c) √13/3
(d) None of these
Solution:
Given points are A(3, 3) and B(7, 6).
The slope of AB, m = (y2 – y1)/(x2 – x1)
= (6 – 3)/(7 – 3)
= 3/4
Equation of line passing through given points is y – y1 = m(x – x1)
=> y – 3 = (¾)(x – 3)
=> y – (¾)x = -9/4 + 3
=> y – (¾)x = 3/4
Divide by ¾
=> (4/3)y – x = 1
Comparing with intercept form, x/a + y/b = 1
we get a = -1, b = 3/4
So the coordinates will be (-1, 0) and (0, 3/4).
The distance between the points = √[(0 +1)2 + (3/4)2] (using distance formula)
= √(25/16)
= 5/4
Hence option (a) is the answer.
Question 10: The equation x2 – 3xy + λy2 + 3x – 5y + 2 = 0 when λ is a real number, represents a pair of straight lines. If θ is the angle between the lines, then cosec2 θ is equal to
(a) 3
(b) 9
(c) 10
(d) 100
Solution:
Given x2 – 3xy + λy2 + 3x – 5y + 2 = 0
Comparing with general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get
a = 1, h = -3/2, b = λ, g = 3/2, f = -5/2, c = 2
For pair of straight lines, abc + 2fgh – af2 – bg2 – ch2 = 0
=> 2λ + 45/4 – 25/4 – 9λ/4 – 9/2 = 0
=> 2λ + 2/4 – 9λ/4 = 0
=> -λ + 2 = 0
=> λ = 2
We know angle between pair of straight lines is given by
tan θ = 2√(h2 – ab)/(a+ b)
= 2√(9/4 – λ)/(1 + λ)
= √(9 – 4λ)/(1 + λ)
= ⅓
cot θ = 3
cosec2θ = 1 + cot2θ
= 1 + 32
= 10
Hence option (c) is the answer.
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