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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Easy

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Showing 1-30 of 201 questions

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1 Easy
1 of 201
Application +4 | -1
If ax2+2hxy+by2+2gx+2fy+c=0a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 represents parallel straight lines, then
View answer and explanation
Correct option: A
hf=bgh f=b g
Option A (hf=bgh f=b g) is correct because for parallel lines, the condition bg2=af2bg^2 = af^2 must hold, which simplifies to hf=bghf = bg when h2=abh^2 = ab.
2 Easy
2 of 201
Application +4 | -1
Image of the point (−1,3)(-1,3) with respect to the line y=2xy=2 x is
View answer and explanation
Correct option: C
(3,1)(3,1)
The correct answer is C (3,1) because it is the reflection of the point (-1,3) across the line y=2x, found by using perpendicular lines and midpoints.
3 Easy
3 of 201
Analyzing +4 | -1
If the pair of lines represented by ax2+2hxy+by2+2gx+2fy+c=0a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 intersect on yy-axis, then
View answer and explanation
Correct option: A
2fgh=bg2+ch22 f g h=b g^2+c h^2
Option A (2fgh=bg2+ch22 f g h=b g^2+c h^2) is correct because it represents the condition for the pair of lines to intersect on the y-axis, derived from the general equation and the intersection point's coordinates.
4 Easy
4 of 201
Application +4 | -1
If the sum of the slopes of the lines given by 4x2+2λxy−7y2=04 x^2+2 \lambda x y-7 y^2=0 is equal to the product of the slopes, then λ\lambda is equal to
View answer and explanation
Correct option: B
−2-2
The correct answer is B (-2) because the condition m1+m2=m1m2m_1 + m_2 = m_1m_2 leads to λ=−2\lambda = -2.
5 Easy
5 of 201
Application +4 | -1
The equation 3x2+2hxy+3y2=03 x^2+2 h x y+3 y^2=0 represents a pair of straight lines passing through the origin. The two lines are
View answer and explanation
Correct option: B
Real and distinct if h2>9h^2>9
Option B (Real and distinct if h2>9h^2>9) is correct because the condition for real and distinct lines is h2>abh^2 > ab, which translates to h2>9h^2 > 9 in this case.
6 Easy
6 of 201
Application +4 | -1
The equation of pair of straight lines joining the point of intersection of the curve x2+y2=4x^2+y^2=4 and y−x=2y-x=2 to the origin, is
View answer and explanation
Correct option: A
x2+y2=(y−x)2x^2+y^2=(y-x)^2
Option A (x2+y2=(y−x)2x^2+y^2=(y-x)^2) is correct because it represents the equation obtained by homogenizing the curve with respect to the line.
7 Easy
7 of 201
Understanding +4 | -1
The lines parallel to the axes and passing through the point (4,−5)(4,-5) are
View answer and explanation
Correct option: C
x=4,y=−5x=4, y=-5
Option C (x=4,y=−5x=4, y=-5) is correct because a line parallel to the x-axis has the equation y = constant, and a line parallel to the y-axis has the equation x = constant.
8 Easy
8 of 201
Application +4 | -1
The angle between the pair of straight lines formed by joining the points of intersection of x2+y2=4{x^2} + {y^2} = 4 and y=3x+cy = 3x + c to the origin is a right angle. Then, c2{c^2} is equal to
View answer and explanation
Correct option: A
20
Option A (20) is correct because homogenizing the equation and applying the condition for perpendicular lines leads to c2=20{c^2} = 20.
9 Easy
9 of 201
Analyzing +4 | -1
The equation x2+22xy+2y2+4x+42  y+1=0{x^2} + 2\sqrt 2 xy + 2{y^2} + 4x + 4\sqrt 2 \;y + 1 = 0 represents a pair of lines which are parallel to each other. The distance between them is
View answer and explanation
Correct option: D
2 units
Option D (2 units) is correct because it is the calculated distance between the two parallel lines represented by the given equation.
10 Easy
10 of 201
Application +4 | -1
If 2x2+7xy+3y2+8x+14y+λ=02 x^2+7 x y+3 y^2+8 x+14 y+\lambda=0 represents a pair of straight lines, find λ\lambda.
View answer and explanation
Correct option: A
8
Option A (8) is correct because it is the value of λ\lambda that makes the given equation represent a pair of straight lines, derived by applying the condition abc+2fgh−af2−bg2−ch2=0abc + 2fgh - af^2 - bg^2 - ch^2 = 0.
11 Easy
11 of 201
Application +4 | -1
The product of perpendiculars let fall from the point (x1,y1)\left(x_1, y_1\right) upon the lines represented by ax2+2hxy+by2a x^2+2 h x y+b y^2, is
View answer and explanation
Correct option: A
∣ax12+2hx1y1+by12∣(a−b)2+4h2\frac{\left|a x_1^2+2 h x_1 y_1+b y_1^2\right|}{\sqrt{(a-b)^2+4 h^2}}
The correct answer is A (∣ax12+2hx1y1+by12∣(a−b)2+4h2\frac{{\left|a x_1^2+2 h x_1 y_1+b y_1^2\right|}}{{\sqrt{{(a-b)^2+4 h^2}}}}) because it represents the product of perpendicular distances from a point to the lines represented by the given equation, derived using the sum and product of slopes.
12 Easy
12 of 201
Application +4 | -1
ABCDA B C D is a rhombus. Its diagonals ACA C and BDB D intersect at the point MM and satisfy BD=2ACB D=2 A C. If the coordinates of DD and MM are (1,1)(1,1) and (2,−1)(2,-1) respectively, the coordinates of AA are
View answer and explanation
Correct option: B
(1,−3/2)(1,-3 / 2)
Option B ((1,-3 / 2)) is correct because it satisfies the geometric conditions of a rhombus, where diagonals are perpendicular and one is twice the length of the other.
13 Easy
13 of 201
Application +4 | -1
The combined equation of the pair of lines through the point (1,0)(1,0) and perpendicular to the lines represented by 2x2−xy−y2=02 x^2-x y-y^2=0, is
View answer and explanation
Correct option: B
2y2+xy−x2+2x−y−1=02 y^2+x y-x^2+2 x-y-1=0
Option B (2y2+xy−x2+2x−y−1=02 y^2+x y-x^2+2 x-y-1=0) is correct because it represents the combined equation of the pair of lines through (1,0) and perpendicular to the given lines.
14 Easy
14 of 201
Application +4 | -1
The angle between the lines represented by the equation 2x2+3xy−5y2=0,2{x^2} + 3xy - 5{y^2} = 0, is
View answer and explanation
Correct option: D
tan−1∣73∣\mathrm{ta}{\mathrm{n}^{ - 1}}\left| {\frac{7}{3}} \right|
Option D (tan−1∣73∣\mathrm{ta}{\mathrm{n}^{ - 1}}\left| {\frac{7}{3}} \right|) is correct because it is calculated using the formula for the angle between lines represented by a homogeneous equation of degree 2.
15 Easy
15 of 201
Application +4 | -1
Two lines are given by the equation (3x−4y)2+k(3x−4y)=0(3 x-4 y)^2+k(3 x-4 y)=0. One of the values of kk, so that the distance between the lines is 3 is
View answer and explanation
Correct option: C
-15
Option C (-15) is correct because it satisfies the condition where the distance between the two lines is 3, derived from the given equation.

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16 Easy
16 of 201
Analyzing +4 | -1
If the slope of one of the lines represented by ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0 be the square of the other, then a+hh+8h2ab\frac{{a + h}}{h} + \frac{{8{h^2}}}{{ab}} is
View answer and explanation
Correct option: D
6
Option C (An equilateral triangle) is correct because the lines form an equilateral triangle with angles of 60 degrees.
17 Easy
17 of 201
Application +4 | -1
A straight line LL through the point (3,−2)(3,-2) is inclined at an angle 60∘60^{\circ} to the line 3x+y=1\sqrt{3} x+y=1. If LL also intersects the xx-axis, then the equation of LL is
View answer and explanation
Correct option: B
y−3x+2+33=0y-\sqrt{3} x+2+3 \sqrt{3}=0
Option B (y−3x+2+33=0y-\sqrt{3} x+2+3 \sqrt{3}=0) is correct because it represents the equation of the line passing through (3,-2) and inclined at 60 degrees to the given line, also intersecting the x-axis.
18 Easy
18 of 201
Application +4 | -1
The equation 8x2+8xy+2y2+26x+13y+15=08{x^2} + 8xy + 2{y^2} + 26x + 13y + 15 = 0 represents a pair of straight lines. The distance between then is
View answer and explanation
Correct option: B
725\frac{7}{{2\sqrt 5 }}
The correct answer is B (725\frac{7}{{2\sqrt 5 }}) because it represents the distance between the parallel lines represented by the given equation, calculated using the formula for the distance between parallel lines.
19 Easy
19 of 201
Application +4 | -1
The equation x2+ky2+4xy=0x^2+k y^2+4 x y=0 represents two coincident lines, if k=nk=n. Find the value of nn.
View answer and explanation
Correct option: A
4
Option A (4) is correct because the equation represents two coincident lines when k = 4, making the expression a perfect square.
20 Easy
20 of 201
Application +4 | -1
The angle between the pair of lines whose equation is 4x2+10xy+my2+5x+10y=04 x^2+10 x y+m y^2+5 x+10 y=0, is
View answer and explanation
Correct option: B
tan⁡−134\tan ^{-1} \frac{3}{4}
Option B (tan⁡−134\tan^{-1} \frac{3}{4}) is correct because the equation represents a pair of straight lines when m=4, and the angle between them is calculated using the formula.
21 Easy
21 of 201
Application +4 | -1
If the equation 2x2+λxy+2y2=02 x^2+\lambda x y+2 y^2=0 represents a pair of real and distinct lines, then
View answer and explanation
Correct option: C
λ∈(−∞,−4)∪(4,∞)\lambda \in(-\infty,-4) \cup(4, \infty)
Option C (λ∈(−∞,−4)∪(4,∞)\lambda \in(-\infty,-4) \cup(4, \infty)) is correct because it represents the range of λ\lambda for which the given equation represents a pair of real and distinct lines.
22 Easy
22 of 201
Application +4 | -1
For what value of λ\lambda, the equation λx2+2xy+λy2+4x+4y+\lambda x^2+2 x y+\lambda y^2+4 x+4 y+ 3=03=0 represents a pair of straight line?
View answer and explanation
Correct option: B
1
Option B (1) is correct because when λ=1\lambda = 1, the determinant of the matrix formed by the coefficients of the given equation is zero, indicating that the equation represents a pair of straight lines.
23 Easy
23 of 201
Analyzing +4 | -1
The equation y2−x2+2x−1=0{y^2} - {x^2} + 2x - 1 = 0 represents
View answer and explanation
Correct option: C
A pair of straight lines
Option C (A pair of straight lines) is correct because the given equation can be factored into two linear equations, representing two straight lines.
24 Easy
24 of 201
Application +4 | -1
If the sum of the slopes of the lines given by 4x2+2kxy−7y2=04 x^2+2 k x y-7 y^2=0 is equal to the product of the slopes, then k=k=
View answer and explanation
Correct option: C
−2-2
Option C (-2) is correct because it satisfies the condition where the sum of the slopes equals the product of the slopes for the given equation.
25 Easy
25 of 201
Analyzing +4 | -1
The combined equation of the lines L1L_1 and L2L_2 is 2x2+6xy+y2=02 x^2+6 x y+y^2=0 and that of the lines L3L_3 and L4L_4 is 4x2+18xy+y2=04 x^2+18 x y+y^2=0. If the angle between L1L_1 and L4L_4 be α\alpha, then the angle between L2L_2 and L3L_3 will be
View answer and explanation
Correct option: D
α\alpha
Option D (α\alpha) is correct because the two pairs of lines have the same angle bisectors, implying that the angle between L1L_1 and L4L_4 is equal to the angle between L2L_2 and L3L_3.
26 Easy
26 of 201
Analyzing +4 | -1
If the lines joining the origin to the points of intersection of the line y=mx+2y=m x+2 and the curve x2+y2=1x^2+y^2=1 are at right-angles, then
View answer and explanation
Correct option: C
m2=7m^2=7
Option C (m2=7m^2=7) is correct because it is derived from the condition that the lines joining the origin to the intersection points are at right angles, leading to the sum of coefficients of x2x^2 and y2y^2 being zero.
27 Easy
27 of 201
Analyzing +4 | -1
The product of the perpendicular distances from the origin on the pair of straight lines 12x2+25xy+12y2+10x+11y+2=012{x^2} + 25xy + 12{y^2} + 10x + 11y + 2 = 0 is
View answer and explanation
Correct option: B
225\frac{2}{{25}}
Option B (225\frac{2}{{25}}) is correct because it is the product of the perpendicular distances from the origin calculated using the formula c(a+b)2+4h2\frac{c}{\sqrt{(a+b)^2 + 4h^2}}.
28 Easy
28 of 201
Application +4 | -1
The equations of the sides of a parallelogram are x=2,x=3x=2, x=3 and y=1,y=5y=1, y=5. Equations to the pair of diagonals are
View answer and explanation
Correct option: D
4x+y=13,4x−y=74 x+y=13,4 x-y=7
Option D (4x+y=13,4x−y=74 x+y=13,4 x-y=7) is correct because these are the equations of the diagonals AC and BD calculated using the coordinates of the parallelogram's vertices.
29 Easy
29 of 201
Application +4 | -1
The value of λ\lambda, for which the equation x2−y2−x+λy−2=0  {x^2} - {y^2} - x + \lambda y - 2 = 0\; represents a pair of straight lines, are
View answer and explanation
Correct option: C
3,  −33,\; - 3
Option C (3, -3) is correct because these values of λ\lambda satisfy the condition for the given equation to represent a pair of straight lines.
30 Easy
30 of 201
Application +4 | -1
The image of the pair of lines represented by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 by the line mirror y=0y=0 is
View answer and explanation
Correct option: D
ax2−2hxy+by2=0a x^2-2 h x y+b y^2=0
The correct answer is D (ax2−2hxy+by2=0a x^2-2 h x y+b y^2=0) because the image of a point (x, y) with respect to the line y=0 is (x, -y), so replacing y with -y in the equation gives the image of the pair of lines.
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