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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Mixed difficulty

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Showing 31-60 of 264 questions
31 Easy
31 of 264
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.
32 Easy
32 of 264
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.
33 Easy
33 of 264
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.
34 Easy
34 of 264
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.
35 Easy
35 of 264
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.
36 Easy
36 of 264
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.
37 Easy
37 of 264
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}}.
38 Easy
38 of 264
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.
39 Easy
39 of 264
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.
40 Easy
40 of 264
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.
41 Easy
41 of 264
Analyzing +4 | -1
The two straight lines given by
x2(tan⁡2θ+cos⁡2θ)−2xytan⁡θ+y2sin⁡2θ=0x^2\left(\tan ^2 \theta+\cos ^2 \theta\right)-2 x y \tan \theta+y^2 \sin ^2 \theta=0
make with the axis of xx angles such that the difference of their tangents, is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the difference of the tangents of the angles the lines make with the x-axis simplifies to 2.
42 Easy
42 of 264
Application +4 | -1
If the angle θ\theta is acute, then the acute angle between x2(cos⁡θ−sin⁡θ)+2xycos⁡θ+y2(cos⁡θ+sin⁡θ)=0{x^2}\left( {\cos \mathrm{\theta } - \sin \mathrm{\theta }} \right) + 2xy\cos \mathrm{\theta } + {y^2}\left( {\cos \mathrm{\theta } + \sin \mathrm{\theta }} \right) = 0 is
View answer and explanation
Correct option: C
θ\theta
Option D (−2,  12- 2,\;\frac{1}{2}) is correct because these are the slopes of the lines that make a 45-degree angle with the given line, calculated using the tangent of the angle between two lines.
43 Easy
43 of 264
Application +4 | -1
The point (a2,a)\left(a^2, a\right) lies between the straight lines x+y=6x+y=6 and x+y=2x+y=2 for
View answer and explanation
Correct option: C
∣2a−3∣<1|2 a-3| < 1
The correct answer is C (∣2a−3∣<1|2 a-3| < 1) because it represents the range of 'a' for which the point (a2, a) lies between the two lines.
44 Medium
44 of 264
Analyzing +4 | -1
The equation x3+y3=0x^3+y^3=0 represents
View answer and explanation
Correct option: C
one real and two imaginary lines
The correct answer is B (tan⁡−1(247)\tan^{-1}(\frac{24}{7})) because it is derived from the condition of equal areas and perpendicular lines.
45 Easy
45 of 264
Application +4 | -1
The lines bisecting the angle between the bisectors of the angles between the lines ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0 are given by
View answer and explanation
Correct option: C
(a−b)(x2−y2)+4  hxy=0\left( {a - b} \right)\left( {{x^2} - {y^2}} \right) + 4\;hxy = 0
Option C ((a−b)(x2−y2)+4  hxy=0\left( {a - b} \right)\left( {{x^2} - {y^2}} \right) + 4\;hxy = 0) is correct because it represents the equation of the lines bisecting the angles between the bisectors of the original lines.

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46 Easy
46 of 264
Analyzing +4 | -1
The orthocentre of the triangle formed by the pair of lines 2x2−xy−y2+x+2y−1=02 x^2-x y-y^2+x+2 y-1=0 and the line x+y+1=0x+y+1=0, is
View answer and explanation
Correct option: A
(−1,0)(-1,0)
Option A ((-1,0)) is correct because the orthocenter of a right-angled triangle is the vertex where the right angle is formed, which is the intersection of the given lines.
47 Medium
47 of 264
Application +4 | -1
If r(1−m2)+m(p−q)=0r\left(1-m^2\right)+m(p-q)=0, then a bisector of the angle between the lines represented by the equation px2−2rxy+qy2=0p x^2-2 r x y+q y^2=0, is
View answer and explanation
Correct option: C
y=mxy=m x
Option C, y=mxy=mx, is correct because substituting this equation into the bisector equation leads to the given condition.
48 Easy
48 of 264
Application +4 | -1
Let the coordinates of PP be (x,y)(x, y) and of QQ be (α,β)(\alpha, \beta) where α\alpha is the geometric and β\beta is the arithmetic mean of the coordinates of PP. If the mid point of PQP Q is (42,31)(42,31) the coordinates of PP are
View answer and explanation
Correct option: B
(49,25)(49,25)
Option B ((49,25)) is correct because it satisfies both the geometric and arithmetic mean conditions derived from the midpoint formula.
49 Easy
49 of 264
Application +4 | -1
If the pair of straight lines given by Ax2+2Hxy+By2=0(H2>AB)A{x^2} + 2Hxy + B{y^2} = 0({H^2} > AB) forms an equilateral triangle with line ax+by+c=0,ax + by + c = 0, then (A+3B)(3A+B)\left( {A + 3B} \right)\left( {3A + B} \right) is equal to
View answer and explanation
Correct option: D
4H24{H^2}
Option D (4H24{H^2}) is correct because it is the result obtained after using the condition for an equilateral triangle formed by the pair of lines and the given line.
50 Easy
50 of 264
Analyzing +4 | -1
The square of the distance between the origin and the point of intersection of the lines given by ax2+2hxy+by2+2gx+2fy+c=0,a x^2+2 h x y+b y^2+2 g x+2 f y+c=0, is
View answer and explanation
Correct option: C
c(a+b)−f2−g2ab−h2\frac{c(a+b)-f^2-g^2}{a b-h^2}
Option B (c(a+b)−f2−g2h2−ab\frac{{c(a+b)-f^2-g^2}}{{h^2-a b}}) is correct because it represents the square of the distance between the origin and the intersection point of the given lines, derived using partial derivatives and algebraic manipulation.
51 Easy
51 of 264
Application +4 | -1
If the pair of lines represented by ax2+2hxy+by2=0,b≠0a x^2+2 h x y+b y^2=0, b \neq 0, are such that the sum of the slopes of the lines is three times the product of their slopes, then
View answer and explanation
Correct option: B
3a+2h=03 a+2 h=0
Option B (3a+2h=03 a+2 h=0) is correct because it represents the relationship between the coefficients when the sum of the slopes is three times the product of the slopes.
52 Easy
52 of 264
Analyzing +4 | -1
Equations (b−c)x+(c−a)y+(a−b)=0(b-c) x+(c-a) y+(a-b)=0 \quad and (b3−c3)x+(c3−a3)y+(a3−b3)=0\left(b^3-c^3\right) x+\left(c^3-a^3\right) y+\left(a^3-b^3\right)=0 \quad (where, a,ba, b and cc are not all equal) will represent the same line, if
View answer and explanation
Correct option: D
a+b+c=0a+b+c=0
Option D (a+b+c=0a+b+c=0) is correct because it satisfies the condition for the two given equations to represent the same line.
53 Easy
53 of 264
Application +4 | -1
One bisector of the angle between the lines given by a(x−1)2+2h(x−1)y+by2=0a(x-1)^2+2 h(x-1) y+b y^2=0 is 2x+y−2=02 x+y-2=0. The equation of the other bisector is
View answer and explanation
Correct option: C
x−2y−1=0x-2 y-1=0
Option C (x−2y−1=0x-2y-1=0) is correct because the angle bisectors are perpendicular, and the other bisector must pass through (1,0) and be perpendicular to the given bisector.
54 Easy
54 of 264
Application +4 | -1
A line through P(1,2)P(1,2) is such that it makes unequal intercepts on the axes, and the intercept between the axes is trisected at PP, an equation of the line is
View answer and explanation
Correct option: B
4x+y=64 x+y=6
Option B (4x+y=64 x+y=6) is correct because it satisfies the condition that the intercept between the axes is trisected at P(1,2)P(1,2), and the intercepts are unequal.
55 Easy
55 of 264
Analyzing +4 | -1
The pairs of straight lines ax2+2hxy−ay2=0a{x^2} + 2hxy - a{y^2} = 0 and hx2−2axy−hy2=0h{x^2} - 2axy - h{y^2} = 0 are such that
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the given pairs of lines are perpendicular, bisect each other's angles, and are equally inclined.
56 Medium
56 of 264
Application +4 | -1
The value of kk for which the lines 7x−8y+5=07 x-8 y+5=0, 3x−4y+5=03 x-4 y+5=0 and 4x+5y+k=04 x+5 y+k=0 are concurrent is given by
View answer and explanation
Correct option: A
−45-45
Option A (-45) is correct because the determinant of the coefficients of the concurrent lines must be zero, leading to k = -45.
57 Easy
57 of 264
Application +4 | -1
The equation 12x2+7xy+ay2+13x−y+3=012{x^2} + 7xy + a{y^2} + 13x - y + 3 = 0 represents a pair of perpendicular lines. Then, the value of 'a' is
View answer and explanation
Correct option: C
-12
The correct answer is -12 because for perpendicular lines, the sum of the coefficients of x² and y² must be zero.
58 Medium
58 of 264
Analyzing +4 | -1
To the lines ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 the lines a2x2+2h(a+b)xy+b2y2=0a^2 x^2+2 h(a+b) x y+b^2 y^2=0 are
View answer and explanation
Correct option: A
equally inclined
Option A (equally inclined) is correct because the angle bisectors of both pairs of lines are the same, indicating they are equally inclined.
59 Medium
59 of 264
Application +4 | -1
The three lines lx+my+n=0,mx+ny+l=0,nx+ly+m=0l x+m y+n=0, m x+n y+l=0, n x+l y+m=0 are concurrent if
View answer and explanation
Correct option: D
l+m+n=0l+m+n=0
Option D (l+m+n=0) is correct because the determinant of the coefficients of the concurrent lines must be zero, which simplifies to l+m+n=0.
60 Medium
60 of 264
Application +4 | -1
The angle between the lines joining the origin to the point of intersection of the line 3x+y=2\sqrt{3} x+y=2 and the circle x2+y2=4x^2+y^2=4, is
View answer and explanation
Correct option: C
π3\frac{\pi}{3}
Option C (π3\frac{\pi}{3}) is correct because homogenizing the equation and using the formula for the angle between the lines yields π3\frac{\pi}{3}.
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