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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Easy

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31 Easy
31 of 201
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.
32 Easy
32 of 201
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.
33 Easy
33 of 201
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.
34 Easy
34 of 201
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.
35 Easy
35 of 201
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.
36 Easy
36 of 201
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.
37 Easy
37 of 201
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.
38 Easy
38 of 201
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.
39 Easy
39 of 201
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.
40 Easy
40 of 201
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.
41 Easy
41 of 201
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.
42 Easy
42 of 201
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.
43 Easy
43 of 201
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.
44 Easy
44 of 201
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.
45 Easy
45 of 201
Application +4 | -1
If the equation 12  x2+7  xy−py2−18  x+qy+6=012\;{x^2} + 7\;xy - p{y^2} - 18\;x + qy + 6 = 0 represents a pair of perpendicular straight lines, then
View answer and explanation
Correct option: A
p=12,q=1p = 12,q = 1
Option A (p=12,q=1p = 12, q = 1) is correct because it satisfies both the condition for perpendicular lines and the general equation for a pair of straight lines.

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46 Easy
46 of 201
Application +4 | -1
If the equation λx2+2y2−5xy+5x−7y+3=0\lambda x^2+2 y^2-5 x y+5 x-7 y+3=0 represents two straight lines, then find the value of λ\lambda
View answer and explanation
Correct option: A
2
Option A (2) is correct because it is the value of λ\lambda that makes the given equation represent two straight lines, derived by applying the condition abc+2fgh−af2−bg2−ch2=0abc + 2fgh - af^2 - bg^2 - ch^2 = 0.
47 Easy
47 of 201
Analyzing +4 | -1
If the pair of lines ax2−2xy+by2=0a x^2-2 x y+b y^2=0 and bx2−2xy+ay2=0b x^2-2 x y+a y^2=0 be such that each pair bisects the angle between the other pair, then ∣a−b∣|a-b| equals to
View answer and explanation
Correct option: B
2
Option B (2) is correct because the condition that each pair of lines bisects the angle between the other pair leads to the equation ∣a−b∣=2|a-b|=2.
48 Easy
48 of 201
Application +4 | -1
If the equation 4x2+4xy+y2−6x−3y−4=04 x^2+4 x y+y^2-6 x-3 y-4=0 represents a pair of parallel line, then the distance between these lines is
View answer and explanation
Correct option: B
5\sqrt{5}
Option B (5\sqrt{5}) is correct because it is the calculated distance between the two parallel lines represented by the given equation, using the appropriate formula.
49 Easy
49 of 201
Application +4 | -1
Area of the triangle formed by the lines 3x2−4xy+y2=0,  2x−y=63{x^2} - 4xy + {y^2} = 0,\;2x - y = 6 is
View answer and explanation
Correct option: C
36 sq units
Option C (36 sq units) is correct because the area of the triangle formed by the given lines is calculated using the determinant formula with the intersection points of the lines.
50 Easy
50 of 201
Application +4 | -1
Equation of a straight line passing through the point of intersection of x−y+1=0x-y+1=0 and 3x+y−5=03 x+y-5=0 are perpendicular to one of them is
View answer and explanation
Correct option: B
x+y−3=0x+y-3=0
Option B (x+y−3=0x+y-3=0) is the correct answer because it represents a line passing through the intersection of the given lines and is perpendicular to one of them.
51 Easy
51 of 201
Understanding +4 | -1
The lines represents by ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0 are perpendicular to each other, if
View answer and explanation
Correct option: B
a+b=0a + b = 0
Option B (a+b=0a + b = 0) is correct because the sum of the coefficients of x2x^2 and y2y^2 is zero for perpendicular lines.
52 Easy
52 of 201
Application +4 | -1
The set of values of hh for which the equation 4x2+hxy−3y2=04 x^2+h x y-3 y^2=0 represents a pair of real and distinct lines is
View answer and explanation
Correct option: A
RR
Option A (RR) is correct because the discriminant of the quadratic equation formed by the given equation is always positive, indicating real and distinct lines for all real values of hh.
53 Easy
53 of 201
Analyzing +4 | -1
The four straight lines given by the equations 12x2+7xy−12y2=012 x^2+7 x y-12 y^2=0 and
12x2+7xy−12y2−x+7y−1=012 x^2+7 x y-12 y^2-x+7 y-1=0 lie along the sides of a
View answer and explanation
Correct option: A
square
Option A (square) is correct because the given equations represent two pairs of parallel lines that are perpendicular to each other, and the distances between the parallel lines are equal.
54 Easy
54 of 201
Application +4 | -1
If the segment of the line between the lines x−y+2=0x-y+2=0 and x+y+4=0x+y+4=0 is bisected at the origin, equation of the line is
View answer and explanation
Correct option: C
y−3x=0y-3 x=0
Option C (y−3x=0y-3 x=0) is correct because it represents the equation of a line passing through the origin and the midpoint of the segment formed by the intersection of the given lines.
55 Easy
55 of 201
Application +4 | -1
Perimeter of the quadrilateral bounded by the coordinate axis and the lines x+y=50x+y=50 and 3x+y=903 x+y=90 is
View answer and explanation
Correct option: C
80+202+101080+20 \sqrt{2}+10 \sqrt{10}
The correct answer is C (80+202+101080+20 \sqrt{2}+10 \sqrt{10}) because it is the sum of the lengths of the sides of the quadrilateral formed by the given lines and axes.
56 Easy
56 of 201
Application +4 | -1
If the lines 3x−y+1=03 x-y+1=0 and x−2y+3x-2 y+3 =0=0 are equally inclined to the line y=mxy=m x, then the value of mm is given by
View answer and explanation
Correct option: C
7m2−2m−7=07 m^2-2 m-7=0
Option C (7m2−2m−7=07 m^2-2 m-7=0) is correct because it is the equation obtained by equating the tangents of the angles between the line y=mx and the two given lines.
57 Easy
57 of 201
Application +4 | -1
PP is a point on xx-axis, QQ is a point on yy-axis. Both are equidistant from the points (7,6)(7,6) and (3,4)(3,4). Distance between PP and QQ is
View answer and explanation
Correct option: C
155/215 \sqrt{5} / 2
Option C (1552\frac{15 \sqrt{5}}{2}) is correct because it is the distance between P and Q calculated after finding their coordinates using the equidistant condition from the given points.
58 Easy
58 of 201
Application +4 | -1
Distance between the parallel lines 4x2+20xy+25y2+2x+5y−12=04 x^2+20 x y+25 y^2+2 x+5 y-12=0 is
View answer and explanation
Correct option: C
729\frac{7}{\sqrt{29}}
The correct answer is C (729\frac{7}{\sqrt{29}}) because the distance between the parallel lines is calculated using the formula after factoring the given equation.
59 Easy
59 of 201
Analyzing +4 | -1
If two pairs of straight lines having equations y2+xy−12x2=0y^2+x y-12 x^2=0 and ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 have one line common, then a=a=
View answer and explanation
Correct option: A
−3(2h+3b)-3(2 h+3 b)
Option A (−3(2h+3b)-3(2 h+3 b)) is correct because substituting y=3xy=3x into the second equation and solving for aa yields this result.
60 Easy
60 of 201
Application +4 | -1
If the lengths of the perpendiculars drawn from origin to xsec⁡θ+ycosec⁡θ=ax \sec \theta+y \operatorname{cosec} \theta=a and xcos⁡θ−ysin⁡θ=acos⁡2θx \cos \theta-y \sin \theta=a \cos 2 \theta are pp and p′p^{\prime} respectively, then 4p2+p′2=……4 p^2+p^{\prime 2}=\ldots \ldots
View answer and explanation
Correct option: A
a2a^2
Option A (a2a^2) is correct because the expression 4p2+p′24p^2 + p'^2 simplifies to a2a^2 using trigonometric identities.
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