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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Mixed difficulty

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Showing 121-150 of 264 questions

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121 Easy
121 of 264
Understanding +4 | -1
The combined equation of bisectors of angles between coordinate axes, is
View answer and explanation
Correct option: B
x2−y2=0x^2-y^2=0
Option B (x2−y2=0x^2-y^2=0) is correct because the bisectors of the coordinate axes are y=x and y=-x, and their combined equation is x²-y²=0.
122 Easy
122 of 264
Analyzing +4 | -1
If one of the lines of my2+(1−m2)xy−mx2=0m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0 is a bisector of the angle between the lines xy=0xy = 0, then mm is/are
View answer and explanation
Correct option: C
±1\pm 1
Option C (±1\pm 1) is correct because when the lines y=±xy = \pm x are substituted into the equation, it leads to m=±1m = \pm 1.
123 Medium
123 of 264
Application +4 | -1
The value of hh, so that the equation 6x2+2hxy+12y2+22x+31y+20=0\quad 6 x^2+2 h x y+12 y^2+22 x+31 y+20=0 represent two straight lines, is
View answer and explanation
Correct option: A
17120\frac{171}{20}
Option A (17120\frac{171}{20}) is correct because it satisfies the condition for the given equation to represent a pair of straight lines.
124 Easy
124 of 264
Application +4 | -1
Equation of line pair perpendicular to ax2+by2+2hxy=0a x^2+b y^2+2 h x y=0
View answer and explanation
Correct option: C
bx2−2hxy+ay2=0b x^2-2 h x y+a y^2=0
Option C (bx2−2hxy+ay2=0b x^2-2 h x y+a y^2=0) is correct because it represents the equation of a line pair perpendicular to the given line pair.
125 Easy
125 of 264
Application +4 | -1
The pair of lines represented by 3ax2+5xy+(a2−2)y2=03 a x^2+5 x y+\left(a^2-2\right) y^2=0 are perpendicular to each other for
View answer and explanation
Correct option: A
Two values of aa
Option A (Two values of aa) is correct because setting the sum of the coefficients of x2x^2 and y2y^2 to zero results in a quadratic equation with two real roots for aa.
126 Easy
126 of 264
Analyzing +4 | -1
If pairs straight lines x2−2pxy−y2=0{x^2} - 2pxy - {y^2} = 0 and x2−2qxy−y2=0{x^2} - 2qxy - {y^2} = 0 be such that each pair bisects the angle between the other pair, then
View answer and explanation
Correct option: B
pq=−1pq = - 1
Option B (91 sq. units) is correct because the area of the triangle ABC, calculated using the coordinates of A, B, and C, is 91 square units.
127 Medium
127 of 264
Application +4 | -1
If the lines represented by the equation ax2−bxy−y2=0a x^2-b x y-y^2=0 make angles α\alpha and β\beta with the xx-axis, then tan⁡(α+β)=\tan (\alpha+\beta)=
View answer and explanation
Correct option: B
−b1+a\frac{-b}{1+a}
Option B (−b1+a\frac{{-b}}{{1+a}}) is correct because it is derived from the formula for the tangent of the sum of two angles, where the slopes of the lines are related to the coefficients of the given equation.
128 Medium
128 of 264
Application +4 | -1
If the acute angles between the pairs of lines 3x2−7xy+4y2=03 x^2-7 x y+4 y^2=0 and 6x2−5xy+y2=06 x^2-5 x y+y^2=0 be θ1\theta_1 and θ2\theta_2 respectively, then
View answer and explanation
Correct option: A
θ1=θ2\theta_1=\theta_2
Option A (θ1=θ2\theta_1=\theta_2) is correct because the acute angles between the pairs of lines are equal.
129 Easy
129 of 264
Application +4 | -1
If the line 5x=y\sqrt{5} x=y meets the lines x=1,x=2x=1, x=2, …x=n\ldots x=n at points A1,A2,…,AnA_1, A_2, \ldots, A_n respectively, then (0A1)2+(0A2)2+…+(0An)2\left(0 A_1\right)^2+\left(0 A_2\right)^2+\ldots+\left(0 A_n\right)^2 is equal to
View answer and explanation
Correct option: B
2n3+3n2+n2 n^3+3 n^2+n
Option B (2n3+3n2+n2 n^3+3 n^2+n) is correct because it is the sum of the squares of the distances from the origin to the points A1, A2, ..., An.
130 Easy
130 of 264
Application +4 | -1
The angle between the lines in x2−xy−6y2−7x+31y−18=0{x^2} - xy - 6{y^2} - 7x + 31y - 18 = 0 is
View answer and explanation
Correct option: B
45∘45^\circ
Option B (45∘45^\circ) is correct because the angle between the lines is calculated using the formula involving the coefficients of the quadratic equation.
131 Medium
131 of 264
Application +4 | -1
The area bounded by the curves x+2∣y∣=1x+2|y|=1 and x=0x=0, is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because the area is formed by two triangles with a base of 1 and a height of 1/2, resulting in a total area of 1/2.
132 Easy
132 of 264
Application +4 | -1
If the slopes of one of the lines given by ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0 is 5 times the other, then
View answer and explanation
Correct option: A
5h2=9ab5{h^2} = 9ab
Option A (5h2=9ab5{h^2} = 9ab) is correct because it represents the condition when one slope is 5 times the other for the given equation of pair of straight lines.
133 Hard
133 of 264
Application +4 | -1
The square of distance between the point of intersection of the lines represented by the equation ax2+2hxy+by2+2gx+2fy+c=0a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 and origin, is
View answer and explanation
Correct option: A
c(a+b)−f2−g2ab−h2\frac{c(a+b)-f^2-g^2}{a b-h^2}
Option A (c(a+b)−f2−g2ab−h2\frac{c(a+b)-f^2-g^2}{a b-h^2}) is correct because it represents the square of the distance between the point of intersection of the lines and the origin.
134 Medium
134 of 264
Application +4 | -1
If one of the lines of the pair ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 bisects the angles between positive directions of the axes, then a,ba, b and hh satisfy the relation
View answer and explanation
Correct option: B
a+b=−2ha+b=-2 h
Option B (a+b=−2ha+b=-2h) is correct because if y=xy=x bisects the angles, substituting it into the equation gives the relationship between a, b, and h.
135 Easy
135 of 264
Analyzing +4 | -1
If two of the straight lines represented by ax3+bx2y+cxy2+dy3=0a x^3+b x^2 y+c x y^2+d y^3=0 are at right angles, then,
View answer and explanation
Correct option: D
a2+ac+bd+d2=0a^2+a c+b d+d^2=0
Option D (a2+ac+bd+d2=0a^2+a c+b d+d^2=0) is correct because it represents the condition when two of the three lines represented by the given cubic equation are at right angles.

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136 Easy
136 of 264
Application +4 | -1
The locus of the point P(h,k)\mathrm{P}(h, k), when the area of the triangle formed by the lines y=x,x+y=2y=x, x+y=2 and the line through P(h,k)P(h, k) and parallel to the xx-axis is 4h24 h^2 is
View answer and explanation
Correct option: B
2x+y−1=02 x+y-1=0
Option B (2x+y−1=02 x+y-1=0) is correct because it represents the locus of the point P(h, k) based on the given area condition.
137 Medium
137 of 264
Application +4 | -1
Area of the triangle formed by the lines y2−9xy+18x2=0y^2-9 x y+18 x^2=0 and y=9y=9, is
View answer and explanation
Correct option: A
274\frac{27}{4} sq units
Option A (274\frac{27}{4} sq units) is correct because it is the area of the triangle calculated using the formula for the area of a triangle formed by a pair of straight lines and a third line.
138 Easy
138 of 264
Application +4 | -1
If one of the lines of the pair ax2+2a x^2+2 hxy +by2=0+b y^2=0 bisects the angle between positive direction of the axes, then a, b and h satisfy the relation.
View answer and explanation
Correct option: B
a+b=−2ha+b=-2 h
Option B (a+b=−2ha+b=-2 h) is correct because substituting y=xy=x into the equation and simplifying leads to the relation a+b=−2ha+b=-2h.
139 Easy
139 of 264
Application +4 | -1
If the lines x+(a−1)y+1=0x+(a-1) y+ 1=0 and 2x+a2y−1=02 x+a^2 y-1=0 are mutually perpendicular, then .......
View answer and explanation
Correct option: D
a=−1a=-1
Option D (a=−1a=-1) is correct because it is the only real solution to the equation derived from the perpendicularity condition of the two lines.
140 Easy
140 of 264
Application +4 | -1
The equation of the bisectors of the angle between the lines represented by the equation x2−y2=0x^2-y^2=0, is
View answer and explanation
Correct option: C
xy=0x y=0
Option C (xy=0x y=0) is correct because it represents the combined equation of the bisectors of the angles between the given lines.
141 Easy
141 of 264
Application +4 | -1
The equation 4x2−24xy+11y2=04 x^2-24 x y+11 y^2=0 represents
View answer and explanation
Correct option: C
Two lines through the origin
Option C (Two lines through the origin) is correct because the given equation is a homogeneous equation of degree 2, which always represents a pair of straight lines passing through the origin.
142 Medium
142 of 264
Understanding +4 | -1
The combined equation of the images of the pair of lines given by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 in the line mirror y=0y=0, is
View answer and explanation
Correct option: A
ax2−2hxy+by2=0a x^2-2 h x y+b y^2=0
Option A (ax2−2hxy+by2=0a x^2-2 h x y+b y^2=0) is correct because reflecting a pair of lines in the line y=0 only changes the sign of the y term.
143 Easy
143 of 264
Application +4 | -1
Let the equation of the pair of lines, y=pxy = px and y=qxy = qx can be written as (y−px)(y−qx)=0(y - px)(y - qx) = 0 Then, the equation of the pair of the angle bisectors of the line x2−4xy−5y2=0{x^2} - 4xy - 5{y^2} = 0 is
View answer and explanation
Correct option: C
x2+3xy−y2=0{x^2} + 3xy - {y^2} = 0
Option C (x2+3xy−y2=0{x^2} + 3xy - {y^2} = 0) is correct because it's the equation of the angle bisectors derived from the given pair of lines using the standard formula.
144 Medium
144 of 264
Application +4 | -1
The area of the triangle formed by joining the origin to the points of intersection of the line 5x+2y=35\sqrt{5} x+2 y=3 \sqrt{5} and circle x2+y2=10x^2+y^2=10, is
View answer and explanation
Correct option: B
5 sq units
Option B (5 sq units) is correct because it is the area of the triangle formed by the origin and the intersection points of the given line and circle.
145 Easy
145 of 264
Application +4 | -1
The equation of the line parallel to lines L1≡x+2y−5=0L_1 \equiv x+2 y-5=0 and L2≡x+2y+9=0L_2 \equiv x+2 y+9=0 and dividing the distance between L1L_1 and L2L_2 in the ratio 1:61: 6 (internally), is
View answer and explanation
Correct option: A
x+2y−3=0x+2 y-3=0
Option A (x+2y−3=0x+2 y-3=0) is correct because it represents a line parallel to the given lines and divides the distance between them in the specified ratio.
146 Easy
146 of 264
Application +4 | -1
If the product of the perpendiculars drawn from the point (1,1) on the lines ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 is 1 , then
View answer and explanation
Correct option: C
h(a+b)+ab=0h(a+b)+a b=0
Option C (h(a+b)+ab=0h(a+b)+a b=0) is correct because it is derived from the condition that the product of perpendiculars from (1,1) to the lines is 1.
147 Easy
147 of 264
Application +4 | -1
Let OO be the origin, A(1,0)A(1,0) and B(0,1)B(0,1) and P(x,y)P(x, y) are points such that xy>0x y > 0 and x+y<1x+y < 1, then
View answer and explanation
Correct option: A
PP lies either inside the triangle OABO A B or in the third quadrant
The correct answer is that PP lies either inside the triangle OABOAB or in the third quadrant because xy>0xy > 0 implies PP is in the first or third quadrant, and x+y<1x + y < 1 restricts PP to below the line x+y=1x + y = 1.
148 Medium
148 of 264
Application +4 | -1
If the line L: x5+yb=1\frac{x}{5}+\frac{y}{b}=1 passes through (13,32)(13,32) and line K:xc+y3=1\mathrm{K}: \frac{x}{c}+\frac{y}{3}=1 is parallel to L\mathrm{L}, then the distance between L\mathrm{L} and K=…….\mathrm{K}=\ldots \ldots ..
View answer and explanation
Correct option: D
2317\frac{23}{\sqrt{17}}
Option D (2317\frac{{23}}{{\sqrt{{17}}}}) is correct because it represents the distance between the two parallel lines L and K.
149 Medium
149 of 264
Application +4 | -1
Area of the parallelogram whose sides are xcos⁡α+ysin⁡α=p  xcos⁡α+ysin⁡α=qx \cos \alpha+y \sin \alpha=p \,\,x \cos \alpha+y \sin \alpha=q, xcos⁡β+ysin⁡β=rx \cos \beta+y \sin \beta=r and xcos⁡β+ysin⁡β=sx \cos \beta+y \sin \beta=s is
View answer and explanation
Correct option: A
±(p−q)(r−s)cosec⁡(α−β)\pm(p-q)(r-s) \operatorname{cosec}(\alpha-\beta)
Option A (x2+y2−hx−ky=0x^2+y^2-hx-ky=0) is correct because it represents the locus of the foot of the perpendicular from the origin to a line passing through (h,k).
150 Easy
150 of 264
Application +4 | -1
The combined equation of the two lines ax+by+ca x+b y+c =0=0 and a′x+b′y+c′=0a^{\prime} x+b^{\prime} y+c^{\prime}=0 can be written as (ax+by(a x+b y +c)(a′x+b′y+c′)=0+c)\left(a^{\prime} x+b^{\prime} y+c^{\prime}\right)=0
The equation of the angle bisectors of the lines represented by the equation 2x2+xy−3y2=02 x^{2}+x y-3 y^{2}=0 is
View answer and explanation
Correct option: D
x2−y2−10xy=0x^{2}-y^{2}-10 x y=0
Option D (x2−y2−10xy=0x^{2}-y^{2}-10 x y=0) is correct because it represents the equation of the angle bisectors derived from the given equation.
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