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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Mixed difficulty

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Showing 1-30 of 264 questions

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1 Easy
1 of 264
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 264
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 264
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 264
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 264
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 Medium
6 of 264
Application +4 | -1
The value of hh for which 3x2−3hxy+4y2=03 x^2-3 h x y+4 y^2=0 represents a pair of coincident lines are
View answer and explanation
Correct option: D
±4/3\pm 4 / \sqrt{3}
Option D (±4/3\pm 4 / \sqrt{3}) is correct because the condition for coincident lines (\h2=ab\h^2 = ab) is satisfied when h equals this value.
7 Medium
7 of 264
Application +4 | -1
The equation of a pair of straight lines is ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0. By what angle must the axes be rotated so that the term containing xyx y in the equation may be removed?
View answer and explanation
Correct option: B
12tan⁡−1(2ha−b)\frac{1}{2} \tan ^{-1}\left(\frac{2 h}{a-b}\right)
Option B (12tan⁡−1(2ha−b)\frac{1}{2} \tan ^{-1}\left(\frac{2 h}{a-b}\right)) is correct because it gives the angle required to rotate the axes to eliminate the xy term in the given equation.
8 Medium
8 of 264
Application +4 | -1
A point moves in such a way that the distance between the feet of the perpendiculars drawn from it to the lines ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 is a constant 2λ2 \lambda. Then, the locus of the point is
View answer and explanation
Correct option: B
(x2+y2)(h2−ab)=λ2[(a−b)2+4h2]\left(x^2+y^2\right)\left(h^2-a b\right)=\lambda^2\left[(a-b)^2+4 h^2\right]
Option B ((x2+y2)(h2−ab)=λ2[(a−b)2+4h2]\left(x^2+y^2\right)\left(h^2-a b\right)=\lambda^2\left[(a-b)^2+4 h^2\right]) is correct because it represents the locus of the point derived from the given condition about the distance between the feet of perpendiculars.
9 Easy
9 of 264
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.
10 Easy
10 of 264
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.
11 Easy
11 of 264
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.
12 Medium
12 of 264
Application +4 | -1
If the angle between the lines xa+yb=1,xa−yb=1\frac{x}{a}+\frac{y}{b}=1, \frac{x}{a}-\frac{y}{b}=1 has measure α\alpha, then tan⁡α=……\tan \alpha=\ldots \ldots
View answer and explanation
Correct option: D
∣2aba2−b2∣\left|\frac{2 a b}{a^2-b^2}\right|
Option D is correct. The angle between two lines is given by tan⁡α=∣m1−m21+m1m2∣\tan \alpha = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|. Substituting the slopes of the given lines, we get tan⁡α=∣2aba2−b2∣\tan \alpha = \left|\frac{2 a b}{a^2 - b^2}\right|.
13 Easy
13 of 264
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.
14 Easy
14 of 264
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.
15 Medium
15 of 264
Application +4 | -1
The equation of a line bisecting the join of (2010, 1600)1600) and (−1340,1080)(-1340,1080) and having intercept on the axes in the ratio 1:21: 2 is
View answer and explanation
Correct option: C
2x+y=20102 x+y=2010
Option C (2x+y=20102 x+y=2010) is correct because the line bisects the join of the given points and has intercepts in the ratio 1:2, satisfying the given conditions.

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16 Easy
16 of 264
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.
17 Medium
17 of 264
Application +4 | -1
Equation of the straight line making equal angles with the straight lines x+y=1x+y=1 and 2x+3y=12 x+3 y=1 and passing through the point (1,2)(1,2) may be given as
View answer and explanation
Correct option: A
13(x+y−3)=±2(2x+3y−8)\sqrt{13}(x+y-3)= \pm \sqrt{2}(2 x+3 y-8)
Option A (13(x+y−3)=±2(2x+3y−8)\sqrt{13}(x+y-3)= \pm \sqrt{2}(2 x+3 y-8)) is correct because it represents the equation of the straight line that makes equal angles with the given lines and passes through the specified point (1,2).
18 Easy
18 of 264
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.
19 Medium
19 of 264
Analyzing +4 | -1
A pair of straight line drawn through the origin form with the line 2x+3y=62 x+3 y=6 an isosceles right angled triangle, then the lines and the area of the triangle, thus formed is
View answer and explanation
Correct option: A
x−5y=0,5x+y=0;Δ=3613x-5 y=0,5 x+y=0 ; \Delta=\frac{36}{13}
Option A (x−5y=0,5x+y=0;Δ=3613x-5 y=0,5 x+y=0 ; \Delta=\frac{{36}}{{13}}) is correct because these lines form an isosceles right-angled triangle with the given line, and the area is calculated correctly.
20 Hard
20 of 264
Application +4 | -1
The equation of the pair of straight lines, each of which makes angle α\alpha with the line y=xy=x, is
View answer and explanation
Correct option: D
x2−2xysec⁡2α+y2=0x^2-2 x y \sec 2 \alpha+y^2=0
Option D (x2−2xysec⁡2α+y2=0x^2-2 x y \sec 2 \alpha+y^2=0) is correct because it represents the equation of the pair of straight lines, each of which makes an angle α\alpha with the line y=xy=x.
21 Easy
21 of 264
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.
22 Hard
22 of 264
Analyzing +4 | -1
If the line lx+my+n=0lx + my + n = 0 intersects the curve ax2+2hxy+by2=1a{x^2} + 2hxy + b{y^2} = 1 at PP and QQ such that the circle with PQPQ as a diameter passes through the origin, then l2+m2={l^2} + {m^2} =
View answer and explanation
Correct option: A
n2(a+b){n^2}\left( {a + b} \right)
Option A n2(a+b){n^2}\left( {a + b} \right) is correct because it is derived from the condition that the lines joining the origin to the intersection points are at right angles.
23 Easy
23 of 264
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.
24 Easy
24 of 264
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.
25 Easy
25 of 264
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.
26 Easy
26 of 264
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.
27 Easy
27 of 264
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.
28 Easy
28 of 264
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.
29 Hard
29 of 264
Analyzing +4 | -1
If ax3+by3+cx2y+dxy2=0a x^3+b y^3+c x^2 y+d x y^2=0 represents three distinct straight lines, such that each bisects the angle between the other two, then which of the following is true?
View answer and explanation
Correct option: A
3b+c=03 b+c=0
Option B (3a+d=03 a+d=0) is correct because it represents the condition for the three distinct lines to bisect the angles between each other, derived from the relationship between the roots and coefficients of the cubic equation.
30 Easy
30 of 264
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.
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