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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Medium

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1 Medium
1 of 59
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.
2 Medium
2 of 59
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.
3 Medium
3 of 59
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.
4 Medium
4 of 59
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|.
5 Medium
5 of 59
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.
6 Medium
6 of 59
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).
7 Medium
7 of 59
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.
8 Medium
8 of 59
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.
9 Medium
9 of 59
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.
10 Medium
10 of 59
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.
11 Medium
11 of 59
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.
12 Medium
12 of 59
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.
13 Medium
13 of 59
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}.
14 Medium
14 of 59
Application +4 | -1
If the bisectors of the lines x2−2pxy−y2=0x^2-2 p x y-y^2=0 be x2−2qxy−y2=0x^2-2 q x y-y^2=0, then
View answer and explanation
Correct option: A
pq+1=0p q+1=0
Option A (pq+1=0p q+1=0) is correct because it represents the relationship between p and q when the bisectors of the lines x2−2pxy−y2=0x^2-2 p x y-y^2=0 are x2−2qxy−y2=0x^2-2 q x y-y^2=0.
15 Medium
15 of 59
Application +4 | -1
The sides of a rectangle are x=0,y=0,x=4x=0, y=0, x=4 and y=3y=3. The equation of the straight line having slope 12\frac{1}{2} that divides the rectangle into two equal halves, is
View answer and explanation
Correct option: A
2y=x+12 y=x+1
Option A (2y=x+12y = x + 1) is correct because the line with the given slope passes through the center of the rectangle, dividing it into two equal halves.

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16 Medium
16 of 59
Application +4 | -1
Pair of straight lines perpendicular to each other represented by
View answer and explanation
Correct option: A
2x2=2y(2x+y)2x^2=2y(2x+y)
Option A (2x2=2y(2x+y)2x^2=2y(2x+y)) is correct because it represents a pair of straight lines that are perpendicular to each other, satisfying the condition that the sum of the coefficients of x² and y² is zero.
17 Medium
17 of 59
Application +4 | -1
Equations of the bisectors of the lines 3x−4y+7=03 x-4 y+7=0 and 12x+5y−2=012 x+5 y-2=0 are given by
View answer and explanation
Correct option: A
21x+77y−101=0,11x−3y+9=021 x+77 y-101=0,11 x-3 y+9=0
Option A (21x+77y−101=0,11x−3y+9=021 x+77 y-101=0,11 x-3 y+9=0) is correct because it represents the equations of the bisectors of the given lines, derived using the angle bisector formula.
18 Medium
18 of 59
Application +4 | -1
The combined equation of the pair of lines through the origin and perpendicular to the pair of lines given by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0, is
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 combined equation of the pair of lines through the origin and perpendicular to the pair of lines given by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0.
19 Medium
19 of 59
Application +4 | -1
If the line x=ay+b,z=cy+dx=a y+b, z=c y+d and the line z=a′y+b′,z′=c′y+d′z=a^{\prime} y+b^{\prime}, z^{\prime}=c^{\prime} y+d^{\prime} are perpendicular, then-
View answer and explanation
Correct option: A
aa′+cc′=−1{aa}^{\prime}+{cc}^{\prime}=-1
Option A ({aa}^{\'}+{cc}^{\'}=-1{aa}^{\'}+{cc}^{\'}=-1) is correct because the lines are perpendicular if the dot product of their direction ratios is zero, leading to the given condition.
20 Medium
20 of 59
Application +4 | -1
The equation of the lines bisecting the angles between the pair of lines 3x2+xy−2y2=03 x^2+x y-2 y^2=0 is
View answer and explanation
Correct option: D
x2−10xy−y2=0x^2-10 x y-y^2=0
Option D (x2−10xy−y2=0x^2-10 x y-y^2=0) is correct because it represents the equation of the angle bisectors calculated using the standard formula.
21 Medium
21 of 59
Application +4 | -1
Three lines 3x−y=2,5x+ay=33 x-y=2,5 x+a y=3 and 2x+y=32 x+y=3 are concurrent, then a=a=
View answer and explanation
Correct option: D
−2-2
Option D (-2) is correct because the determinant formed by the coefficients of the concurrent lines must be zero, leading to a = -2.
22 Medium
22 of 59
Application +4 | -1
If hh denotes the arithmetic mean and kk denotes the geometric mean of the intercepts made on the coordinate axes by the lines passing through the point (1,1)(1,1), then the point (h,k)(h, k) lies on
View answer and explanation
Correct option: B
a parabola
Option D (all values of λ\lambda) is correct because the resulting homogeneous equation represents lines that are always perpendicular, regardless of the value of λ\lambda.
23 Medium
23 of 59
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.
24 Medium
24 of 59
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.
25 Medium
25 of 59
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.
26 Medium
26 of 59
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.
27 Medium
27 of 59
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.
28 Medium
28 of 59
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.
29 Medium
29 of 59
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.
30 Medium
30 of 59
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.
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