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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Mixed difficulty

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91 Easy
91 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 m1m_1 and m2m_2 is 4x2+18xy+y2=04 x^2+18 x y+y^2=0. If the angle between l1l_1 and m2m_2 is α\alpha, then the angle between l2l_2 and m1m_1 will be
View answer and explanation
Correct option: C
α\alpha
Option C (α) is correct because the pairs of lines are equally inclined to each other, implying the angle between l2 and m1 is the same as the angle between l1 and m2.
92 Easy
92 of 264
Analyzing +4 | -1
Given the four lines with the equations

x+2y−3=0,3x+4y−7=0,x+2 y-3=0,3 x+4 y-7=0,

2x+3y−4=0,4x+5y−6=0, then 2 x+3 y-4=0,4 x+5 y-6=0, \text { then }
View answer and explanation
Correct option: C
three lines are concurrent
Option C (three lines are concurrent) is correct because the intersection of x+2y−3=0x+2y-3=0 and 2x+3y−4=02x+3y-4=0 lies on the line 4x+5y−6=04x+5y-6=0.
93 Easy
93 of 264
Application +4 | -1
The angle between the lines represented by x2+2xysec⁡θ+y2=0x^2+2 x y \sec \theta+y^2=0, is
View answer and explanation
Correct option: C
θ\theta
Option C (θ\theta) is correct because applying the formula for the angle between the lines represented by the given equation simplifies to θ\theta.
94 Easy
94 of 264
Application +4 | -1
A line perpendicular to the line segment joining the points (7,3)(7,3) and (3,7)(3,7) divides it in the ratio 1:31: 3, the equation of the line is
View answer and explanation
Correct option: D
x−y−2=0x-y-2=0
Option D (x−y−2=0x-y-2=0) is correct because it represents the equation of a line perpendicular to the given line segment and dividing it in the ratio 1:3.
95 Easy
95 of 264
Application +4 | -1
The product of the perpendiculars drawn from the point (1,2)(1,2) to the pair of lines x2+4xy+y2=0x^2+4 x y+y^2=0 is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the product of perpendiculars is calculated using the formula and substituting the given values, resulting in 13/4.
96 Easy
96 of 264
Application +4 | -1
The distance between the pair of parallel lines x2+2xy+y2−8ax−8ay−9a2=0{x^2} + 2xy + {y^2} - 8ax - 8ay - 9{a^2} = 0 is
View answer and explanation
Correct option: D
52a5\sqrt 2 a
Option D (52a5\sqrt 2 a) is correct because it is the distance calculated using the formula for the distance between parallel lines represented by the given equation.
97 Easy
97 of 264
Application +4 | -1
If the equation 6x2−5xy+y2=06 x^2-5 x y+y^2=0 represents a pair of distinct straight lines each passing through the origin. Then, the separate equation of these lines are
View answer and explanation
Correct option: A
y−3x=0y-3 x=0 and y−2x=0y-2 x=0
Option A (y−3x=0y-3 x=0 and y−2x=0y-2 x=0) is correct because the given homogeneous equation can be factored into these two linear equations representing lines passing through the origin.
98 Easy
98 of 264
Application +4 | -1
If the slope of one of the lines given by 6x2+axy+6 x^2+a x y+ y2=0y^2=0 exceeds the slope of the other by one, then aa is equal to
View answer and explanation
Correct option: D
±5\pm 5
Option D (±5\pm 5) is correct because by relating the sum and product of the slopes to the coefficients of the given equation and using the condition that the difference of the slopes is 1, we find that a=±5a = \pm 5.
99 Easy
99 of 264
Application +4 | -1
If the distance of a point (x1,y1)\left(x_1, y_1\right) from each of the two straight lines, which pass through the origin of coordinates, is δ\delta, then the two lines are given by
View answer and explanation
Correct option: A
(x1y−xy1)2=δ2(x2+y2)\left(x_1 y-x y_1\right)^2=\delta^2\left(x^2+y^2\right)
Option A ((x1y−xy1)2=δ2(x2+y2)x_1 y-x y_1)^2=\delta^2(x^2+y^2)) is correct because it is derived from the distance formula of a point from a line passing through the origin.
100 Medium
100 of 264
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.
101 Easy
101 of 264
Application +4 | -1
The angle between the pair of lines represented by 2x2−7xy+3y2=02 x^2-7 x y+3 y^2=0, is
View answer and explanation
Correct option: B
45∘45^{\circ}
Option B (45∘45^{\circ}) is correct because using the formula for the angle between lines, tan⁡θ=2h2−aba+b\tan \theta = \frac{{2 \sqrt{{h^2 - ab}}}}{{a + b}}, yields θ=45∘\theta = 45^{\circ}.
102 Easy
102 of 264
Application +4 | -1
The combined equation of the bisectors of the angle between the lines represented by (x2+y2)3=4xy\left(x^2+y^2\right) \sqrt{3}=4 x y is
View answer and explanation
Correct option: A
y2−x2=0y^2-x^2=0
Option A, y2−x2=0y^2-x^2=0, is correct because it represents the combined equation of the bisectors derived from the given equation.
103 Easy
103 of 264
Analyzing +4 | -1
The triangle formed by the lines whose combined equation is (y2−4xy−x2)(x+y−1)=0\left(y^2-4 x y-x^2\right)(x+y-1)=0, is
View answer and explanation
Correct option: B
Right angled
Option B (Right angled) is correct because the equation (y2−4xy−x2)(x+y−1)=0\left(y^2-4 x y-x^2\right)(x+y-1)=0 represents a pair of perpendicular lines and another line, forming a right-angled triangle.
104 Easy
104 of 264
Application +4 | -1
If the bisectors of angles represented by ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0 and a′x2+2h′xy+b′y2=0a'{x^2} + 2h'xy + b'{y^2} = 0 are same, then
View answer and explanation
Correct option: A
(a−b)h′=(a′−b′)h\left( {a - b} \right)h' = \left( {a' - b'} \right)h
The correct answer is A ((a - b)h' = (a' - b')h) because the condition for the bisectors of two pairs of lines to be the same is used.
105 Easy
105 of 264
Application +4 | -1
If the equation 3x2+xy−y2−3x+6y+k=03{x^2} + xy - {y^2} - 3x + 6y + k = 0 represents a pair of straight lines, then the values of kk is
View answer and explanation
Correct option: C
-9
Option B (6x+y−19=06x + y - 19 = 0) is correct because it represents the equation of the line that bisects the segment joining the points (-4, 6) and (8, 8) perpendicularly.

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106 Easy
106 of 264
Understanding +4 | -1
The equation of a line in parametric form is given by
View answer and explanation
Correct option: B
x−x1cos⁡θ=y−y1sin⁡θ=r\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r
Option B (x−x1cos⁡θ=y−y1sin⁡θ=r\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r) is correct because it represents the parametric form of a line, where r is the distance from the point (x1,y1)(x_1, y_1) to (x,y)(x, y).
107 Easy
107 of 264
Analyzing +4 | -1
If two of the lines given by the equation ay4+bxy3+cx2y2+dx3y+ex4=0a{y^4} + bx{y^3} + c{x^2}{y^2} + d{x^3}y + e{x^4} = 0 are at right angle, then
View answer and explanation
Correct option: A
(c+a+e)(e−a)2+(b+d)(ad+be)=0\left( {c + a + e} \right){\left( {e - a} \right)^2} + \left( {b + d} \right)\left( {ad + be} \right) = 0
Option A (c+a+e)(e−a)2+(b+d)(ad+be)=0\left( {c + a + e} \right){\left( {e - a} \right)^2} + \left( {b + d} \right)\left( {ad + be} \right) = 0 is correct because the condition for two lines to be at right angles leads to this relation between the coefficients of the given equation.
108 Easy
108 of 264
Application +4 | -1
If the sum of the slopes of the lines given by 3x2−2cxy−5y2=03 x^2-2 c x y-5 y^2=0 is twice their product, then the value of cc is
View answer and explanation
Correct option: B
3
The correct answer is 3 because the sum of the slopes of the lines is twice their product, which leads to the equation 2c/3 = 2, solving for c gives c=3.
109 Easy
109 of 264
Analyzing +4 | -1
If the equation x4+bx3y+cx2y2+dxy3+ey4=0x^4+b x^3 y+c x^2 y^2+d x y^3+e y^4=0 represents two pairs of perpendicular lines, then
View answer and explanation
Correct option: B
b+d=0b+d=0 and e=−1e=-1
Option B (b+d=0b+d=0 and e=−1e=-1) is correct because it represents the condition when the given equation represents two pairs of perpendicular lines, derived from comparing coefficients.
110 Easy
110 of 264
Application +4 | -1
The point of intersection of the pair of straight lines given by 6x2+5xy−4y2+7x+13y−3=06 x^2+5 x y-4 y^2+7 x+13 y-3=0, is
View answer and explanation
Correct option: C
(−1,1)(-1,1)
Option C ((-1,1)) is correct because it is the point of intersection obtained by solving the partial derivatives of the given equation with respect to x and y.
111 Medium
111 of 264
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.
112 Easy
112 of 264
Application +4 | -1
If the slopes of the lines given by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 are in the ratio 3:13: 1, then h2=h^2=
View answer and explanation
Correct option: B
4ab3\frac{4 a b}{3}
Option B (4ab3\frac{4 a b}{3}) is correct because it represents the relationship between h2h^2 and abab when the slopes of the lines are in the ratio 3:1.
113 Easy
113 of 264
Application +4 | -1
If the equation 4x2+hxy+y2=04{x^2} + hxy + {y^2} = 0 represent coincident lines, then hh is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because for the equation 4x2+hxy+y2=04{x^2} + hxy + {y^2} = 0 to represent coincident lines, the discriminant must be zero, leading to h=±4h = \pm 4.
114 Easy
114 of 264
Application +4 | -1
The gradient of one of the lines given by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 is twice that of the other, then
View answer and explanation
Correct option: C
8h2=9ab8 h^2=9 a b
Option D (9h2=8ab9 h^2=8 a b) is correct because using the given condition about the gradients and the relationships between the coefficients and the gradients leads to 9h2=8ab9h^2 = 8ab.
115 Medium
115 of 264
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.
116 Easy
116 of 264
Analyzing +4 | -1
If θ1\theta_1 and θ2\theta_2 be the angle which the lines given by (x2+y2)(cos⁡2θsin⁡2α+sin⁡2θ)=(xtan⁡α−ysin⁡θ)2\left(x^2+y^2\right)\left(\cos ^2 \theta \sin ^2 \alpha+\sin ^2 \theta\right)=(x \tan \alpha-y \sin \theta)^2 make with the axis of xx, then for θ=π6\theta=\frac{\pi}{6}, the value of tan⁡θ1+tan⁡θ2=\tan \theta_1+\tan \theta_2=
View answer and explanation
Correct option: B
−83cosec⁡2α-\frac{8}{3} \operatorname{cosec} 2 \alpha
Option B (-83\frac{{8}}{{3}} cosec⁡2α\operatorname{cosec} 2 \alpha) is correct because it represents the sum of the tangents of the angles the lines make with the x-axis, calculated using trigonometric identities and the given equation.
117 Medium
117 of 264
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.
118 Easy
118 of 264
Application +4 | -1
A straight line passes through the point (1,1)(1,1) and the portion of the line intercepted between xx and yy axes is divided at the point in the ratio 3:43: 4. An equation of the line is
View answer and explanation
Correct option: B
4x+3y=74 x+3 y=7
Option B (4x+3y=74 x+3 y=7) is correct because it satisfies the condition that the line passes through (1,1) and the intercepted portion is divided in the ratio 3:4.
119 Easy
119 of 264
Application +4 | -1
Reflection of the line x+y+1=0x+y+1=0 in the line lx+my+n=0l x+m y+n=0 is
View answer and explanation
Correct option: B
(x+y+1)(l2+m2)−2(l+m)(lx+my+n)=0(x+y+1)\left(l^2+m^2\right)-2(l+m)(l x+m y+n)=0
Option B ((x+y+1)(l2+m2)−2(l+m)(lx+my+n)=0(x+y+1)(l^2+m^2)-2(l+m)(l x+m y+n)=0) is correct because it represents the reflection of the line x+y+1=0 in the line lx+my+n=0.
120 Easy
120 of 264
Application +4 | -1
If one of the lines of my2+(1−m2)xy−mx2=0m y^2+\left(1-m^2\right) x y-m x^2=0 is a bisector of the angle between the lines xy=0x y=0, then mm is
View answer and explanation
Correct option: C
1
The correct answer is 1 because substituting y=xy = x into the equation and simplifying leads to m2=1m^2 = 1, and considering the geometry, m=1m = 1.
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