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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-90 of 139 questions
61 Medium
61 of 139
Application +4 | -1
If arithmetic mean of reciprocals of intercepts of line y=mx+cy=m x+c on coordinate axes is 1/41 / 4, then this line will pass through the point
View answer and explanation
Correct option: B
(2,2)(2,2)
Option B is correct. If the arithmetic mean of reciprocals of intercepts of the line y=mx+cy=mx+c on coordinate axes is 1/41/4, then this line will pass through the point (2,2)(2,2).
62 Medium
62 of 139
Application +4 | -1
If the image of point P(2,3)\mathrm{P}(2,3) in a line L\mathrm{L} is Q(4,5)\mathrm{Q}(4,5), then the image of point R(0,0)R(0,0) in the same line is:
View answer and explanation
Correct option: D
(7,7)(7,7)
The correct answer is D ((7,7)) because it involves finding the equation of the line L using the midpoint and slope of PQ, then finding the image of R in that line.
63 Medium
63 of 139
Application +4 | -1
The distance of the point (2,3)(2,3) from the line 2x−3y2 x-3 y +28=0+28=0, measured parallel to the line 3x−y+1=0\sqrt{3 x}-y+1=0, is equal to
View answer and explanation
Correct option: D
4+634+6 \sqrt{3}
The correct answer is D (4+634+6 \sqrt{3}) because it represents the distance calculated using parametric coordinates and the given line equations.
64 Medium
64 of 139
Application +4 | -1
The line x+2y=4x + 2y = 4 is translated parallel to itself by 3 units in the sense of increasing xx and then rotated by 30∘30^\circ in the clockwise direction about the point where the shifted line cuts the x-axis. The equation of the line in the new position is
View answer and explanation
Correct option: A
y=tan⁡(θ−30∘)(x−4−35)y = \tan \left( {\theta - 30^\circ } \right)\left( {x - 4 - 3\sqrt 5 } \right)
The correct answer is A (y=tan⁡(θ−30∘)(x−4−35)y = \tan \left( {\theta - 30^\circ } \right)\left( {x - 4 - 3\sqrt 5 } \right)) because it represents the equation of the translated and rotated line, considering the shift and the clockwise rotation about the new x-intercept.
65 Medium
65 of 139
Application +4 | -1
The equation to the bisecting the join of (3,−4)\mathrm{(3, - 4)} and (5,2) and having its intercepts on the x-axis and the y-axis in the ratio 2 : 1 is
View answer and explanation
Correct option: C
x+2y=2x + 2y = 2
Option C (x+2y=2x + 2y = 2) is correct because it bisects the join of the points and has the specified intercepts ratio.
66 Medium
66 of 139
Application +4 | -1
The equation of a straight line equally inclined to the axes and equidistant from the points (1,−2)(1,-2) and (3,4)(3,4) is
View answer and explanation
Correct option: C
x−y−1=0x-y-1=0
Option C (x−y−1=0x-y-1=0) is correct because it represents a line equally inclined to the axes and equidistant from the given points, which is found by solving for the constant term in the line equation.
67 Medium
67 of 139
Application +4 | -1
If 3a+2b+6c=03 a+2 b+6 c=0, then family of straight lines ax+by+c=0a x+b y+c=0 passes through a fixed point whose coordinates are given by
View answer and explanation
Correct option: A
(1/2,1/3)(1 / 2,1 / 3)
Option A (1/2, 1/3) is correct because it represents the fixed point through which the family of lines passes, derived by setting up and solving simultaneous equations.
68 Medium
68 of 139
Application +4 | -1
The equation of bisector of that angle between the lines x+y+1=0x+y+1=0 and 2x−3y−5=02 x-3 y-5=0 which contains the point (10,−20)(10,-20) is
View answer and explanation
Correct option: A
x(13+22)+y(13−32)+(13−52)=0x(\sqrt{13}+2 \sqrt{2})+y (\sqrt{13}-3 \sqrt{2}) +(\sqrt{13}-5 \sqrt{2})=0
Option A (x(13+22)+y(13−32)+(13−52)=0x(\sqrt{{13}}+2 \sqrt{{2}})+y (\sqrt{{13}}-3 \sqrt{{2}}) +(\sqrt{{13}}-5 \sqrt{{2}})=0) is correct because it represents the angle bisector that contains the given point (10, -20).
69 Medium
69 of 139
Understanding +4 | -1
If A and B are two fixed points, then the locus of a point which moves in such a way that the angle APB is a right angle is
View answer and explanation
Correct option: A
A circle
Option A (A circle) is correct because the locus of a point P such that angle APB is a right angle is a circle with AB as the diameter.
70 Medium
70 of 139
Application +4 | -1
The angle between the lines represented by the equation (x2+y2)sin⁡θ+2xy=0\left(x^2+y^2\right) \sin \theta+2 x y=0 is
View answer and explanation
Correct option: C
π2−θ\frac{\pi}{2}-\theta
Option C (π2−θ\frac{\pi}{2}-\theta) is correct because it is calculated by using the formula for the angle between the lines and trigonometric identities.
71 Medium
71 of 139
Application +4 | -1
If OO is the origin and QQ is a variable point on x2=4yx^2=4 y, then the locus of the mid-point of OQO Q, is
View answer and explanation
Correct option: D
x2=2yx^2=2 y
Option D is correct. If OO is the origin and QQ is a variable point on x2=4yx^2=4y, then the locus of the mid-point of OQOQ is x2=2yx^2=2y.
72 Medium
72 of 139
Application +4 | -1
The line which is parallel to x-axis and crosses the curve y=x  y = \sqrt x \; at an angle 45∘45^\circ, is
View answer and explanation
Correct option: B
y=12y = \frac{1}{2}
Option B (y=12y = \frac{1}{2}) is correct because it represents a line parallel to the x-axis that intersects the curve at a 45-degree angle.
73 Medium
73 of 139
Application +4 | -1
A line having slope '1' is drawn from a point A(−3,0)A(-3,0) cuts a curve y=x2+x+1y=x^2+x+1 at P & QP \,\&\, Q. Find ℓ(AP)ℓ(AQ)\ell({AP}) \ell({AQ}).
View answer and explanation
Correct option: A
14
Option A (14) is correct because it is the product of the distances from the point A to the intersection points P and Q, found by solving the quadratic equation and applying Vieta's formulas.
74 Medium
74 of 139
Application +4 | -1
The line x−2y=0x-2 y=0 will be a bisector of the angle between the lines represented by the equation x2−2hxy−2y2=0x^2-2 h x y-2 y^2=0, if h=h=
View answer and explanation
Correct option: C
−2-2
Option C (−2-2) is correct because the line x - 2y = 0 is a bisector, and using the properties of angle bisectors and the given equation, we can solve for h.
75 Medium
75 of 139
Application +4 | -1
If the line px−qy=rpx - qy = r intersects the coordinate axes at (a,0) and (0,b), thyen the value of a+b is equal to
View answer and explanation
Correct option: B
r(q−ppq)r\left( {\frac{{q - p}}{{pq}}} \right)
Option A (r(q+ppq)r\left( {\frac{{{{q + p}}}}{{{{pq}}}}} \right)) is correct because it represents the sum of the x and y intercepts of the line px - qy = r.
76 Medium
76 of 139
Application +4 | -1
A line through the point A(2,4)A(2,4) intersects the line x+y=9x+y=9 at the point P.P. The minimum distance of AP,A P, is
View answer and explanation
Correct option: C
32\frac{3}{\sqrt{2}}
Option C (32\frac{3}{\sqrt{2}}) is correct because it represents the minimum distance AP, which is the perpendicular distance from A to the line x+y=9.
77 Medium
77 of 139
Application +4 | -1
Family of a lines xsec⁡2θ+ytan⁡2θ−2=0x \sec ^2 \theta+y \tan ^2 \theta-2=0, for different real θ\theta, is
View answer and explanation
Correct option: C
concurrent at (2,−2)(2,-2)
Option C (concurrent at (2,−2)(2,-2)) is correct because all lines in the family pass through this specific point, regardless of the value of θ\theta.
78 Medium
78 of 139
Application +4 | -1
Consider a family of straight lines (x+y)+λ(2x−y+1)=0(x+y)+\lambda(2 x-y+1)=0. Equation of the straight line belonging to this family that is farthest from (1,−3)(1,-3) is
View answer and explanation
Correct option: D
15y−6x=715 y-6 x=7
Option D (15y−6x=715y - 6x = 7) is correct because it represents the line in the family that is perpendicular to the line connecting the point of concurrency and the given point.
79 Medium
79 of 139
Application +4 | -1
A line AB makes zero intercepts on x-axis and y-axis and it is perpendicular to another line CD which is 3x+4y+6=03x + 4y + 6 = 0. The equation of line AB is
View answer and explanation
Correct option: C
4x−3y=04x - 3y = 0
Option C (4x−3y=04x - 3y = 0) is correct because the line passes through the origin and is perpendicular to the given line.
80 Medium
80 of 139
Application +4 | -1
Number of lines drawn from the point (4,−5)(4,-5) so that its distance from (−2,3)(-2,3) will be equal to 12 are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the discriminant of the resulting quadratic equation is negative, indicating no real solutions for the slope 'm', and thus no such lines exist.
81 Medium
81 of 139
Application +4 | -1
If the parametric coordinates of a point are (tan⁡θ−sin⁡θ,tan⁡θ+sin⁡θ)(\tan \theta-\sin \theta, \tan \theta+\sin \theta), then the locus of the point is
View answer and explanation
Correct option: B
(x2−y2)2=16xy\left(x^2-y^2\right)^2=16 x y
Option B ((x2−y2x^2-y^2)2=16xy^2=16xy) is correct because it represents the locus of the point derived from the given parametric coordinates.
82 Medium
82 of 139
Analyzing +4 | -1
Let θ1\theta_{1} be the angle between two lines 2x+3y+c1=02 x+3 y+c_{1}=0 and −x+5y+c2=0-x+5 y+c_{2}=0 and θ2\theta_{2} be the angle between two lines 2x+3y+c1=02 x+3 y+c_{1}=0 and −x+5y+c3=0-x+5 y+c_{3}=0, where c1,c2,c3c_{1}, c_{2}, c_{3} are any real numbers :
Statement-1: If c2c_{2} and c3c_{3} are proportional, then θ1=θ2\theta_{1}=\theta_{2}. Statement-2: θ1=θ2\theta_{1}=\theta_{2} for all c2c_{2} and c3c_{3}.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation of Statement-1.
Option A (Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation of Statement-1) is correct because the angles are equal when the lines are parallel, and this parallelism is implied when the constants are proportional.
83 Medium
83 of 139
Analyzing +4 | -1
If the points (−2,0),(−1,13)(-2,0),\left(-1, \frac{1}{\sqrt{3}}\right) and (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta) are collinear, then the number of values of θ∈[0,2π]\theta \in[0,2 \pi] is
View answer and explanation
Correct option: B
1
Option B (1) is correct because solving the collinearity condition using the determinant results in a single valid value for θ\theta within the given interval.
84 Medium
84 of 139
Understanding +4 | -1
The parallelism condition for two straight lines one of which is specified by the equation ax+by+c=0ax + by + c = 0 and the other being represented parametrically by x=αt+β,  y=γt+δ,x = \mathrm{\alpha }t + \mathrm{\beta },\mathrm{\;}y = \mathrm{\gamma }t + \mathrm{\delta }, is given by
View answer and explanation
Correct option: C
aα+bγ=0a\mathrm{\alpha } + b\mathrm{\gamma } = 0
Option C (aα+bγ=0a\mathrm{\alpha } + b\mathrm{\gamma } = 0) is correct because it represents the condition where the slopes of the two lines are equal.
85 Medium
85 of 139
Application +4 | -1
The measure of the angle between the lines xcos⁡β+ysin⁡β=5π2x \cos \beta+y \sin \beta=\frac{5 \pi}{2} and xcos⁡α+ysin⁡α=π2x \cos \alpha+y \sin \alpha=\frac{\pi}{2} is
View answer and explanation
Correct option: D
∣β−α∣|\beta-\alpha|
Option D (∣β−α∣|\beta-\alpha|) is correct because it represents the absolute difference between the angles, which gives the acute angle between the lines.
86 Medium
86 of 139
Application +4 | -1
If P=(1,0),Q=(−1,0)P=(1,0), Q=(-1,0) and R=(2,0)R=(2,0) are three given points, then locus of the point SS satisfying the relation SQ2+SR2=2SP2S Q^2+S R^2=2 S P^2, is
View answer and explanation
Correct option: D
a straight line parallel to YY-axis
Option D (a straight line parallel to YY-axis) is correct because the locus simplifies to an equation of the form x=cx = c, which represents a vertical line.
87 Medium
87 of 139
Analyzing +4 | -1
The value of λ\lambda for which the lines joining the point of intersection of curves C1C_1 and C2C_2 to the origin are equally inclined to the XX-axis, where C1:λx2+3y2−2λxy+9x=0C_1: \lambda x^2+3 y^2-2 \lambda x y+9 x=0 and C2:3x2−4y2+8xy−3x=0C_2: 3 x^2-4 y^2+8 x y-3 x=0, is
View answer and explanation
Correct option: B
12
Option B (12) is correct because it is the value of λ\lambda that makes the lines joining the intersection points of the curves to the origin equally inclined to the X-axis.
88 Medium
88 of 139
Application +4 | -1
The image of the point (4,-3) with respect to the line y=x is
View answer and explanation
Correct option: D
(−3,  4)\left( { - 3,\;4} \right)
Option D ((-3,4)) is correct because it represents the reflection of the point (4,-3) across the line y=x.
89 Medium
89 of 139
Application +4 | -1
If x2−y2+2hxy+2gx+2fy+c=0x^2-y^2+2 h x y+2 g x+2 f y+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x+2y+7=0x+2 y+7=0 and 2x−y+8=02 x-y+8=0, then the value of g+c+h−fg+c+h-f equals
View answer and explanation
Correct option: C
14
Option C (14) is correct because it involves finding the equation of the angle bisectors, comparing it to the given equation, and then calculating g+c+h-f.
90 Medium
90 of 139
Application +4 | -1
The acute angle formed between the lines joining the origin to the points of intersection of the curves x2+y2−2x−1=0x^2+y^2-2 x-1=0 and x+y=1x+y=1, is
View answer and explanation
Correct option: B
tan⁡−12\tan ^{-1} 2
Option B (tan⁡−12\tan ^{-1} 2) is correct because it is calculated by homogenizing the equation and then applying the formula for the angle between the lines.
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