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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Mixed difficulty

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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 1-30 of 555 questions
1 Easy
1 of 555
Analyzing +4 | -1
If the three lines x−3y=p,ax+2y=qx-3 y=p, a x+2 y=q and ax+y=ra x+y=r form a right-angled triangle then :
View answer and explanation
Correct option: A
a2−9a+18=0a^{2}-9 a+18=0
Option A (a2−9a+18=0a^{2}-9 a+18=0) is correct because it represents the condition where the product of the slopes of two lines is -1, indicating a right-angled triangle.
2 Medium
2 of 555
Application +4 | -1
The orthocentre of the triangle whose vertices are

{at1t2,  a(t1+t2)},{at2t3,a(t2+t3)},  {at3t1,  a(t3+t1)}\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\} is
View answer and explanation
Correct option: B
{−a,a(t1+t2+t3+t1t2t3)}\left\{ { - a,a\left( {{t_1} + {t_2} + {t_3} + {t_1}{t_2}{t_3}} \right)} \right\}
Option B ({−a,a(t1+t2+t3+t1t2t3)}\left\{ { - a,a\left( {{t_1} + {t_2} + {t_3} + {t_1}{t_2}{t_3}} \right)} \right\}) is correct because it represents the orthocenter of the triangle formed by the given vertices on a parabola.
3 Easy
3 of 555
Application +4 | -1
The distance between the parallel lines y=2  x+4y = 2\;x + 4 and 6x=3y+56x = 3y + 5 is
View answer and explanation
Correct option: D
175/1517\sqrt 5 /15
Option D (175/1517\sqrt 5 /15) is correct because it is the distance between the two parallel lines calculated using the formula.
4 Easy
4 of 555
Application +4 | -1
The distance between the pair of lines given by x2+y2+2xy−8ax−8ay−9a2=0{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0 is
View answer and explanation
Correct option: D
52 a5\sqrt 2 \,a
Option D (52 a5\sqrt 2 \,a) is correct because the distance between the parallel lines is calculated by factoring the given equation and applying the distance formula.
5 Easy
5 of 555
Application +4 | -1
Find the image of the point (3,8)(3,8) with respect of the line x+3y=7x+3 y=7 assuming the line to be plane mirror:
View answer and explanation
Correct option: B
(−1,−4)(-1,-4)
Option C ((-1,-4)) is correct because it is the correct image point calculated using the mirror formula.
6 Easy
6 of 555
Application +4 | -1
The centroid of the triangle whose three sides are given by the combined equation (x2+7xy+2y2)(y−1)=0\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0 is
View answer and explanation
Correct option: C
(−73,23)\left( { - \frac{7}{3},\frac{2}{3}} \right)
Option C ((-7/3, 2/3)) is correct because it is the centroid calculated from the vertices of the triangle formed by the given lines.
7 Easy
7 of 555
Application +4 | -1
A line LL passes through the points (1,1)(1,1) and (2,0)(2,0) and another line L′L^{\prime} passes through (12,0)\left(\frac{1}{2}, 0\right) and perpendicular to LL. Then the area of the triangle formed by the lines L,L′L, L^{\prime} and y−y- axis, is NN. Find 16N16 N
View answer and explanation
Correct option: A
25
The correct answer is A (25) because it is the value of 16N after finding the area of the triangle formed by the lines L, L', and the y-axis.
8 Easy
8 of 555
Application +4 | -1
The orthocenter of the triangle with vertices (−2,  −6),  (−2,  4)\left( { - 2,\; - 6} \right),\;\left( { - 2,\;4} \right) and (1, 3) is
View answer and explanation
Correct option: C
(1,3)
Option C (1,3) is correct because the orthocenter of a right-angled triangle is the vertex at the right angle.
9 Easy
9 of 555
Application +4 | -1
Points A(1,3) and C(5,1) are opposite vertices of a rectangle ABCD. If the slope of BD is 2, then its equation is
View answer and explanation
Correct option: A
2  x−y=42\;x - y = 4
Option A (2x−y=42x - y = 4) is correct because the equation of BD is found using the midpoint of AC and the given slope of BD.
10 Easy
10 of 555
Application +4 | -1
If the lines kx−2y−1=0kx - 2y - 1 = 0 and 6x−4y−m=26x - 4y - m = 2 are identical (coincident ) lines, then the values of kk and mm are
View answer and explanation
Correct option: A
k=3,  m=2k = 3,\;m = 2
Option A (k=3,  m=2k = 3,\;m = 2) is correct because for identical lines, the ratios of the coefficients must be equal.
11 Medium
11 of 555
Application +4 | -1
If (a+3b)(3a+b)=4h2(a+3 b)(3 a+b)=4 h^2, then the angle between the lines represented by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 is
View answer and explanation
Correct option: C
60∘60^{\circ}
Option C (60∘60^{\circ}) is correct because it is calculated using the given condition (a+3b)(3a+b)=4h2(a+3b)(3a+b)=4h^2 and the formula for the angle between the lines.
12 Easy
12 of 555
Application +4 | -1
All points lying inside the triangle formed by the points (1, 3), (5, 0) and (−1,  2)\left( { - 1,\;2} \right) satisfy
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the given inequalities hold true for points inside the triangle formed by the given points.
13 Easy
13 of 555
Application +4 | -1
The equation of a straight line which cut off an intercept of 5 units on negative direction of YY-axis and make an angle of 120∘120^{\circ} with positive direction of XX-axis, is
View answer and explanation
Correct option: B
y+3x+5=0y+\sqrt{3} x+5=0
Option B (y+3x+5=0y+\sqrt{3} x+5=0) is correct because the equation of a line is y=mx+cy = mx + c, where mm is the slope and cc is the y-intercept. Given the angle and y-intercept, we can find the equation.
14 Easy
14 of 555
Application +4 | -1
If the line passing through (3,4)(3,4) and (x,5)(x, 5) makes 135∘135^{\circ} angle with the positive direction of xx-axis, then x=x=
View answer and explanation
Correct option: B
2
Option B (2) is correct because the slope calculated from the two points must equal the tangent of the angle (135 degrees), leading to x=2.
15 Medium
15 of 555
Application +4 | -1
The point (a2,a+1)\left(a^2, a+1\right) lies in the angle between the lines 3x−y+1=03 x-y+1=0 and x+2y−5=0x+2 y-5=0 containing the origin, if
View answer and explanation
Correct option: A
a∈(−3,0)∪(13,1)a \in(-3,0) \cup\left(\frac{1}{3}, 1\right)
Option A (a∈(−3,0)∪(13,1)a \in(-3,0) \cup(\frac{1}{3}, 1)) is correct because the point and the origin must lie on the same side of both lines, leading to two inequalities that define the range of 'a'.

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16 Easy
16 of 555
Application +4 | -1
If the points (1, 2) and (3, 4) were to be on the same side of the line 3x−5y+a=0,  3x - 5y + a = 0,\;, then
View answer and explanation
Correct option: D
a<7  or  a>11a < 7\;\mathrm{or\;}a > 11
Option D (a<7  or  a>11a < 7\;\mathrm{or\;}a > 11) is correct because the points (1, 2) and (3, 4) must yield the same sign when plugged into the line equation, leading to the inequality (a−7)(a−11)>0(a - 7)(a - 11) > 0.
17 Medium
17 of 555
Application +4 | -1
The incentre of the triangle formed by (0, 0), (5, 12), (16, 12) is
View answer and explanation
Correct option: A
(7, 9)
Option A (7, 9) is correct because the incenter is calculated using the formula (ax1+bx2+cx3a+b+c,  ay1+by2+cy3a+b+c)\left( {\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\;\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right), where a,b,ca, b, c are the side lengths of the triangle.
18 Easy
18 of 555
Application +4 | -1
If the area of the triangle with vertices (x,0),(1,1)\left( {x,0} \right),\left( {1,1} \right) and (0, 2) is 4 sq unit, then the value of xx is
View answer and explanation
Correct option: C
−6- 6
Option C (-6) is correct because the area of the triangle is given as 4, and solving the determinant equation yields x = -6.
19 Easy
19 of 555
Application +4 | -1
Let PSP S be the median of the triangle with vertices P(2,2),Q(6,−1)P(2,2), Q(6,-1) and R(7,3)R(7,3). The equation of the line passing through (1,−1)(1,-1) and parallel to PSP S is 2x+by+c=0.2 x+b y+c=0. Find b+cb+c.
View answer and explanation
Correct option: A
16
The correct answer is B (16) because it is the value of b+c after finding the equation of the line parallel to PS and passing through (1,-1).
20 Medium
20 of 555
Application +4 | -1
The equation of the bisectors of angle between the lines represented by equation (y−mx)2=(x+my)2(y-m x)^2=(x+m y)^2 is
View answer and explanation
Correct option: A
mx2+(m2−1)xy−my2=0m x^2+\left(m^2-1\right) x y-m y^2=0
Option A (mx2+(m2−1)xy−my2=0m x^2+\left(m^2-1\right) x y-m y^2=0) is correct because it is calculated using the formula for the equation of the bisectors of the angle between the lines.
21 Easy
21 of 555
Application +4 | -1
If PP and QQ are the lengths of the perpendiculars from the origin to the lines xcos⁡θ+ysin⁡θ=kcos⁡2θx \cos \theta+y \sin \theta=k \cos 2 \theta and xsec⁡θ−ycosec⁡θ=kx \sec \theta-y \operatorname{cosec} \theta=k respectively then
View answer and explanation
Correct option: B
p2+4q2=k2p^2+4 q^2=k^2
Option B (p2+4q2=k2p^2+4 q^2=k^2) is correct because it is derived by substituting the expressions for p and q into the equation and simplifying using trigonometric identities.
22 Easy
22 of 555
Application +4 | -1
The equation of the straight line which is perpendicular to y=x and passes through (3, 2) is
View answer and explanation
Correct option: B
x+y=5x + y = 5
Option B (x+y=5x + y = 5) is correct because it represents a line perpendicular to y=x and passes through the point (3, 2).
23 Easy
23 of 555
Application +4 | -1
If a,  b,  ca,\;b,\;c are in AP, then the straight line ax+by+c=0ax + by + c = 0 will always pass through the point
View answer and explanation
Correct option: B
(1,  −2)\left( {1,\; - 2} \right)
Option B ((1, -2)) is correct because the line always passes through this point, which satisfies the condition 2b = a + c for a, b, c in AP.
24 Easy
24 of 555
Application +4 | -1
The angle between the line joining the points (1,  −2),  (3,  2)\left( {1,\; - 2} \right),\;\left( {3,\;2} \right) and the line x+2y−7=0x + 2y - 7 = 0 is
View answer and explanation
Correct option: B
π/2\pi /2
Option B (π/2\pi /2) is correct because the product of the slopes of the two lines is -1, indicating they are perpendicular.
25 Medium
25 of 555
Application +4 | -1
The reflection of the curve xy=1x y=1 in the line y=2xy=2 x is the curve 12x2+rxy+sy2+t=012 x^2+r x y+s y^2+t=0, then the value of rr is-
View answer and explanation
Correct option: A
−7-7
Option A (-7) is correct because it is the coefficient of the xy term in the reflected curve equation, obtained by finding the reflection of a point on the original curve and substituting it back into the equation.
26 Medium
26 of 555
Application +4 | -1
The line 3x+2y=243 x+2 y=24 meets the yy-axis at AA and the xx-axis at B;CB ; C is a point on the perpendicular bisector of ABA B such that the area of the triangle ABCA B C is 91sq91 \mathrm{sq}. units. The coordinates of CC can be
View answer and explanation
Correct option: B
(29/2,13)(29 / 2,13)
Option B ((29 / 2, 13)) is correct because it satisfies both the perpendicular bisector equation and the area condition for triangle ABC.
27 Medium
27 of 555
Analyzing +4 | -1
If t1,t2{t_1},{t_2} and t3{t_3} are distinct points (t1,2at1+at13),(t2,2at2+at23)\left( {{t_1},2a{t_1} + at_1^3} \right),\left( {{t_2},2a{t_2} + at_2^3} \right) and t3,2at3+at33){t_3},2a{t_3} + at_3^3) are collinear, if
View answer and explanation
Correct option: C
t1+t2+t3=0{t_1} + {t_2} + {t_3} = 0
Option C (t1+t2+t3=0{t_1} + {t_2} + {t_3} = 0) is correct because the collinearity condition of the three points leads to this relationship between t1,t2,t_1, t_2, and t3t_3.
28 Easy
28 of 555
Application +4 | -1
The equation of straight line through the intersection of the lines x−2y=1x - 2y = 1 and x+3y=2x + 3y = 2 and parallel to 3x+4y=03x + 4y = 0, is
View answer and explanation
Correct option: C
3x+4y−5=03x + 4y - 5 = 0
Option C (3x+4y−5=03x + 4y - 5 = 0) is correct because it represents the equation of a line parallel to 3x+4y=03x + 4y = 0 and passing through the intersection of the given lines.
29 Medium
29 of 555
Analyzing +4 | -1
The equation λ(x3−3xy2)+y3−3x2y=0\lambda\left(x^3-3 x y^2\right)+y^3-3 x^2 y=0 represents
View answer and explanation
Correct option: A
three lines equally inclined to the other
Option A (three lines equally inclined to the other) is correct because the equation represents three lines passing through the origin, equally inclined to each other.
30 Easy
30 of 555
Application +4 | -1
If the xx-intercept of some line LL is double as that of the line, 3x+4y=123 x+4 y=12 and the yy-intercept of LL is half as that of the same line, then the slope of LL is :
View answer and explanation
Correct option: D
−3/16-3 / 16
Option D (−3/16-3 / 16) is correct because it is derived from finding the intercepts of line L and calculating the slope.
Revision notes and key takeaways

Review STRAIGHT LINES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 555 single-correct-answer MCQs on STRAIGHT LINES.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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