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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Mixed difficulty

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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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451 Easy
451 of 555
Application +4 | -1
One possible condition for the three points (a,b),(b,a)(a, b),(b, a) and (a2,−b2)\left(a^2,-b^2\right) to be collinear is
View answer and explanation
Correct option: C
a=1+ba=1+b
Option C (a=1+b) is correct because the collinearity condition leads to the factor (a-b)(a+b)(1-a+b) = 0, which gives a = 1+b as a possible condition.
452 Easy
452 of 555
Analyzing +4 | -1
If the lines ax+2  y+1=0,  bx+3  y+1=0,  cx+4  y+1=0ax + 2\;y + 1 = 0,\;bx + 3\;y + 1 = 0,\;cx + 4\;y + 1 = 0 are concurrent, then a,b,c are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the given condition leads to a, b, and c forming an arithmetic progression.
453 Easy
453 of 555
Application +4 | -1
The point on the line 3x+4y=53x + 4y = 5, which is equidistant from (1, 2) and (3, 4) is
View answer and explanation
Correct option: B
(15,  −10)\left( {15,\; - 10} \right)
Option B ((15, -10)) is correct because it satisfies both the equation of the line and the equidistant condition from the two given points.
454 Easy
454 of 555
Analyzing +4 | -1
If the roots of the equation ax2+bx+10=0a x^2+b x+10=0 are not real and distinct where a,b∈Ra, b \in R and α\alpha and β\beta are the values of aa and bb for which 5a+b5 a+b is minimum, then the family of lines α(4x+2y+3)+β(x−y−1)=0\alpha(4 x+2 y+3)+\beta(x-y-1)=0 are concurrent at
View answer and explanation
Correct option: C
(−1/6,−7/6)(-1 / 6,-7 / 6)
Option C ((−1/6,−7/6)(-1 / 6,-7 / 6)) is correct because it is the intersection point of the family of lines derived from the minimum value condition.
455 Easy
455 of 555
Application +4 | -1
If the points (1,1),(−1,−1),(−3,3)\left( {1,1} \right),\left( { - 1, - 1} \right),\left( { - \sqrt 3 ,\sqrt 3 } \right) are the vertices of a triangle, then this triangle is
View answer and explanation
Correct option: C
Equilateral
Option C (Equilateral) is correct because all three sides of the triangle have equal lengths, which is 222\sqrt{2}.
456 Medium
456 of 555
Application +4 | -1
The points (1,1),(−2,7)(1,1),(-2,7) and (3,−3)(3,-3)
View answer and explanation
Correct option: C
are collinear
Option C (are collinear) is correct because the sum of two distances between the points equals the third distance, indicating they lie on the same line.
457 Easy
457 of 555
Application +4 | -1
The equation of the image of the lines y=∣x∣y = \left| x \right| in the line mirror x=2x = 2 is
View answer and explanation
Correct option: A
y=∣x−4∣y = \left| {x - 4} \right|
Option A (y=∣x−4∣y = \left| {x - 4} \right|) is correct because the reflection of y=∣x∣y = |x| across the line x=2x = 2 results in a horizontal shift of 4 units.
458 Easy
458 of 555
Application +4 | -1
If the straight line ax+by+c=0ax + by + c = 0 always passes through (1,-2), then a,b,c are
View answer and explanation
Correct option: A
in AP
Option A ((0,am)\left( {0,\frac{a}{m}} \right)) is correct because it represents the coordinates of the foot of the perpendicular from (a,0) on the given line.
459 Medium
459 of 555
Application +4 | -1
The equation of line passing through (−1,π2)\left(-1, \frac{\pi}{2}\right) and perpendicular to 3sin⁡θ+2cos⁡θ=4r\sqrt{3} \sin \theta+2 \cos \theta=\frac{4}{r} is
View answer and explanation
Correct option: A
2=3rcos⁡θ−2rsin⁡θ2=\sqrt{3} r \cos \theta-2 r \sin \theta
Option A (2=3rcos⁡θ−2rsin⁡θ2=\sqrt{3} r \cos \theta-2 r \sin \theta) is correct because it represents the equation of a line passing through the given point and perpendicular to the given equation.
460 Easy
460 of 555
Application +4 | -1
If the sum of distances of a point from two perpendicular lines is unity, then its locus is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the locus is a square, not a circle, ellipse, or hyperbola.
461 Easy
461 of 555
Analyzing +4 | -1
The points P(a,b),Q(c,d),R(a,d)P(a, b), Q(c, d), R(a, d) and S(c,b),S(c, b), where a,b,c,da, b, c, d are distinct real numbers, are
View answer and explanation
Correct option: B
Vertices of a square
The correct answer is Vertices of a rectangle. The slopes of the lines PS and QR are 0, and the slopes of PR and QS are infinite, indicating that the points form a rectangle.
462 Easy
462 of 555
Application +4 | -1
The equation of the line passing through the intersection of x−3  y+3−1=0x - \sqrt 3 \;y + \sqrt 3 - 1 = 0 and x+y−2=0x + y - 2 = 0 and making an angle of 15∘{15^ \circ } with the first line is
View answer and explanation
Correct option: A
x−y=0x - y = 0
Option A (x−y=0x - y = 0) is correct because it represents a line passing through the intersection of the given lines and making a 15-degree angle with the first line.
463 Easy
463 of 555
Application +4 | -1
The distance between the parallel lines 3x−4y+7=03 x-4 y+7=0 and 3x−4y+5=03 {x}-4 {y}+5=0 is ab\frac{{a}}{{b}}. Value of a+b{a}+{b} is
View answer and explanation
Correct option: A
7
Option A (7) is correct because the distance between parallel lines is calculated using the formula, and the sum of the numerator and denominator of the simplified distance is 7.
464 Medium
464 of 555
Analyzing +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 a pair of straight lines equidistant from the origin, then
View answer and explanation
Correct option: B
f4−g4=c(bf2−ag2)f^4-g^4=c\left(b f^2-a g^2\right)
Option B (f4−g4=c(bf2−ag2)f^4-g^4=c(b f^2-a g^2)) is correct because it represents the condition for the given equation to represent a pair of straight lines equidistant from the origin.
465 Medium
465 of 555
Application +4 | -1
If a>0,  b>0a > 0,\;b > 0 the maximum area of the triangle formed by the points O(0,  0),  A(acos⁡θ,  bsin⁡θ)O\left( {0,\;0} \right),\;A\left( {a\cos \theta ,\;b\sin \theta } \right) and B(acos⁡θ,  −bsin⁡θ)B\left( {a\cos \theta ,\; - b\sin \theta } \right) is (in sq unit)
View answer and explanation
Correct option: A
ab2 when  θ=π4\frac{{ab}}{2}\,\mathrm{when}\;\mathrm{\theta } = \frac{\pi }{4}
Option A (ab2 when  θ=π4\frac{{ab}}{2}\,\mathrm{when}\;\mathrm{\theta } = \frac{\pi }{4}) is correct because the maximum area is achieved when sin⁡2θ=1\sin 2\theta = 1, which occurs at θ=π4\theta = \frac{\pi}{4}.
466 Easy
466 of 555
Application +4 | -1
If P=(1,0),Q=(−1,0)P = \left( {1,0} \right),Q = \left( { - 1,0} \right) and R=(2,0)R = \left( {2,0} \right) are three given points, then the locus of the point S(x,y)S\left( {x,y} \right) satisfying the relation SQ2++SR2=2SP2S{Q^2} + + S{R^2} = 2S{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 equation simplifies to a linear equation in x, indicating a vertical line.
467 Easy
467 of 555
Application +4 | -1
The line L given by x5+yb=1\frac{x}{5} + \frac{y}{b} = 1 passes through the point (13, 32). The line K is parallel to L and has the equation xc+y3−1\frac{x}{c} + \frac{y}{3} - 1. Then, the distance between L and K is
View answer and explanation
Correct option: D
2317\frac{{23}}{{\sqrt {17} }}
Option D (2317\frac{{23}}{{\sqrt {17} }}) is correct because it represents the distance between the two parallel lines L and K after finding their equations and applying the distance formula.
468 Medium
468 of 555
Analyzing +4 | -1
The line 3x+2y=243x + 2y = 24 meets y-axis at A and x-axis at B. The perpendicular bisector of AB meets the line through (0,-1) parallel to x-axis at C. The area of the triangle ABC is
View answer and explanation
Correct option: B
91 sq. units
Option B (91 sq. units) is correct because it is the area of the triangle formed by the given line and the perpendicular bisector, calculated using determinant formula.
469 Easy
469 of 555
Application +4 | -1
If area of triangle with vertices (0, 0), (0, 6) and (α,  β)\left( {\mathrm{\alpha },\mathrm{\;\beta }} \right) is 15 sq unit, then
View answer and explanation
Correct option: D
α=±5,  β  \mathrm{\alpha } = \pm 5,\mathrm{\;\beta \;} can take any real value
Option D (α=±5\alpha = \pm 5, β\beta can take any real value) is correct because the area of the triangle depends only on the x-coordinate of the third vertex, and the y-coordinate can be any real number.
470 Easy
470 of 555
Application +4 | -1
If a straight line L is perpendicular to the line 5x−y=15x - y = 1 such that the area of the Δ\Delta formed by the line L and the coordinate axes is 5, then the equation of the line L is
View answer and explanation
Correct option: D
x+5  y±52=0x + 5\;y \pm 5\sqrt 2 = 0
Option D (x+5  y±52=0x + 5\;y \pm 5\sqrt 2 = 0) is correct because it represents the equation of a line perpendicular to the given line and forms a triangle with the coordinate axes with an area of 5.
471 Easy
471 of 555
Understanding +4 | -1
Consider the following statements.

The three given points A,B,CA, B, C are collinear i.e., lie on the same straight line, if

I. area of △ABC\triangle A B C is zero.

II. slope of AB=A B= Slope of BCB C.

III. any one of the three points lie on the straight line joining the other two points.

Choose the correct option.
View answer and explanation
Correct option: D
All are true
Option D is correct because all three statements are valid conditions for three points to be collinear.
472 Hard
472 of 555
Analyzing +4 | -1
The equation x3−6x2y+11xy2−6y3=0{x^3} - 6{x^2}y + 11x{y^2} - 6{y^3} = 0 represents three straight lines passing through the origin, the slopes of which form an
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the slopes of the three lines form a Harmonic Progression.
473 Easy
473 of 555
Application +4 | -1
The image of a point A(3,8)A(3,8) in the line x+3y−7=0x+3 y-7=0, is (a,b)(a, b). Find a−ba-b.
View answer and explanation
Correct option: A
3
Option A (3) is correct because the image of the point A(3,8) in the given line is (-1,-4), and a-b = -1 - (-4) = 3.
474 Easy
474 of 555
Application +4 | -1
The area of triangle formed by the points (a,  b+c),  (b,  c+a),  (c,  a+b)\left( {a,\;b + c} \right),\;\left( {b,\;c + a} \right),\;\left( {c,\;a + b} \right) is equal to
View answer and explanation
Correct option: D
0
Option B (11.11) is correct because the weight percentage of deuterium in heavy water (D2O) is calculated as (2*2 / 20) * 100 = 20%.
475 Easy
475 of 555
Application +4 | -1
Equation of the straight line making equal intercepts on the axes and passing through the point (2, 4), is
View answer and explanation
Correct option: C
x+y-6=0
Option C (x+y-6=0) is correct because it represents a line with equal intercepts that passes through the point (2, 4).
476 Easy
476 of 555
Application +4 | -1
Equation of the straight line cutting off an intercept 2 from the negative direction of the axis of y and inclined at 30∘{30^ \circ } to the positive direction of x-axis, is
View answer and explanation
Correct option: D
3y−x+23=0\sqrt 3 y - x + 2\sqrt 3 = 0
Option D (3y−x+23=0\sqrt 3 y - x + 2\sqrt 3 = 0) is correct because it represents the equation of a line with a slope of 13\frac{1}{{\sqrt 3 }} and a y-intercept of -2.
477 Easy
477 of 555
Remembering +4 | -1
Consider the following statements.

I. The equation of a straight line passing through the point (x1,y1)\left(x_1, y_1\right) and having slope m{m} is given by y−y1=m(x−x1)y-y_1=m\left(x-x_1\right)

II. Equation of the yy-axis is x=0x=0.

Choose the correct option.
View answer and explanation
Correct option: C
Both are true
Option C (Both are true) is correct because the point-slope form of a line and the equation of the y-axis are correctly stated.
478 Easy
478 of 555
Application +4 | -1
A(−5,0)A\left( { - 5,0} \right) and B(3,0)B\left( {3,0} \right) are two of the vertices of a triangle ABC. Its area is 20 square cms. The vertex C lies on the line x−y=2.x - y = 2.. The coordinates of C are
View answer and explanation
Correct option: D
(−3,−5)\left( { - 3, - 5} \right) or (7,5)\left( {7,5} \right)
Option D ((−3,−5)\left( { - 3, - 5} \right) or (7,5)\left( {7,5} \right)) is correct because these coordinates satisfy both the area condition and the line equation.
479 Easy
479 of 555
Application +4 | -1
A line passes through the point of intersection of the lines 3x+y+1=03x + y + 1 = 0 and 2x−y+3=02x - y + 3 = 0 and makes equal intercepts with axes. Then, equation of the line is
View answer and explanation
Correct option: A
5x+5y−3=05x + 5y - 3 = 0
Option D (5x+5y+3=05x + 5y + 3 = 0) is correct because it passes through the intersection point and has equal intercepts.
480 Easy
480 of 555
Application +4 | -1
The point (4,1)(4,1) undergoes the following two successive transformation
(i) Reflection about the line y=xy=x
(ii) Translation through a distance 2 units along the positive xx-axis
Then the final coordinates of the point are (x,y)(x, y). Find x+yx+y.
View answer and explanation
Correct option: A
7
The correct answer is A (7) because it is the sum of the final coordinates after performing the given transformations.
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