Skip to main content
Back to STRAIGHT LINES

Your Progress

0 / 408 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Easy

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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.

Take a chapter quiz →
Showing 31-60 of 408 questions
31 Easy
31 of 408
Analyzing +4 | -1
If the lines x+2ay+a=0,x+3by+b=0x+2 a y+a=0, x+3 b y+b=0 and x+4cy+c=0x+4 c y+c=0 are concurrent, then a,ba, b and cc are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the concurrency condition leads to the relationship b=2aca+cb = \frac{{2ac}}{{a+c}}, which is the definition of Harmonic Progression.
32 Easy
32 of 408
Application +4 | -1
The vertices of a Δ  OBC\Delta \;OBC are (0,  0),  B(−3,−1)\left( {0,\;0} \right),\;B\left( { - 3, - 1} \right) and C(−1,−3).C\left( { - 1, - 3} \right). The equation of a line parallel to BC and intersecting sides OB and OC whose distance from the origin is 1/2, is
View answer and explanation
Correct option: D
x+y+12=0x + y + \frac{1}{{\sqrt 2 }} = 0
Option D (x+y+12=0x + y + \frac{1}{{\sqrt 2 }} = 0) is correct because it represents a line parallel to BC at a distance of 1/2 from the origin.
33 Easy
33 of 408
Application +4 | -1
If (−4,  5)\left( { - 4,\;5} \right) is one vertex and 7x−y+8=07x - y + 8 = 0 is one diagonal of a square, then the equation of the second diagonal is
View answer and explanation
Correct option: C
x+7y=31x + 7y = 31
Option C (x+7y=31x + 7y = 31) is correct because it represents the equation of the second diagonal which is perpendicular to the given diagonal and passes through the given vertex.
34 Easy
34 of 408
Application +4 | -1
Let LL denote the line in the xy−xy- plane with xx and yy intercepts as 3 and 1 respectively. Then, the image of the point (−1, −4)\left( { - 1,\, - 4} \right) in this line is
View answer and explanation
Correct option: A
(115, 285)\left( {\frac{{11}}{5},\,\frac{{28}}{5}} \right)
Option A ((115, 285)\left( {\frac{{11}}{5},\,\frac{{28}}{5}} \right)) is correct because it is the result of applying the formula for finding the image of a point with respect to a line.
35 Easy
35 of 408
Application +4 | -1
A line is drawn from the point P(α,β)P(\alpha, \beta) making an angle θ\theta with the positive direction of XX-axis to meet the line ax+by+c=0a x+b y+c=0 at QQ. The length of PQP Q is
View answer and explanation
Correct option: A
−aα+bβ+cacos⁡θ+bsin⁡θ-\frac{a \alpha+b \beta+c}{a \cos \theta+b \sin \theta}
Option A (−aα+bβ+cacos⁡θ+bsin⁡θ-\frac{{a \alpha+b \beta+c}}{{a \cos \theta+b \sin \theta}}) is correct because it represents the length PQ derived from the equation of a line and its intersection with another line.
36 Easy
36 of 408
Application +4 | -1
The equations of the lines through (1, 1) and making angle of 45∘{45^ \circ } with the line x+y=0x + y = 0 are
View answer and explanation
Correct option: D
x−1=0,  y−1=0x - 1 = 0,\;y - 1 = 0
Option D (x−1=0,y−1=0x - 1 = 0, y - 1 = 0) is correct because these are the equations of lines passing through (1,1) at a 45-degree angle to x + y = 0.
37 Easy
37 of 408
Analyzing +4 | -1
The pairs of straight lines x2−3xy+2y2=0{x^2} - 3xy + 2{y^2} = 0 and x2−3xy+2y2+x−2=0{x^2} - 3xy + 2{y^2} + x - 2 = 0 form a
View answer and explanation
Correct option: C
Parallelogram
Option C (Parallelogram) is correct because the pairs of lines are parallel, but the angle between them is not 90 degrees.
38 Easy
38 of 408
Application +4 | -1
The centroid of an equilateral triangle is (0, 0). If two vertices of the triangle lie on x+y=22,x + y = 2\sqrt 2 , then one of them will have its coordinates
View answer and explanation
Correct option: A
(2+6,  2−6)\left( {\sqrt 2 + \sqrt 6 ,\;\sqrt 2 - \sqrt 6 } \right)
Option A ((2+6,  2−6)\left( {\sqrt 2 + \sqrt 6 ,\;\sqrt 2 - \sqrt 6 } \right)) is correct because it represents one of the vertices of the equilateral triangle that lies on the given line.
39 Easy
39 of 408
Application +4 | -1
If the foot of the perpendicular from the origin to a straight line is at the point(3,-4). Then, the equation of the line is
View answer and explanation
Correct option: A
3x−4y=253x - 4y = 25
Option A (3x−4y=253x - 4y = 25) is correct because it represents the equation of the line with the foot of the perpendicular from the origin at (3,-4).
40 Easy
40 of 408
Application +4 | -1
The inclination of the straight line passing through the point (−3,  6)\left( { - 3,\;6} \right) and the mid point of the line joining the points (4,-5) and (-2,9) is
View answer and explanation
Correct option: D
3π4\frac{{3\mathrm{\pi }}}{4}
Option D (3π4\frac{{3\mathrm{\pi }}}{{4}}) is correct because the inclination is calculated using the midpoint formula and the arctangent function.
41 Easy
41 of 408
Analyzing +4 | -1
Area of quadrilateral whose vertices are (2, 3), (3, 4), (4, 5) and(5, 6), is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the area of the quadrilateral with the given vertices is zero, indicating that the points are collinear.
42 Easy
42 of 408
Application +4 | -1
A square of area 25 sq unit is formed by taking two sides as 3x+4y=k13x + 4y = {k_1} and 3x+4y=k23x + 4y = {k_2}, then ∣k1−k2∣\left| {{k_1} - {k_2}} \right| is
View answer and explanation
Correct option: C
25
Option D (25) is correct because the distance between the parallel lines, which forms the side of the square, is equal to the side length of the square, and the absolute difference between the constants is 25.
43 Easy
43 of 408
Application +4 | -1
If A(−5,  0)A\left( { - 5,\;0} \right) and B(3,  0)B\left( {3,\;0} \right) are two vertices of a triangle ABCABC. Its area is 20 sq cm. The vertex CC lies on the line x−y=2x - y = 2. The coordinates of CC are
View answer and explanation
Correct option: D
(−3,  −5)or  (7,  5)\left( { - 3,\; - 5} \right)\mathrm{or}\;\left( {7,\;5} \right)
Option D ((-3, -5) or (7, 5)) is correct because these coordinates satisfy both the area condition and the line equation.
44 Easy
44 of 408
Application +4 | -1
The distance of the line 2x+3y−5=02 x+3 y-5=0 from the point (3,5)(3,5) along the line 5x−3y=05 x-3 y=0 in units is
View answer and explanation
Correct option: C
43421\frac{4 \sqrt{34}}{21}
Option C (43421\frac{4 \sqrt{34}}{21}) is correct because it represents the distance of the line from the point along the given line.
45 Easy
45 of 408
Application +4 | -1
If the pair of straight lines xy−x−y+1=0xy - x - y + 1 = 0 and the line ax+2y−3=0ax + 2y - 3 = 0 are concurrent, then aa is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because the concurrency condition requires the intersection point of the first two lines to lie on the third line, leading to a=1.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
46 Easy
46 of 408
Analyzing +4 | -1
If the lines represented by the equation ax2+2hxy+by2+2gx+2fy+c=0a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 are equidistant from the origin, then
View answer and explanation
Correct option: A
f4−g4=c(bf2−ag2){f^4} - {g^4} = c\left( {b{f^2} - a{g^2}} \right)
The correct answer is f4−g4=c(bf2−ag2){f^4} - {g^4} = c\left( {b{f^2} - a{g^2}} \right) because the condition for the lines to be equidistant from the origin leads to this relationship between the coefficients.
47 Easy
47 of 408
Understanding +4 | -1
Condition that the two lines represented by the equation ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 to be perpendicular is
View answer and explanation
Correct option: B
a=−ba=-b
Option B (a=−ba=-b) is correct because for perpendicular lines, the sum of the coefficients of x² and y² must be zero.
48 Easy
48 of 408
Application +4 | -1
The equation of a line which is equidistant from the lines x=−4x=-4 and x=8,x=8, is
View answer and explanation
Correct option: A
x=2x=2
Option A (x=2) is correct because the line equidistant from x=-4 and x=8 is a vertical line located at the average of the two x-values, which is x=2.
49 Easy
49 of 408
Application +4 | -1
If the lines x=a+m,  y=−2  x = a + m,\;y = - 2\; and y=mxy = mx are concurrent, then least value of ∣a∣\left| a \right| is
View answer and explanation
Correct option: C
222\sqrt 2
Option C (222\sqrt 2) is correct because it is the least value of |a| that ensures the existence of real solutions for m.
50 Easy
50 of 408
Understanding +4 | -1
If ∣x1y11x2y21x3y31∣=0\left| {\begin{array}{cccccccccccccccccccc} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0, then the points (x1,y1),  (x2,  y2)\left( {{x_1},{y_1}} \right),\;\left( {{x_2},\;{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right) are
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because if the determinant is zero, the points are collinear, not vertices of any triangle.
51 Easy
51 of 408
Application +4 | -1
The image of the point (−1,  3)\left( { - 1,\;3} \right) by the line x−y=0x - y = 0 is
View answer and explanation
Correct option: A
(3,−1)\left( {3, - 1} \right)
Option A ((3, -1)) is correct because the image of a point (x, y) across the line x - y = 0 (or y = x) is (y, x).
52 Easy
52 of 408
Analyzing +4 | -1
Straight lines xa+yb=1,xb+ya=1,xa+yb=2\frac{x}{a}+\frac{y}{b}=1, \frac{x}{b}+\frac{y}{a}=1, \frac{x}{a}+\frac{y}{b}=2 and xb+ya=2\frac{x}{b}+\frac{y}{a}=2 form a rhombus of area (in square units)
View answer and explanation
Correct option: D
a2b2∣a2−b2∣\frac{a^2 b^2}{\left|a^2-b^2\right|}
Option D (a2b2∣a2−b2∣\frac{{a^2 b^2}}{{\left|a^2-b^2\right|}}) is correct because it represents the area of the rhombus formed by the given lines, calculated using the determinant method.
53 Easy
53 of 408
Application +4 | -1
The number of integral values of m, for which the x-coordinate of the point of intersection of the lines 3x+4y=93x + 4y = 9 and y=mx+1y = mx + 1 is also an integer, is
View answer and explanation
Correct option: A
2
The correct answer is A (2) because there are two integer values of m that make the x-coordinate of the intersection point an integer.
54 Easy
54 of 408
Application +4 | -1
The bisector of the acute angle formed between the lines 4x−3y+7=04x - 3y + 7 = 0 and 3x−4y+14=03x - 4y + 14 = 0 has the equation
View answer and explanation
Correct option: C
x−y+3=0x - y + 3 = 0
Option C (x - y + 3 = 0) is correct because it represents the bisector of the acute angle between the two lines, determined by checking the sign.
55 Easy
55 of 408
Application +4 | -1
If a line joining two points A(2,0)A(2,0) and B(3,1)B(3,1) is rotaled about AA in anticlockwise direction through an angle 15∘15^{\circ} such that the point BB goes to CC in the new position, then the coordinates of CC are
View answer and explanation
Correct option: D
(2−12,−32)\left(2-\frac{1}{\sqrt{2}},-\frac{\sqrt{3}}{2}\right)
Option D (2−12,−32)\left(2-\frac{1}{\sqrt{2}},-\frac{\sqrt{3}}{2}\right) is correct because it represents the coordinates of the point C after rotating point B about A by 15 degrees in the anticlockwise direction.
56 Easy
56 of 408
Application +4 | -1
The equation of line bisecting perpendicularly the segment joining the points (-4,6) and (8, 8), is
View answer and explanation
Correct option: B
6x+y−19=06x + y - 19 = 0
Option A (20 sq. units) is correct because it is the area of the rectangle calculated using the distance between the given vertices and the perpendicular distance from a vertex to the given side.
57 Easy
57 of 408
Application +4 | -1
Find the image of the point P(4,−13)P(4,-13) in the line mirror 5x+y+6=05 x+y+6=0, is
View answer and explanation
Correct option: C
(−1,−14)(-1,-14)
Option C is correct. The image of the point P(4,−13)P(4, -13) in the line 5x+y+6=05x + y + 6 = 0 is found using the formula for the image of a point in a line, resulting in (−1,−14)(-1, -14).
58 Easy
58 of 408
Application +4 | -1
If the slope of one line is double the other in line pair x2+6xy+by2=0x^2+6 x y+b y^2=0, then b=b=?
View answer and explanation
Correct option: D
8
Option D (8) is correct because it satisfies the condition that one slope is double the other in the given pair of lines.
59 Easy
59 of 408
Application +4 | -1
If the sum of the distance of a point PP from two perpendicular lines in a plane is 1, then the locus of PP is a
View answer and explanation
Correct option: A
Rhombus
Option A (Rhombus) is correct because the locus of a point with a constant sum of distances from two perpendicular lines forms a rhombus.
60 Easy
60 of 408
Understanding +4 | -1
Angles made by the lines represented by the equation xy+y=0x y+y=0 with yy-axis are
View answer and explanation
Correct option: A
0∘0^{\circ} and 90∘90^{\circ}
Option A (0∘0^{\circ} and 90∘90^{\circ}) is correct because the lines represented are y=0 and x=-1, making angles of 90 and 0 degrees with the y-axis respectively.
Revision notes and key takeaways

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

Showing 31-60 of 408 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - STRAIGHT LINES MCQs (All) 555 MCQs → ENGINEERING MATHEMATICS - STRAIGHT LINES MCQs (Hard) 8 MCQs → ENGINEERING MATHEMATICS - STRAIGHT LINES MCQs (Medium) 139 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs
📚 Switch Set (EASY)
Chapter Sets
ALL 555 MCQs EASY 408 MCQs HARD 8 MCQs MEDIUM 139 MCQs
0 /408
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →