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

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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 151-180 of 408 questions

Preparing practice and saved progress…

151 Easy
151 of 408
Analyzing +4 | -1
(i) The points (−1,0),(4,−2)(-1,0),(4,-2) and (cos⁡2θ,sin⁡2θ)(\cos 2 \theta, \sin 2 \theta) are collinear

(ii) The points (−1,0),(4,−2)(-1,0),(4,-2) and (1−tan⁡2θ1+tan⁡2θ,2tan⁡θ1+tan⁡2θ)\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}, \frac{2 \tan \theta}{1+\tan ^2 \theta}\right) are collinear
View answer and explanation
Correct option: B
statemetn (i) has more solution than statement (ii) for θ\theta
Option A (statement (i) has more solution than statement (ii) for θ\theta) is correct because statement (i) yields more solutions for θ\theta than statement (ii) due to the different trigonometric forms.
152 Easy
152 of 408
Application +4 | -1
If the line 2x+y=k2 x+y=k passes through the point which divides the line segment joining the points (1,1)(1,1) and (2,4)(2,4) in the ratio 3:23: 2, then kk equals :
View answer and explanation
Correct option: C
6
The correct answer is C (6) because the value of k is found by substituting the coordinates of the point dividing the line segment in the given ratio into the equation of the line.
153 Easy
153 of 408
Application +4 | -1
Bisector of angle between x+2y−1=0x+2 y-1=0 and 2x+y+1=02 x+y+1=0 containing (2,1)(2,1) is
View answer and explanation
Correct option: D
x−y+2=0x-y+2=0
Option B, x-y+2=0, is the correct bisector of the angle between the two lines that contains the point (2,1).
154 Easy
154 of 408
Application +4 | -1
The area of the parallelogram formed by the lines xcos⁡α+ysin⁡α=p,  xcos⁡α+ysin⁡α=q,x\cos \alpha + y\sin \alpha = p,\;x\cos \alpha + y\sin \alpha = q, xcos⁡β+ysin⁡β=rx\cos \beta + y\sin \beta = r and xcos⁡β+ysin⁡β=sx\cos \beta + y\sin \beta = s for given values of p,q,rp,q,r and s is least, if (α−β)=\left( {\alpha - \beta } \right) =
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option A (±π2\pm \frac{\pi }{2}) is incorrect. The area of the parallelogram is minimized when the angle between the lines is maximized, which occurs at ±π2\pm \frac{\pi }{2}. The question asks when the area is LEAST, not maximum. The correct answer should be when the area is largest, which is when the angle is smallest.
155 Easy
155 of 408
Application +4 | -1
Equation of the line equidistant from the parallel lines 9x+6y−7=09 x+6 y-7=0 and 3x+2y+6=03 x+2 y+6=0 is
View answer and explanation
Correct option: B
18x+12y+11=018 x+12 y+11=0
Option B (18x+12y+11=018 x+12 y+11=0) is correct because it represents the line equidistant from the two given parallel lines.
156 Easy
156 of 408
Analyzing +4 | -1
If a,b,ca, b, c be in A.P., then ax+by+c=0a x+b y+c=0 represents
View answer and explanation
Correct option: B
a family of concurrent lines
Option B (a family of concurrent lines) is correct because when a, b, c are in A.P., the line equation can be rearranged to represent lines passing through a fixed point.
157 Easy
157 of 408
Application +4 | -1
The distance between the lines 5x−12y+65=05x - 12y + 65 = 0 and 5x−12y−39=05x - 12y - 39 = 0 is
View answer and explanation
Correct option: D
8
The correct answer is D (8) because it is calculated using the formula for the distance between two parallel lines.
158 Easy
158 of 408
Application +4 | -1
If a,b,c∈Ra, b, c \in \mathrm{R} and 1 is a root of equation ax2+bx+c=0a x^{2}+b x+c=0, then the curve y=4ax2+3bx+2c,a≠0y=4 a x^{2}+3 b x+2 c, a \neq 0 intersect xx-axis at
View answer and explanation
Correct option: D
exactly one point
The correct answer is D (exactly one point) because by substituting the first equation into the second, we get a linear equation, which intersects the x-axis at only one point.
159 Easy
159 of 408
Application +4 | -1
The line segment joining the points (1,2)(1,2) and (−2,1)(-2,1) is divided by the line 3x+4y=73 x+4 y=7 in the ratio:
View answer and explanation
Correct option: D
4:94: 9
Option D (4:94: 9) is correct because it represents the ratio in which the line divides the line segment, calculated using the section formula.
160 Easy
160 of 408
Application +4 | -1
The straight line 3x+y=93x + y = 9 divided the line segment joining the points (1, 3) and (2,7) in the ratio
View answer and explanation
Correct option: B
3:4 internally
Option B (3:4 internally) is correct because the ratio is calculated by substituting the points into the line equation and taking the negative ratio.
161 Easy
161 of 408
Analyzing +4 | -1
If a variable line drown through the point of intersection of straight lines xα+yβ=1\frac{x}{\mathrm{\alpha }} + \frac{y}{\mathrm{\beta }} = 1 and xβ+yα=1\frac{x}{\mathrm{\beta }} + \frac{y}{\mathrm{\alpha }} = 1 meets the coordinate axes in A and B,then the locus of the mid point of AB is
View answer and explanation
Correct option: B
αβ(x+y)=2xy(α+β)\mathrm{\alpha \beta }\left( {x + y} \right) = 2xy\left( {\mathrm{\alpha } + \mathrm{\beta }} \right)
Option B (αβ(x+y)=2xy(α+β)\mathrm{\alpha \beta }\left( {x + y} \right) = 2xy\left( {\mathrm{\alpha } + \mathrm{\beta }} \right)) is correct because it represents the locus of the midpoint of AB based on the given conditions.
162 Easy
162 of 408
Analyzing +4 | -1
If A(1,1),B(3+1,  2)A\left( {1,1} \right),B\left( {\sqrt 3 + 1,\;2} \right) and C(3,3+2)C\left( {\sqrt 3 ,\sqrt 3 + 2} \right) be three vertices of a square, then the diagonal through B is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated equation of the diagonal through B does not match any of the given options.
163 Easy
163 of 408
Application +4 | -1
If orthocenter and circumcentre of a triangle are respectively (1, 1) and (3, 2), then the coordinates of its centroid are
View answer and explanation
Correct option: A
(73,53)\left( {\frac{7}{3},\frac{5}{3}} \right)
Option A ((73,53)\left( {\frac{7}{3},\frac{5}{3}} \right)) is correct because the centroid divides the line segment joining the orthocenter and circumcenter in the ratio 2:1.
164 Easy
164 of 408
Application +4 | -1
If a straight line through the point P(3,5)P(3,5) makes an angle π6\frac{\pi}{6} with xx-axis and meets the lines 2x+y+5=0 & 3x−2y+7=02 x+y+5=0 \,\& \,3 x-2 y+7=0 at Q & RQ \,\&\, R then PQPR=\frac{P Q}{P R}=
View answer and explanation
Correct option: C
2(123−8)63+3\frac{2(12 \sqrt{3}-8)}{6 \sqrt{3}+3}
Option C, 2(123−8)63+3\frac{{2(12 \sqrt{{3}}-8)}}{{6 \sqrt{{3}}+3}}, is the correct ratio of PQ to PR, calculated by finding the intersection points of the line with the given lines and using the distance formula.
165 Easy
165 of 408
Application +4 | -1
The distance of point (−1,1)(-1,1) from the line 12x−5y+9=012 x-5 y+9=0 is
View answer and explanation
Correct option: B
813\frac{8}{13}
Option B (813\frac{{8}}{{13}}) is correct because it is the correct distance calculated using the point-to-line distance formula.
166 Easy
166 of 408
Application +4 | -1
The incentre of the triangle formed by the lines x=0,y=0x = 0,y = 0 and 3x  4y=123x\;4y = 12 is at
View answer and explanation
Correct option: B
(1,1)
Option B (1,1) is correct because the incenter of the triangle formed by the given lines is calculated using the formula for the incenter of a right-angled triangle.
167 Easy
167 of 408
Application +4 | -1
The coordinates of foot of the perpendicular drawn from the point (2,4)(2,4) on the line x+y=1x+y=1, are
View answer and explanation
Correct option: B
(−12,32)\left(-\frac{1}{2}, \frac{3}{2}\right)
Option B ((−12,32)\left(-\frac{1}{2}, \frac{3}{2}\right)) is correct because it represents the coordinates of the point on the line x+y=1x+y=1 that is closest to (2,4).
168 Easy
168 of 408
Application +4 | -1
The orthocentre of the triangle formed by the lines x=2,y=3x=2, y=3 and 3x+2y=63 x+2 y=6 is at the point
View answer and explanation
Correct option: C
(2,3)(2,3)
Option C (2,3) is correct because the orthocentre of a right triangle is the vertex where the right angle is formed, which is the intersection of x=2 and y=3.
169 Easy
169 of 408
Analyzing +4 | -1
The distance between the two points AA and A′A^{\prime} which lie on y=2y=2 such that both the line segments ABA B and A′BA^{\prime} B (where BB is the point (2,3)(2,3) ) subtend angle π4\frac{\pi}{4} at the origin, is equal to
View answer and explanation
Correct option: C
525\frac{52}{5}
The correct answer is B (525\frac{{52}}{{5}}) because it represents the distance between points A and A' calculated using the tangent of the angle and the given point B.
170 Easy
170 of 408
Application +4 | -1
Two lines are drawn through (3,4)(3,4), each of which makes angle of 45∘45^{\circ} with the line x−y=2x-y=2, then area of the triangle formed by these lines is NN. Find 2N2 N.
View answer and explanation
Correct option: A
9
The correct answer is A (9) because it is the value of 2N after finding the area of the triangle formed by the given lines.
171 Easy
171 of 408
Analyzing +4 | -1
If the straight lines x+3y=4,3x+y=4x+3 y=4,3 x+y=4 and x+y=0x+y=0 form a triangle, then the triangle is
View answer and explanation
Correct option: D
right angled isosceles
Option D (right angled isosceles) is correct because the triangle formed by the lines has two equal sides and a right angle.
172 Easy
172 of 408
Application +4 | -1
Let A(1,1),B(6,5),C(3/2,2)A(1,1), B(6,5), C(3 / 2,2) be the vertices of △ABC\triangle A B C. If a point PP is such that areas of △PAB,△PAC\triangle P A B, \triangle P A C and △PBC\triangle P B C are equal and Q(−76,−13)Q\left(-\frac{7}{6},-\frac{1}{3}\right) is a point, then PQ=P Q=
View answer and explanation
Correct option: D
5
Option D (5) is correct because point P is the centroid of the triangle, and the distance PQ is calculated using the distance formula.
173 Easy
173 of 408
Application +4 | -1
If a point P(4,  3)P\left( {4,\;3} \right) is shifted by a distance2\sqrt 2 unit parallel to the line y=xy = x, then coordinates of p in new position are
View answer and explanation
Correct option: A
(5, 4)
Option D (30.36) is correct because the strength of 10 volume H2O2 solution is calculated to be approximately 30.36 g/L.
174 Easy
174 of 408
Analyzing +4 | -1
The equation of the sides of a triangle are x−3y=0x - 3y = 0, 4x+3y=54x + 3y = 5 and 3x+y=03x + y = 0. The line 3x−4y=03x - 4y = 0 passes through
View answer and explanation
Correct option: C
The orthocentre
Option C (The orthocentre) is correct because the intersection of the perpendicular lines x - 3y = 0 and 3x + y = 0 is the orthocenter (0,0), and the line 3x - 4y = 0 passes through it.
175 Easy
175 of 408
Analyzing +4 | -1
The straight line x+2y−9=0,3x+5y−5=0x+2 y-9=0,3 x+5 y-5=0 and ax+by−1=0a x+b y-1=0 are concurrent if the straight line 35x−22y+1=035 x-22 y+1=0 passes through the point
View answer and explanation
Correct option: A
(a,b)(a, b)
Option A ((a,b)(a, b)) is correct because the concurrency condition leads to an equation that the line 35x−22y+1=035x - 22y + 1 = 0 satisfies at the point (a,b)(a, b).
176 Easy
176 of 408
Application +4 | -1
If the point (a,a) falls between the lines ∣x+y∣=4\left| {x + y} \right| = 4, then
View answer and explanation
Correct option: A
∣a∣=2\left| a \right| = 2
Option C (∣a∣<2|a| < 2) is correct because the point (a, a) must lie between the lines x + y = 4 and x + y = -4.
177 Easy
177 of 408
Application +4 | -1
The circumcentre of the triangle with vertices (0, 30), (4, 0) and (30, 0) is
View answer and explanation
Correct option: D
(17, 17)
Option D (17, 17) is correct because the circumcenter is equidistant from all vertices of the triangle, and solving the equations based on this condition yields the coordinates (17, 17).
178 Easy
178 of 408
Application +4 | -1
The equation of a straight line parallel to 2x+3y+11=02 x+3 y+11=0 and which is such that the sum of its intercept on the axes is 15 , is
View answer and explanation
Correct option: A
2x+3y−18=02 x+3 y-18=0
Option A (2x+3y−18=02 x+3 y-18=0) is correct because it represents a line parallel to the given line with intercepts summing to 15.
179 Easy
179 of 408
Understanding +4 | -1
Consider the following statements about straight lines:

I. Slope of horizontal line is zero and slope of vertical line is undefined.

II. Two lines are parallel if and only if their slopes are equal.

III. Two lines are perpendicular if and only if product of their slope is −1.-1 .

Which of the above statements are true?
View answer and explanation
Correct option: D
All the above
Option D is correct because all three statements about straight lines are fundamental and well-established geometric principles.
180 Easy
180 of 408
Application +4 | -1
Let the base of a triangle lie along the line x=a and be of length 2a. The area of this triangle is a2{a^2} if the vertex lies on the lines
View answer and explanation
Correct option: B
x=0,  x=ax = 0,\;x = a
Option A (x=−a,  x=2ax = - a,\;x = 2a) is incorrect. The correct answer should be x=0,  x=2ax = 0,\;x = 2a because these lines satisfy the area condition.
Revision notes and key takeaways

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

Showing 151-180 of 408 questions

Your practice

0 / 408 answered · 408 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

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.