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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Easy

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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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Showing 241-270 of 408 questions

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241 Easy
241 of 408
Application +4 | -1
The slopes of the line which passes through the origin, and the mid-point of the line segment joining the points. P(3,−4)P(3,-4) and Q(−5,−2)Q(-5,-2) is
View answer and explanation
Correct option: A
3
Option A (3) is correct because it is the slope of the line passing through the origin and the midpoint of the line segment joining P and Q.
242 Easy
242 of 408
Application +4 | -1
Given a family of lines a(2x+y+4)+b(x−2y−3)=0a\left( {2x + y + 4} \right) + b\left( {x - 2y - 3} \right) = 0, the number of lines belonging to the family at a distance 10\sqrt {10} from P(2,  −3)P\left( {2,\; - 3} \right) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because only one line in the family is at the specified distance from the point P.
243 Easy
243 of 408
Analyzing +4 | -1
For a>0,b>0,c>0a>0, b>0, c>0, the distance between (1,1)(1,1) and the point of intersection of the lines ax+by+c=0a x+b y+c=0 and bx+ay+c=0b x+a y+c=0 is less than 222 \sqrt{2}. Then,
View answer and explanation
Correct option: A
a+b−c>0a + b - c > 0
Option A (a+b−c>0a + b - c > 0) is correct because it is derived from the distance condition between the point of intersection and (1,1) being less than 222\sqrt{2}.
244 Easy
244 of 408
Application +4 | -1
The number of integer values of mm for which the xx-coordinate of the point of intersection of the lines 3x+4y=93 x+4 y=9 and y=mx+1y=m x+1 is also an integer; is
View answer and explanation
Correct option: A
2
Option A (2) is correct because there are two integer values of m that make the x-coordinate of the intersection point an integer.
245 Easy
245 of 408
Application +4 | -1
The straight line whose sum of the intercepts on the axes is equal to half of the product of the intercepts, passes through the points
View answer and explanation
Correct option: B
(2, 2)
Option B ((4,2)) is correct because it satisfies the conditions of the right-angled triangle ABC, where A and B are given, and BC = 5 units.
246 Easy
246 of 408
Analyzing +4 | -1
If the quadrilateral formed by the lines ax+by+c=0,a′x+b′y+c=0,ax+by+c′=0a x+b y+c=0, a^{\prime} x+b^{\prime} y+c=0, a x+b y+c^{\prime}=0,
a′x+b′y+c′=0a^{\prime} x+b^{\prime} y+c^{\prime}=0 have perpendicular diagonals, then
View answer and explanation
Correct option: C
a2+b2=a′2+b′2{a^2} + {b^2} = {a'^2} + {b'^2}
Option C (a2+b2=a′2+b′2{a^2} + {b^2} = {a'^2} + {b'^2}) is correct because the quadrilateral is a rhombus, implying equal distances between parallel sides.
247 Easy
247 of 408
Application +4 | -1
If a,ba, b and cc are any three terms of an AP, then the ax+by+c=0a x+b y+c=0
View answer and explanation
Correct option: B
always passes through a fixed point
The correct answer is that the line always passes through a fixed point because the coefficients satisfy the condition for an arithmetic progression, allowing us to find a point that always satisfies the equation.
248 Easy
248 of 408
Application +4 | -1
The equation of the straight lines passing through the point (4, 3) and making intercepts on the coordinate axes whose sum is -1, is
View answer and explanation
Correct option: D
x2−y3=1  \frac{x}{2} - \frac{y}{3} = 1\; and x−2+y1=1\frac{x}{{ - 2}} + \frac{y}{1} = 1
The correct answer is x2−y3=1  \frac{x}{2} - \frac{y}{3} = 1\; and x−2+y1=1\frac{x}{{ - 2}} + \frac{y}{1} = 1 because these are the equations of lines passing through (4, 3) with intercepts summing to -1.
249 Easy
249 of 408
Application +4 | -1
The equation of the line ABA B is y=xy=x. If AA and BB lie on
the same side of the line mirror 2x−y=12 x-y=1, then the equation of the
image of ABA B, is
View answer and explanation
Correct option: C
7x−y−6=07 x-y-6=0
Option C (7x−y−6=07 x-y-6=0) is correct because it represents the image of the line AB after reflection across the line mirror 2x-y=1.
250 Easy
250 of 408
Application +4 | -1
The centroid of the triangle formed by the pair of straight lines 12x2−20xy+7y2=012{x^2} - 20xy + 7{y^2} = 0 and the line 2x−3y+4=02x - 3y + 4 = 0 is
View answer and explanation
Correct option: C
(83,83)\left( {\frac{8}{3},\frac{8}{3}} \right)
Option C ((83,83)\left( {\frac{8}{3},\frac{8}{3}} \right)) is correct because it is the centroid calculated from the vertices of the triangle formed by the intersection of the given lines.
251 Easy
251 of 408
Analyzing +4 | -1
If a≠b≠c,a \ne b \ne c, then the equations (b−c)x+(c−a)y+(a−b)=0\left( {b - c} \right)x + \left( {c - a} \right)y + \left( {a - b} \right) = 0 and, (b3−c3)x+(c3−a3)y+(a3−b3)=0\left( {{b^3} - {c^3}} \right)x + \left( {{c^3} - {a^3}} \right)y + \left( {{a^3} - {b^3}} \right) = 0
will represent the same line, if
View answer and explanation
Correct option: D
a+b+c=0a + b + c = 0
Option D (a+b+c=0a + b + c = 0) is correct because it's the condition that makes the two given equations represent the same line.
252 Easy
252 of 408
Application +4 | -1
Area of the parallelogram formed by the lines y=mx,y=mx+1,y=nxy=m x, y=m x+1, y=n x and y=nx+1y=n x+1 equals
View answer and explanation
Correct option: D
1∣m−n∣\frac{1}{|m-n|}
Option D (1∣m−n∣\frac{{1}}{{|m-n|}}) is correct because the area of a parallelogram formed by given lines is calculated using the formula involving the determinant of the coefficients.
253 Easy
253 of 408
Application +4 | -1
If the three points (3q,0),(0,3p)\left( {3q,0} \right),\left( {0,3p} \right) and (1,1)\left( {1,1} \right) are collinear then which one is true?
View answer and explanation
Correct option: C
1p+1q=3\frac{1}{p} + \frac{1}{q} = 3
Option C 1p+1q=3\frac{1}{p} + \frac{1}{q} = 3 is correct because it is the condition derived from the collinearity of the three points.
254 Easy
254 of 408
Application +4 | -1
A rectangle ABCD,A≡(0,0),B≡(4,0),C≡(4,2)A B C D, A \equiv(0,0), B \equiv(4,0), C \equiv(4,2) and D≡(0,2)D \equiv(0,2) undergoes the following transformations successively:

(i) f1(x,y)→(y,x)f_1(x, y) \rightarrow(y, x)

(ii) f2(x,y)→(x+3y,y)f_2(x, y) \rightarrow(x+3 y, y)

(iii) f3(x,y)→(x−y2,x+y2)f_3(x, y) \rightarrow\left(\frac{x-y}{2}, \frac{x+y}{2}\right)

The final figure will be
View answer and explanation
Correct option: D
a parallelogram
The final figure will be a parallelogram because after the transformations, opposite sides remain parallel, but the angles are no longer right angles.
255 Easy
255 of 408
Analyzing +4 | -1
Given four lines whose equations are x+2y−3=0,  2x+3y−4=0,3x+4y−7=0x + 2y - 3 = 0,\;2x + 3y - 4 = 0,3x + 4y - 7 = 0 and 4x+5y−6=0,4x + 5y - 6 = 0, then the lines are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because only three of the four lines are concurrent.
256 Easy
256 of 408
Application +4 | -1
Coordinate of the foot of the perpendicular drawn from (0, 0) to the line joining (a  cosα,  asin⁡α)(a\mathrm{\;cos\alpha },\mathrm{\;}a\sin \mathrm{\alpha }) and (acos⁡β,  asin⁡β)  (a\cos \mathrm{\beta },\mathrm{\;}a\sin \mathrm{\beta })\; are
View answer and explanation
Correct option: B
[a2(cos⁡α+cos⁡β),a2(sin⁡α+sin⁡β)]\left[ {\frac{a}{2}(\cos \mathrm{\alpha } + \cos \mathrm{\beta }),\frac{a}{2}(\sin \mathrm{\alpha } + \sin \mathrm{\beta })} \right]
Option B [a2(cos⁡α+cos⁡β),a2(sin⁡α+sin⁡β)]\left[ {\frac{a}{2}(\cos \mathrm{\alpha } + \cos \mathrm{\beta }),\frac{a}{2}(\sin \mathrm{\alpha } + \sin \mathrm{\beta })} \right] is correct because it represents the coordinates of the foot of the perpendicular from (0,0) to the line joining the two given points.
257 Easy
257 of 408
Application +4 | -1
The equation of the bisector of the acute angle between the lines 2x−y+4=02 x-y+4=0 and x−2y=1x-2 y=1, is
View answer and explanation
Correct option: B
x−y+1=0x-y+1=0
Option B is correct. The equation of the bisector of the acute angle between the lines 2x−y+4=02x-y+4=0 and x−2y=1x-2y=1 is x−y+1=0x-y+1=0.
258 Easy
258 of 408
Application +4 | -1
The angle between the straight line x−y3=5x - y\sqrt 3 = 5 and 3x+y=7\sqrt 3 x + y = 7 is
View answer and explanation
Correct option: A
90∘90^\circ
Option A (90∘90^\circ) is correct because the dot product of the direction vectors of the two lines is zero, indicating perpendicularity.
259 Easy
259 of 408
Application +4 | -1
The lines ax+by+c=0,  bx+cy+a=0ax + by + c = 0,\;bx + cy + a = 0 and cx+ay+b=0cx + ay + b = 0 (a≠b≠c)\left( {a \ne b \ne c} \right) are concurrent, if
View answer and explanation
Correct option: C
a+b+c=0a + b + c = 0
Option C (a+b+c=0a + b + c = 0) is correct because the determinant of the coefficients must be zero for concurrency, which simplifies to a+b+c=0a + b + c = 0 given a≠b≠ca \ne b \ne c.
260 Easy
260 of 408
Analyzing +4 | -1
Consider three points

P=(−sin⁡(β−α),  −cos⁡β)P = \left( { - \sin \left( {\mathrm{\beta } - \mathrm{\alpha }} \right),\mathrm{\;} - \cos \mathrm{\beta }} \right)

Q=(cos⁡(β−α),  sin⁡β){\cal Q} = \left( {\cos \left( {\mathrm{\beta } - \mathrm{\alpha }} \right),\mathrm{\;}\sin \mathrm{\beta }} \right)

And R=(cos⁡(β−α+θ),  sin⁡(β−θ))R = \left( {\cos \left( {\mathrm{\beta } - \mathrm{\alpha } + \mathrm{\theta }} \right),\mathrm{\;}\sin (\mathrm{\beta } - \mathrm{\theta }} \right)),

Where 0<α,  β,  θ<π40 < \alpha ,\;\beta ,\;\theta < \frac{\pi }{4}. Then,
View answer and explanation
Correct option: D
P,  Q,  RP,\;{\cal Q},\;R are non-collinear
Option D (P,  Q,  RP,\;{\cal Q},\;R are non-collinear) is correct because the determinant formed by the coordinates of the points is non-zero, indicating non-collinearity.
261 Easy
261 of 408
Understanding +4 | -1
Consider the following statements.

I. Centroid of a triangle is a point where angle bisectors meet.

II. If value of area after calculations is negative then we take its negative value.

Choose the correct option.
View answer and explanation
Correct option: C
Both are false
Option C is correct because both statements are false; the centroid is where medians meet, and area is taken as an absolute value.
262 Easy
262 of 408
Application +4 | -1
A triangle ABC, right angled at A, has points A and B as (2,3) and (0,-1) respectively. If BC=5 units, then the point C, is
View answer and explanation
Correct option: B
(4,2)
Option C (θ\theta) is correct because it is the acute angle between the pair of straight lines represented by the given equation, derived using the formula involving the coefficients of the equation.
263 Easy
263 of 408
Application +4 | -1
The coordinates of the incentre of the triangle having sides

3x−4y=0,  5x+12y=03x - 4y = 0,\;5x + 12y = 0

and y−15=0y - 15 = 0 are
View answer and explanation
Correct option: A
−1,  8- 1,\;8
Option A ( - 1, 8) is correct because it is the incentre calculated using the formula with the vertices and side lengths of the triangle.
264 Easy
264 of 408
Application +4 | -1
Normal form of line x+y+1=0x+y+1=0 is
View answer and explanation
Correct option: C
xcos⁡(225∘)+ysin⁡(225∘)=12x \cos \left(225^{\circ}\right)+y \sin \left(225^{\circ}\right)=\frac{1}{\sqrt{2}}
Option C, xcos⁡(225°)+ysin⁡(225°)=12x \cos{(225°)} + y \sin{(225°)} = \frac{{1}}{{\sqrt{{2}}}} is the correct normal form because it represents the given line with the correct orientation and distance from the origin.
265 Easy
265 of 408
Understanding +4 | -1
Consider the following statements.

I. The equation of a straight line making intercepts aa and bb on xx and yy-axis respectively is given by xa+yb=1\frac{x}{a}+\frac{y}{b}=1

II. If ax+by+c1=0a x+b y+c_1=0 and ax+by+c2=0a x+b y+c_2=0 be two parallel lines, then distance between two parallel lines, d=∣c1−c2a2+b2∣d=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|.

Choose the correct option.
View answer and explanation
Correct option: C
Both are true
Option C is correct because both statements are standard formulas and definitions in coordinate geometry.
266 Easy
266 of 408
Application +4 | -1
A line has slope m and y-intercept 4. The distance between the origin and the line is equal to
View answer and explanation
Correct option: C
4m2+1\frac{4}{{\sqrt {{m^2} + 1} }}
Option C (4m2+1\frac{4}{{\sqrt {{{{m^2} + 1}}} }}) is correct because it is the perpendicular distance from the origin to the line expressed in terms of the slope m.
267 Easy
267 of 408
Application +4 | -1
If (−2,6)\left( { - 2,6} \right) is the image of the point (4,2) with respect to the line L=0, then L=
View answer and explanation
Correct option: A
3x−2y+53x - 2y + 5
Option A (3x−2y+53x - 2y + 5) is correct because the line L=0 is the perpendicular bisector of the segment joining the point and its image.
268 Easy
268 of 408
Analyzing +4 | -1
If the three distinct lines x+2ay+a=0,x+3by+b=0x+2 a y+a=0, x+3 b y+b=0 and x+4ay+a=0x+4 a y+a=0 are concurrent, then the point (a,b)(a, b) lies on a:
View answer and explanation
Correct option: C
straight line
Option C (straight line) is correct because the condition for concurrency leads to a linear relationship between a and b.
269 Easy
269 of 408
Application +4 | -1
If AA and BB are two points having coordinates (3, 4) and (5,−2)\left( {5, - 2} \right) respectively and PP is a point such that PA=PBPA = PB and area of triangle PAB=10PAB = 10 sq unit, then the coordinates of PP are
View answer and explanation
Correct option: B
(7, 2) and (1, 0)
Option B ((7, 2) and (1, 0)) is correct because these coordinates satisfy the conditions PA = PB and the area of triangle PAB = 10 sq unit.
270 Easy
270 of 408
Application +4 | -1
The image of the point (3,5) in the line x−y+1=0x - y + 1 = 0, lies on
View answer and explanation
Correct option: D
(x−2)2+(y−4)2=4{\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} = 4
Option D is correct. The image of the point lies on the circle with the given equation.
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