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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 121-150 of 408 questions

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121 Easy
121 of 408
Application +4 | -1
The inclination of the straight line passing through the point A(−3,6)A(-3,6) and the mid-point of the line joining the point B(4,−5)B(4,-5) and C(−2,9),C(-2,9), with xx-axis is
View answer and explanation
Correct option: D
3π4\frac{3 \pi}{4}
Option D (3π4\frac{3 \pi}{4}) is correct because the slope of the line passing through A and the midpoint is -1, corresponding to an angle of 135 degrees or 3π4\frac{3 \pi}{4} radians.
122 Easy
122 of 408
Application +4 | -1
The equation of the line is passing through P(4,5)P(4,5) and making 30∘30^{\circ} with XX-axis. Then, coordinates of point which is at a distance 4 units on either side of PP, are
View answer and explanation
Correct option: C
(4+23,7),(4−23,3)(4+2 \sqrt{3}, 7),(4-2 \sqrt{3}, 3)
The correct answer is C ((4+23,7),(4−23,3)(4+2 \sqrt{{3}}, 7),(4-2 \sqrt{{3}}, 3)) because it correctly calculates the coordinates of points at a distance of 4 units on either side of P using the parametric form of a line.
123 Easy
123 of 408
Application +4 | -1
In a triangle ABCA B C, right angled at the vertex AA, if the position vectors of A,BA, B and CC are respectively 3i^+j^−k^,−i^+3j^+pk^3 \hat{i}+\hat{j}-\hat{k},-\hat{i}+3 \hat{j}+p \hat{k} and 5i^+qj^−4k^5 \hat{i}+q \hat{j}-4 \hat{k}, then the point (p,q)(p, q) lies on a line
View answer and explanation
Correct option: B
making an acute angle with the positive direction of xx-axis
Option B (making an acute angle with the positive direction of x-axis) is correct because the equation of the line on which (p, q) lies has a positive slope, indicating an acute angle with the x-axis.
124 Easy
124 of 408
Application +4 | -1
In a triangle ABC the equation of the bisector of angle BB is y=xy=x. If A=(2,6)A=(2,6) and B=(1,1)B=(1,1), then equation of side BCB C, is
View answer and explanation
Correct option: B
x−5y+4=0x-5 y+4=0
Option B (x−5y+4=0x-5 y+4=0) is correct because it represents the equation of side BC given the bisector of angle B and vertex A.
125 Easy
125 of 408
Application +4 | -1
If the line px2−qxy−y2=0p{x^2} - qxy - {y^2} = 0 makes an angles α\alpha and β\beta with x-axis, then the value of tan (α+β)\left( {\alpha + \beta } \right) is
View answer and explanation
Correct option: A
−q1+p\frac{{ - q}}{{1 + p}}
The correct answer is A ((-q)/(1+p)) because the slopes of the lines are used to find the tangent of the sum of the angles.
126 Easy
126 of 408
Application +4 | -1
Equation of the line passing through the point ( a−1a-1, a+1)a+1) and making zero intercept on both axes is
View answer and explanation
Correct option: D
(a+1)x−(a−1)y=0(a+1) x-(a-1) y=0
Option D ((a+1)x−(a−1)y=0(a+1) x-(a-1) y=0) is correct because the line passes through the origin and the given point, and this equation satisfies both conditions.
127 Easy
127 of 408
Application +4 | -1
The locus of the point of intersection of lines xcos⁡α+ysin⁡α=ax\cos \alpha + y\sin \alpha = a and xsin⁡α−ycos⁡α=bx\sin \alpha - y\cos \alpha = b is (α\alpha is a variable)
View answer and explanation
Correct option: C
x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2}
Option C (x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2}) is correct because it represents the locus of the point of intersection of the given lines after eliminating the variable α\alpha.
128 Easy
128 of 408
Analyzing +4 | -1
If t1,t2t_1, t_2 and t3t_3 are distinct, the points (t1,2at1+at13)\left(t_1, 2 a t_1+a t_1^3\right), (t2,2at2+at23)\left(t_2, 2 a t_2+a t_2^3\right) and (t3,2at3+at33)\left(t_3, 2 a t_3+a t_3^3\right) are collinear if
View answer and explanation
Correct option: C
t1+t2+t3=0t_1+t_2+t_3=0
Option C is correct. The points are collinear if the determinant formed by their coordinates is zero, which simplifies to t1 + t2 + t3 = 0.
129 Easy
129 of 408
Application +4 | -1
A line through the point A(2,0) which makes an angle of 30∘{30^ \circ } with the positive direction of x-axis is rotated about A in clockwise direction through an angle of 15∘{15^ \circ }. Then, the equation of the straight line in the new position is
View answer and explanation
Correct option: B
(2−3)x−y−4+23=0\left( {2 - \sqrt 3 } \right)x - y - 4 + 2\sqrt 3 = 0
Option B ((2−3)x−y−4+23=0\left( {2 - \sqrt 3 } \right)x - y - 4 + 2\sqrt 3 = 0) is correct because it represents the equation of the line after rotation by 15 degrees clockwise.
130 Easy
130 of 408
Application +4 | -1
If non-zero numbers a,b,c are in HP, then the straight line xa+yb+1c=0  \frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0\; always passes through a fixed point. That point is
View answer and explanation
Correct option: B
(1,  −2)\left( {1,\; - 2} \right)
Option B ((1, -2)) is correct because substituting x=1 and y=-2 into the equation satisfies the condition for a, b, and c being in Harmonic Progression.
131 Easy
131 of 408
Application +4 | -1
The image of the point (3,8) in the line x+3y=7,x + 3y = 7, is
View answer and explanation
Correct option: C
(−1,−4)\left( { - 1, - 4} \right)
The correct answer is C ((-1, -4)) because it is the reflection of the point (3,8) across the given line, found using perpendicularity and midpoint conditions.
132 Easy
132 of 408
Analyzing +4 | -1
The lines (lx+my)2−3(mx−ly)2=0{(lx + my)^2} - 3{(mx - ly)^2} = 0 and lx+my+n=0lx + my + n = 0 form
View answer and explanation
Correct option: C
An equilateral triangle
Option C (An equilateral triangle) is correct because the angles between the lines are 60 degrees, indicating an equilateral triangle.
133 Easy
133 of 408
Application +4 | -1
The equation of the line passing through the origin and the point of intersection of the lines xa+yb=1\frac{x}{a} + \frac{y}{b} = 1 and xb+ya=1\frac{x}{b} + \frac{y}{a} = 1 is
View answer and explanation
Correct option: D
x−y=0x - y = 0
Option D (x−y=0x - y = 0) is correct because the line passing through the origin and the intersection of the given lines is y=x.
134 Easy
134 of 408
Application +4 | -1
The set of lines ax+by+c=0a x+b y+c=0, where 3a+2b+4c=03 a+2 b+4 c=0 is concurrent at the point
View answer and explanation
Correct option: A
(3/4,1/2)(3 / 4,1 / 2)
Option A is correct. The set of lines ax+by+c=0ax+by+c=0, where 3a+2b+4c=03a+2b+4c=0, is concurrent at the point (3/4,1/2)(3/4, 1/2).
135 Easy
135 of 408
Application +4 | -1
(i) Find the integral values of 'hh' for which hx2−5xy+4hy2+x+2y−2=0h x^2-5 x y+4 {hy}^2+{x}+2 {y}-2=0 represents two real straight lines.
(ii) If the pair of lines represented by equation k(k−3)x2+16xy+(k+1)y2=0k(k-3) x^2+16 x y+(k+1) y^2=0 are perpendicular to each other, then find kk.
View answer and explanation
Correct option: A
h=1;k=1{h}=1;{k}=1
Option A (h=1; k=1) is correct because the integral value of h is found using the determinant condition for a pair of lines, and k is found by setting the sum of the coefficients of x2 and y2 to zero for perpendicular lines.
136 Easy
136 of 408
Application +4 | -1
If the image of a point P(2,3)P(2,3) in the line mirror y=xy=x is the point QQ and the image QQ in the line mirror y=0y=0 is the point R(x,y)R(x, y), then the coordinates of RR are
View answer and explanation
Correct option: D
(3,−2)(3,-2)
Option D is correct. The image of P(2,3) in y=x is Q(3,2), and the image of Q(3,2) in y=0 is R(3,-2).
137 Easy
137 of 408
Application +4 | -1
The equation of a line passing through (−2,  −4)\left( { - 2,\; - 4} \right) and perpendicular to the line 3x−y+5=03x - y + 5 = 0 is
View answer and explanation
Correct option: C
x+3y+14=0x + 3y + 14 = 0
Option C (-9) is correct because it is the value of k that makes the given equation represent a pair of straight lines, which is determined by setting the determinant of the coefficients to zero.
138 Easy
138 of 408
Application +4 | -1
If the angle between the lines 3y−x+4=0\sqrt{3 y}-x+4=0 and x+y−6=0x+y-6=0 is θ\theta, then tan⁡θ\tan \theta is equal to.
View answer and explanation
Correct option: C
2+32+\sqrt{3}
Option C (2+32+\sqrt{3}) is correct because it is the result of applying the tangent of the angle between two lines formula, given their slopes.
139 Easy
139 of 408
Analyzing +4 | -1
The straight lines x+y=0,  3x+y−4=0x + y = 0,\;3x + y - 4 = 0 and x+3y−4=0x + 3y - 4 = 0 form a triangle which is
View answer and explanation
Correct option: C
Isosceles
Option C (Isosceles) is correct because two sides of the triangle formed by the given lines have equal length.
140 Easy
140 of 408
Application +4 | -1
The image of the point (1,3) in the line x+y−6=0,x + y - 6 = 0, is
View answer and explanation
Correct option: A
(3,5)
Option A ((3,5)) is correct because the image point is found using the formula for reflection across a line.
141 Easy
141 of 408
Application +4 | -1
The nearest point on the line 3x−4y=253x - 4y = 25 from the origin is
View answer and explanation
Correct option: B
(3,  −4)\left( {3,\; - 4} \right)
Option B ((3, -4)) is correct because it is found by determining the foot of the perpendicular from the origin to the given line.
142 Easy
142 of 408
Analyzing +4 | -1
The diagonals of the parallelogram whose sides are lx+my+n=0,  lx+my+n′=0,  mx+ly+n=0,  mx+ly+n′=0lx + my + n = 0,\;lx + my + n' = 0,\;mx + ly + n = 0,\;mx + ly + n' = 0 include an angle
View answer and explanation
Correct option: B
π/2\pi /2
Option B (π/2\pi /2) is correct because the parallelogram formed by the given lines is a rhombus, and the diagonals of a rhombus are perpendicular.
143 Easy
143 of 408
Application +4 | -1
A triangle with vertices (4,  0),  (−1,  −1),  (3,  5)\left( {4,\;0} \right),\;\left( { - 1,\; - 1} \right),\;\left( {3,\;5} \right) is
View answer and explanation
Correct option: A
Isosceles and right angled
Option D (0) is correct because the determinant of the matrix formed by the given points is zero, indicating that the points are collinear and the area of the triangle is zero.
144 Easy
144 of 408
Understanding +4 | -1
The equations ax+by+c=0ax + by + c = 0 and dx+ey+f=0dx + ey + f = 0 represent the same straight line if and only if
View answer and explanation
Correct option: C
ad=be=cf\frac{a}{d} = \frac{b}{e} = \frac{c}{f}
Option C (ad=be=cf\frac{a}{d} = \frac{b}{e} = \frac{c}{f}) is correct because for two equations to represent the same line, their coefficients must be proportional.
145 Easy
145 of 408
Application +4 | -1
If the point P(a,a2)P\left(a, a^2\right) lies inside the triangle formed by the lines x=0,y=0x=0, y=0 and x+y=2x+y=2, then exhaustive range of 'aa' is
View answer and explanation
Correct option: A
(0,1)(0,1)
Option A (0,1)(0,1) is correct because it represents the range of 'a' for which the point P lies inside the triangle.
146 Easy
146 of 408
Application +4 | -1
The value of kk for which the lines 2x−3y+k=0,3x−4y−13=02x - 3y + k = 0,3x - 4y - 13 = 0 and 8x−11y−33=08x - 11y - 33 = 0 are concurrent, is
View answer and explanation
Correct option: B
-7
The correct answer is D (-20) because the condition for concurrency of three lines is applied by setting the determinant of the coefficients to zero and solving for k.
147 Easy
147 of 408
Application +4 | -1
The distance of the x+y−8=0x + y - 8 = 0 and (4,1)\left( {4,1} \right) measured along the direction whose slope is -2 is
View answer and explanation
Correct option: A
3  53\;\sqrt 5
Option A (353\sqrt{5}) is correct because it is the distance calculated along the line with the given slope from the point to the line.
148 Easy
148 of 408
Application +4 | -1
The length of perpendicular from the point (acos⁡α,  asin⁡α)\left( {a\cos \mathrm{\alpha },\;a\sin \mathrm{\alpha }} \right) upon the straight line y=xtan⁡α+c,  c>0,y = x\tan \mathrm{\alpha } + c,\;c > 0, is
View answer and explanation
Correct option: C
ccos⁡αc\cos \mathrm{\alpha }
Option C (ccos⁡αc\cos \mathrm{\alpha }) is correct because it represents the length of the perpendicular from the given point to the line, calculated using the perpendicular distance formula.
149 Easy
149 of 408
Application +4 | -1
If the points (a,  b),  (a′,  b′)\left( {a,\;b} \right),\;\left( {a',\;b'} \right) and (a−a′,  b−b′)\left( {a - a',\;b - b'} \right) are collinear, then
View answer and explanation
Correct option: A
ab′=a′bab' = a'b
Option A (ab′=a′bab' = a'b) is correct because if the points are collinear, the determinant of the matrix formed by the coordinates must be zero, leading to the given relationship.
150 Easy
150 of 408
Application +4 | -1
The locus of the point (x,y)\left( {x,y} \right) which is equidistant from the points (a+b,b−a)\left( {a + b,b - a} \right) and (a−b,a+b)\left( {a - b,a + b} \right) is
View answer and explanation
Correct option: D
bx−ay=0bx - ay = 0
Option D (bx−ay=0bx - ay = 0) is correct because it represents the locus of points equidistant from the two given points, derived by equating the squared distances.
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