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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Mixed difficulty

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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.

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Showing 181-210 of 555 questions

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181 Easy
181 of 555
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.
182 Easy
182 of 555
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.
183 Medium
183 of 555
Application +4 | -1
Let ax+by+c=0{ax}+{by}+{c}=0 be a variable straight line, where a,b{a}, {b} and c{c} are 1st,3rd1^{\text {st}}, 3^{\text {rd}} and 7th7^{\text {th}} terms of an increasing A.P. respectively. Then the variable straight line always passes through a fixed point which lies on-
View answer and explanation
Correct option: D
x2+y2=13x^2+y^2=13
Option D (x2+y2=13x^2+y^2=13) is correct because the variable line always passes through a fixed point (2,-3), which lies on the circle x2+y2=13x^2+y^2=13.
184 Hard
184 of 555
Application +4 | -1
The algebraic sum of the perpendicular distances from A(x1,y1),B(x2,y2)A\left(x_1, y_1\right), B\left(x_2, y_2\right) and C(x3,y3)C\left(x_3, y_3\right) to a variable line is zero, then the line passes through
View answer and explanation
Correct option: B
the centroid of △ABC\triangle A B C
Option B is correct. If the algebraic sum of the perpendicular distances from A(x1,y1)A(x_1, y_1), B(x2,y2)B(x_2, y_2), and C(x3,y3)C(x_3, y_3) to a variable line is zero, then the line passes through the centroid of △ABC\triangle ABC.
185 Easy
185 of 555
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.
186 Easy
186 of 555
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.
187 Easy
187 of 555
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.
188 Easy
188 of 555
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.
189 Easy
189 of 555
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.
190 Medium
190 of 555
Application +4 | -1
The equations of sides of a triangle ABCA B C are AB:x+y−1=0,BC:7x−y−15=0,AC:x−y−1=0A B: x+y-1=0, B C: 7 x-y-15=0, A C: x-y-1=0 the equation of angular bisector of internal angle BB of the triangle is 3x3 x +y−k=0+y-k=0. The positive integer kk must be.
View answer and explanation
Correct option: A
5
Option A (5) is correct because solving for the angular bisector of the internal angle B leads to the equation 3x + y - 5 = 0, thus k = 5.
191 Hard
191 of 555
Application +4 | -1
The image of the point (−1,3,4)(-1,3,4) in the plane x−2y=0x-2 y=0 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated image point does not match any of the provided options.
192 Medium
192 of 555
Application +4 | -1
Let a variable line passing through the centre of the circle x2+y2−16x−4y=0x^2+y^2-16 x-4 y=0, meet the positive coordinate axes at the points AA and BB. Then the minimum value of OA+OBO A+O B, where OO is the origin, is equal to
View answer and explanation
Correct option: B
18
Option B (18) is correct because the minimum value of OA + OB is found by minimizing the function representing the sum of the intercepts.
193 Easy
193 of 555
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.
194 Easy
194 of 555
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.
195 Easy
195 of 555
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).
196 Easy
196 of 555
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.
197 Easy
197 of 555
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).
198 Easy
198 of 555
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.
199 Easy
199 of 555
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.
200 Easy
200 of 555
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.
201 Easy
201 of 555
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.
202 Medium
202 of 555
Analyzing +4 | -1
One possible condition for the three points (a,  b),  (b,  a)\left( {a,\;b} \right),\;\left( {b,\;a} \right) 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+ba = 1 + b) is correct because it is one of the conditions derived from setting the determinant of the matrix formed by the points to zero.
203 Easy
203 of 555
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.
204 Easy
204 of 555
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.
205 Easy
205 of 555
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.
206 Easy
206 of 555
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.
207 Easy
207 of 555
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.
208 Easy
208 of 555
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.
209 Easy
209 of 555
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.
210 Medium
210 of 555
Application +4 | -1
y=x+ry=x+r and y=−x+ry=-x+r where rr takes all decimal digits. Then the number of squares in xyx y plane formed by these lines with diagonals of 2 units length are
View answer and explanation
Correct option: C
64
The correct answer is D (a+b1−ab\frac{a+b}{1-a b}) because it is derived from the trigonometric identity for the tangent of a sum of angles.
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