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JEE • MATHEMATICS • 2D GEOMETRY SYSTEM

2d geometry system practice · Mixed difficulty

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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 31-60 of 187 questions

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31 Medium
31 of 187
Application +4 | -1
Let A(a,b)A(a, b) be a fixed point and OO be the origin of coordinates. If A1A_1 is the mid-point of OA,A2O A, A_2 is the mid-point of AA1,A3A A_1, A_3 is the mid-point of AA2A A_2 and so on. Then the coordinates of AnA_n are
View answer and explanation
Correct option: A
(a(1−2−n),b(1−2−n))\left(a\left(1-2^{-n}\right), b\left(1-2^{-n}\right)\right)
Option A (a(1-2⁻ⁿ), b(1-2⁻ⁿ)) is correct because the coordinates of Aₙ converge to (a, b) as n approaches infinity, and the given expression reflects this convergence.
32 Easy
32 of 187
Application +4 | -1
The xx-coordinates of the incentre of the triangle where the mid-points of the sides are (0,1),(1,1)(0,1),(1,1) and (1,0)(1,0) is
View answer and explanation
Correct option: C
2−22-\sqrt{2}
Option C 2−22-\sqrt{2} is correct because it is the x-coordinate of the incenter of the triangle formed by the given midpoints.
33 Medium
33 of 187
Application +4 | -1
Let A be the set of all points (α,β)\left( {\alpha ,\beta } \right) such that the area of triangle formed by the points (5, 6), (3, 2) and (α,β)\left( {\alpha ,\beta } \right) is 12 sq units. Then, the least possible length of a line segment joining the origin to a point in A, is
View answer and explanation
Correct option: C
85\frac{8}{{\sqrt 5 }}
Option C (85\frac{8}{{\sqrt 5 }}) is correct because it represents the minimum distance from the origin to a point on the line formed by the given area condition.
34 Medium
34 of 187
Application +4 | -1
If the vertices of a triangle be (am12,2am1),(am22,2am2)\left(a m_{1}^{2}, 2 a m_{1}\right),\left(a m_{2}^{2}, 2 a m_{2}\right) and (am32,2am3)\left(a m_{3}^{2}, 2 a m_{3}\right), then the area of the triangle is
View answer and explanation
Correct option: C
a2(m2−m3)(m3−m1)(m1−m2)a^{2}\left(m_{2}-m_{3}\right)\left(m_{3}-m_{1}\right)\left(m_{1}-m_{2}\right)
Option C (a2(m2−m3)(m3−m1)(m1−m2)a^{2}\left(m_{2}-m_{3}\right)\left(m_{3}-m_{1}\right)\left(m_{1}-m_{2}\right)) is correct because it is the result of calculating the area of the triangle using the determinant formula.
35 Medium
35 of 187
Application +4 | -1
The polar coordinates of a point having cartesian coordinates (−1,−1)(-1,-1) are
View answer and explanation
Correct option: C
(2,5π4)\left(\sqrt{2}, \frac{5 \pi}{4}\right)
Option C ((2,5π4)\left(\sqrt{{2}}, \frac{{5 \pi}}{{4}}\right)) is correct because it represents the polar coordinates corresponding to the given Cartesian coordinates.
36 Easy
36 of 187
Application +4 | -1
If area of ΔABC\Delta A B C 5 sq. unit where A(1,1),B(2,2)A(1,1), B(2,2) and CC lies on y=2xy=2 x then co-ordinate of CC can be
View answer and explanation
Correct option: A
(−10,−20)(-10,-20)
Option A, (-10, -20), is correct because it is one of the possible coordinates for point C on the line y=2x such that the area of triangle ABC is 5 sq. units.
37 Medium
37 of 187
Application +4 | -1
The medians ADA D and BEB E of a triangle with vertices at A(0,b),B(0,0)A(0, b), B(0,0) and C(a,0)C(a, 0) are perpendicular to each other if
View answer and explanation
Correct option: D
a=±2 b\mathrm{a}= \pm \sqrt{2} \mathrm{~b}
Option D (a=±2 b\mathrm{a}= \pm \sqrt{2} \mathrm{~b}) is correct because it represents the condition where the slopes of the medians AD and BE are perpendicular, leading to the equation a2=2b2a^2 = 2b^2.
38 Medium
38 of 187
Application +4 | -1
The locus of the mid-point of the distance between the axes of the variable line xcos⁡α+ysin⁡α=px \cos \alpha+y \sin \alpha=p, where pp is constant, is
View answer and explanation
Correct option: B
1x2+1y2=4p2\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{p^{2}}
Option B (1x2+1y2=4p2\frac{{1}}{{x^{2}}}+\frac{{1}}{{y^{2}}}=\frac{{4}}{{p^{2}}}) is correct because it represents the locus of the midpoint of the segment between the axes of the given line.
39 Easy
39 of 187
Application +4 | -1
If OO is the origin and P(x1,y1),Q(x2,y2)P\left(x_1, y_1\right), Q\left(x_2, y_2\right) are two points OP×OQsin⁡∠POQ=O P \times O Q \sin \angle P O Q=
View answer and explanation
Correct option: C
∣x1y2−x2y1∣\left|x_1 y_2-x_2 y_1\right|
Option C (∣x1y2−x2y1∣\left|x_1 y_2-x_2 y_1\right|) is correct because it represents twice the area of the triangle formed by the origin and the two points P and Q.
40 Easy
40 of 187
Application +4 | -1
Coordinates of the orthocentre of the triangle whose sides are x=3,y=4x=3, y=4 and 3x+4y=63 x+4 y=6 is
View answer and explanation
Correct option: D
(3,4)(3,4)
Option D ((3,4)(3,4)) is correct because the orthocentre of a right-angled triangle is the vertex at the right angle, which is the intersection of the lines x=3x=3 and y=4y=4.
41 Easy
41 of 187
Application +4 | -1
If the centroid and circumcentre of aa triangle are (3,3)(3,3) and (6,2)(6,2) respectively, then the orthocentre, is
View answer and explanation
Correct option: A
(−3,5)(-3,5)
Option A (-3, 5) is correct because the centroid divides the line joining the circumcenter and orthocenter in the ratio 1:2.
42 Easy
42 of 187
Application +4 | -1
Two vertices of a triangle are (5,4)(5,4) and (−2,4)(-2,4). If its centroid is (5,6)(5,6) then the third vertex has the coordinates
View answer and explanation
Correct option: A
(12,10)(12,10)
Option A ((12,10)(12,10)) is correct because using the centroid formula, we can solve for the coordinates of the third vertex.
43 Easy
43 of 187
Application +4 | -1
If the points (k,2−2k),(1−k,2k)(k, 2-2 k),(1-k, 2 k) and (−k−4,6−2k)(-k-4,6-2 k) be collinear, then the possible values of kk are
View answer and explanation
Correct option: A
12,−1\frac{1}{2},-1
Option A (12,−1\frac{1}{2},-1) is correct because these values of k make the determinant of the matrix formed by the points equal to zero, indicating collinearity.
44 Easy
44 of 187
Application +4 | -1
A point moves in such a way that its distance from (1,−2)(1,-2) is always the twice from (−3,5)(-3,5), the locus of the point is
View answer and explanation
Correct option: C
3(x2+y2)+26x−44y+131=03\left(x^{2}+y^{2}\right)+26 x-44 y+131=0
Option C (3(x2+y2)+26x−44y+131=03\left(x^{2}+y^{2}\right)+26 x-44 y+131=0) is correct because it represents the locus of the point satisfying the given distance condition.
45 Easy
45 of 187
Application +4 | -1
If the area of the triangle with vertices (x,0),(1,1)(x, 0),(1,1) and (0,2)(0,2) is 4 square units then a value of xx is
View answer and explanation
Correct option: A
-2
Option A (-2) is correct because the value of x is found by setting up the determinant equation for the area of a triangle and solving for x.
46 Easy
46 of 187
Analyzing +4 | -1
If α1,α2,α3,β1,β2,β3\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3 are the values of nn for which ∑r=0n−1x2r\sum\limits_{r = 0}^{n - 1} {{x^{2r}}} is divisible by ∑r=0n−1xr,\sum\limits_{r = 0}^{n - 1} {{x^r}} , then the triangle having vertices (α1,β1),(α2,β2)\left(\alpha_1, \beta_1\right),\left(\alpha_2, \beta_2\right) and (α3,β3)\left(\alpha_3, \beta_3\right) cannot be
View answer and explanation
Correct option: D
an equilateral triangle
Option D (an equilateral triangle) is correct because the vertices derived from the given condition do not form an equilateral triangle.
47 Hard
47 of 187
Application +4 | -1
If equation of three sides of a triangle are x=2,y+1=0x=2, y+1=0 and x+2y=4x+2 y=4 then co-ordinates of circumcentre of this triangle are
View answer and explanation
Correct option: A
(4,0)(4,0)
Option A ((4,0)(4,0)) is correct because it is the circumcenter found by solving the equations formed by the perpendicular bisectors of the sides of the triangle.
48 Easy
48 of 187
Application +4 | -1
If the points (a,0),(0,b)(a, 0),(0, b) and (1,1)(1,1) are collinear, then
View answer and explanation
Correct option: C
1a+1b=1\frac{1}{a}+\frac{1}{b}=1
Option C (1a+1b=1\frac{1}{a}+\frac{1}{b}=1) is correct because it represents the condition derived from setting the determinant of the collinear points to zero.
49 Easy
49 of 187
Application +4 | -1
If the line through the points (h,7)(h, 7) and (2,3)(2,3) intersects the line 3x−4y−5=03 x-4 y-5=0 at right angles, then the value of hh is
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because the product of the slopes of the two perpendicular lines is -1, which allows us to solve for h.
50 Medium
50 of 187
Application +4 | -1
If H(132,112)H\left(\frac{13}{2}, \frac{11}{2}\right) is the orthocentre and P(72,52)P\left(\frac{\sqrt{7}}{2}, \frac{\sqrt{5}}{2}\right) is the circumcentre of a triangle, then coordinates of the centroid are ...... .
View answer and explanation
Correct option: C
(13+276,11+256)\left(\frac{13+2 \sqrt{7}}{6}, \frac{11+2 \sqrt{5}}{6}\right)
Option C is correct. The centroid divides the line segment joining the circumcentre and orthocentre in the ratio 1:2. Using the section formula, the coordinates of the centroid are (13+276,11+256)\left(\frac{13+2 \sqrt{7}}{6}, \frac{11+2 \sqrt{5}}{6}\right).
51 Medium
51 of 187
Application +4 | -1
If the vertices of a triangle ABCA B C are A(−4,−1)A(-4,-1), B(1,2)B(1,2) and C(4,−3)C(4,-3), then the coordinates of the circumcentre of the triangle are,
View answer and explanation
Correct option: C
(0,−2)(0,-2)
Option C ((0,-2)) is correct because the triangle is right-angled at B, making the midpoint of AC the circumcenter.
52 Medium
52 of 187
Application +4 | -1
The foot of the coordinates drawn from (2,4)(2,4) to the line x+y=1x+y=1 is
View answer and explanation
Correct option: B
(−12,32)\left(-\frac{1}{2}, \frac{3}{2}\right)
Option B ((-1/2), (3/2)) is correct because it represents the coordinates of the foot of the perpendicular from the given point to the line.
53 Medium
53 of 187
Application +4 | -1
Area of the parallelogram formed by the lines
a1x+b1y+c1=0,a1x+b1y+d1=0a_1 x+b_1 y+c_1=0, a_1 x+b_1 y+d_1=0 and
a2x+b2y+c2=0,a2x+b2y+d2=0a_2 x+b_2 y+c_2=0, a_2 x+b_2 y+d_2=0 is
View answer and explanation
Correct option: D
(d1−c1)(d2−c2)a1b2−a2b1\frac{\left(d_1-c_1\right)\left(d_2-c_2\right)}{a_1 b_2-a_2 b_1}
Option B ((3/5), (4/5)) is correct because it represents the coordinates of the foot of the perpendicular from the origin to the line.
54 Easy
54 of 187
Application +4 | -1
The area of the triangle formed by the points (k,2−2k),(−k+1,2k)(k, 2-2 k),(-k+1,2 k) and (−4−k,6−2k)(-4-k, 6-2 k) is 70 units. For
View answer and explanation
Correct option: D
only one integral value of kk
Option D (only one integral value of kk) is correct because the area equation simplifies to a quadratic, yielding only one integer solution for k.
55 Easy
55 of 187
Application +4 | -1
The coordinates of the point AA and BB are (ak,0)(a k, 0) and (ak,0),(k=±1)\left(\frac{a}{k}, 0\right),(k= \pm 1). If a point PP moves so that PA=kPBP A=k P B, then the equation to the locus of PP is
View answer and explanation
Correct option: D
x2+y2−a2=0x^{2}+y^{2}-a^{2}=0
Option D (x2+y2−a2=0x^2 + y^2 - a^2 = 0) is correct because the equation represents a circle with radius 'a' centered at the origin, derived from the condition PA=kPBPA = kPB.
56 Easy
56 of 187
Application +4 | -1
The area of the triangle formed by the points (a,b+c),(b,c+a),(c,a+b)(a, b+c),(b, c+a),(c, a+b) is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the points are collinear, resulting in a triangle with zero area, which can be shown by determinant properties.
57 Easy
57 of 187
Application +4 | -1
The sum of squares of the distances of a moving point from two fixed points (a,0)(a, 0) and (−a,0)(-a, 0) is equal to a constant. Then, the equation of its locus is
View answer and explanation
Correct option: B
x2+y2=C−a2x^2+y^2=C-a^2
Option B (x2+y2=C−a2x^2+y^2=C-a^2) is correct because it represents the locus of a point whose sum of squared distances from two fixed points is constant.
58 Easy
58 of 187
Analyzing +4 | -1
Let kk be an integer such that the triangle with vertices (k,−3k),(5,k)(k,-3 k),(5, k) and (−k,2)(-k, 2) has area 28 sq units. Then, the orthocentre of this triangle is at the point
View answer and explanation
Correct option: A
(2,12)\left(2, \frac{1}{2}\right)
Option A (2,12)\left(2, \frac{{1}}{{2}}\right) is correct because after finding the value of k and the vertices, the orthocenter is calculated using properties of circumcenter and centroid.
59 Medium
59 of 187
Analyzing +4 | -1
If t1,t2t_1, t_2 and t3t_3 are distinct and the points (t1,2at1+at13),(t2,2at2+at23)\left(t_1, 2 a t_1+a t_1^3\right),\left(t_2, 2 a t_2+a t_2^3\right),
(t3,2at3+at33)\left(t_3, 2 a t_3+a t_3^3\right) are collinear, then
View answer and explanation
Correct option: C
t1+t2+t3=0t_1+t_2+t_3=0
Option C (t1+t2+t3=0t_1+t_2+t_3=0) is correct because the collinearity condition of the three points leads to this relationship between the parameters.
60 Easy
60 of 187
Application +4 | -1
The area of the triangle with vertices
[(a+1)(a+2),(a+2)][(a+2)(a+3),(a+3)] and [(a+3)(a+4),(a+4)] is \begin{aligned} & {[(a+1)(a+2),(a+2)][(a+2)(a+3),(a+3)]} \\ & \text { and }[(a+3)(a+4),(a+4)] \text { is } \end{aligned}
View answer and explanation
Correct option: A
independent of aa
Option A (independent of a) is correct because the area of the triangle simplifies to a constant value (1) regardless of the value of 'a'.
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