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JEE • MATHEMATICS • 2D GEOMETRY SYSTEM
2d geometry system 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 31-60 of 187 questions
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31
Medium
31 of 187
Application
+4 | -1
Let be a fixed point and be the origin of coordinates. If is the mid-point of is the mid-point of is the mid-point of and so on. Then the coordinates of are
View answer and explanation
Correct option: A
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 -coordinates of the incentre of the triangle where the mid-points of the sides are and is
View answer and explanation
Correct option: C
Option C 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 such that the area of triangle formed by the points (5, 6), (3, 2) and 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
Option C () 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 and , then the area of the triangle is
View answer and explanation
Correct option: C
Option C () 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 are
View answer and explanation
Correct option: C
Option C () 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 5 sq. unit where and lies on then co-ordinate of can be
View answer and explanation
Correct option: A
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 and of a triangle with vertices at and are perpendicular to each other if
View answer and explanation
Correct option: D
Option D () is correct because it represents the condition where the slopes of the medians AD and BE are perpendicular, leading to the equation .
38
Medium
38 of 187
Application
+4 | -1
The locus of the mid-point of the distance between the axes of the variable line , where is constant, is
View answer and explanation
Correct option: B
Option B () 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 is the origin and are two points
View answer and explanation
Correct option: C
Option C () 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 and is
View answer and explanation
Correct option: D
Option D () is correct because the orthocentre of a right-angled triangle is the vertex at the right angle, which is the intersection of the lines and .
41
Easy
41 of 187
Application
+4 | -1
If the centroid and circumcentre of triangle are and respectively, then the orthocentre, is
View answer and explanation
Correct option: A
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 and . If its centroid is then the third vertex has the coordinates
View answer and explanation
Correct option: A
Option A () 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 and be collinear, then the possible values of are
View answer and explanation
Correct option: A
Option A () 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 is always the twice from , the locus of the point is
View answer and explanation
Correct option: C
Option C () 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 and is 4 square units then a value of 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.
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46
Easy
46 of 187
Analyzing
+4 | -1
If are the values of for which is divisible by then the triangle having vertices and 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 and then co-ordinates of circumcentre of this triangle are
View answer and explanation
Correct option: A
Option A () 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 and are collinear, then
View answer and explanation
Correct option: C
Option C () 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 and intersects the line at right angles, then the value of 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 is the orthocentre and is the circumcentre of a triangle, then coordinates of the centroid are ...... .
View answer and explanation
Correct option: C
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 .
51
Medium
51 of 187
Application
+4 | -1
If the vertices of a triangle are , and , then the coordinates of the circumcentre of the triangle are,
View answer and explanation
Correct option: C
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 to the line is
View answer and explanation
Correct option: B
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
and
is
and
is
View answer and explanation
Correct option: D
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 and is 70 units. For
View answer and explanation
Correct option: D
only one integral value of
Option D (only one integral value of ) 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 and are and . If a point moves so that , then the equation to the locus of is
View answer and explanation
Correct option: D
Option D () is correct because the equation represents a circle with radius 'a' centered at the origin, derived from the condition .
56
Easy
56 of 187
Application
+4 | -1
The area of the triangle formed by the points 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 and is equal to a constant. Then, the equation of its locus is
View answer and explanation
Correct option: B
Option B () 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 be an integer such that the triangle with vertices and has area 28 sq units. Then, the orthocentre of this triangle is at the point
View answer and explanation
Correct option: A
Option A 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 and are distinct and the points ,
are collinear, then
are collinear, then
View answer and explanation
Correct option: C
Option C () 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
View answer and explanation
Correct option: A
independent of
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'.
Revision notes and key takeaways
Review 2D GEOMETRY SYSTEM in your textbook or class notes, then use this set to check your understanding.
Showing 31-60 of 187 questions
