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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 151-180 of 187 questions
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151
Easy
151 of 187
Application
+4 | -1
The area of a triangle, two of whose vertices are and is 5 . The coordinates of the third vertex can not be
View answer and explanation
Correct option: D
Option D (2,9) is correct because it does not satisfy the area condition of the triangle being 5.
152
Medium
152 of 187
Application
+4 | -1
The figure formed by four points (1,-2), (3, 6), (5,10) and is a
View answer and explanation
Correct option: D
parallelogram
Option D (parallelogram) is correct because the opposite sides are equal, and the diagonals are not equal, which are the properties of a parallelogram.
153
Easy
153 of 187
Application
+4 | -1
A point moves so that the sum of squares of its distances from the points and is 14 . Let be the locus of , which intersects the -axis at the points and the -axis at the points . Then the area of the quadrilateral is equal to
View answer and explanation
Correct option: B
Option B () is correct because it represents the area of the quadrilateral formed by the intersection points of the locus with the x and y axes.
154
Medium
154 of 187
Application
+4 | -1
If is is and lies on , then the locus of the centroid of the is a straight line parallel to the straight line is
View answer and explanation
Correct option: C
Option C () is correct because the locus of the centroid is parallel to the given line.
155
Easy
155 of 187
Application
+4 | -1
A point moves so that its distance from the point is always three times its distance from the point . The locus of the point is
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the given condition leads to an equation of the form , which is the general equation of a circle.
156
Easy
156 of 187
Application
+4 | -1
If the circumcentre of a triangle lies at the origin and the centroid is the middle point of the line joining the points and , then the orthocentre satisfies the equation
View answer and explanation
Correct option: D
Option D is correct because it represents the equation satisfied by the orthocenter, derived using the centroid and circumcenter relationship.
157
Easy
157 of 187
Application
+4 | -1
A straight line passes through a fixed point . The locus of the foot of perpendicular on it drawn from the origin is
View answer and explanation
Correct option: A
Option A () is correct because it represents the locus of the foot of the perpendicular from the origin to a line passing through (h,k).
158
Easy
158 of 187
Application
+4 | -1
If the coordinates of the points , be and respectively, then the area of the triangle is
View answer and explanation
Correct option: D
7
The correct answer is D (7) because it is the area of the triangle calculated using the determinant formula with the given coordinates.
159
Medium
159 of 187
Application
+4 | -1
The equation of the locus of foot of perpendiculars drawn from the origin to the line passing through a fixed point , is
View answer and explanation
Correct option: A
Option A () is correct because it represents the locus of the foot of the perpendiculars drawn from the origin to a line passing through a fixed point (a, b).
160
Easy
160 of 187
Application
+4 | -1
If the point fall between the lines , then
View answer and explanation
Correct option: C
Option C () is correct because substituting (a, a) into the inequality derived from gives , which simplifies to .
161
Easy
161 of 187
Application
+4 | -1
The orthocentre of the triangle formed by the lines and is
View answer and explanation
Correct option: A
Option A ((0,0)) is correct because the triangle formed is right-angled at the origin, making the origin the orthocenter.
162
Easy
162 of 187
Application
+4 | -1
If the points and are the vertices of triangle, then the triangle is
View answer and explanation
Correct option: C
equilateral
Option C (equilateral) is correct because all three sides of the triangle have equal length.
163
Easy
163 of 187
Application
+4 | -1
The equation of the locus of all points equidistant from the point and the -axis, is
View answer and explanation
Correct option: B
Option B () is the correct equation of the locus of points equidistant from (4,2) and the x-axis.
164
Medium
164 of 187
Application
+4 | -1
A point moves such that its distance from the point is half that of its distance from the line . The locus of this point is
View answer and explanation
Correct option: A
Option A () is correct because it represents the locus of the point derived from the given distance condition.
165
Easy
165 of 187
Application
+4 | -1
The area of a parallelogram formed by the lines , is
View answer and explanation
Correct option: B
Option D () is correct because it represents the area of a parallelogram formed by two pairs of parallel lines.
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166
Medium
166 of 187
Application
+4 | -1
The distance of any point from the point is given by . If and then locus of is
View answer and explanation
Correct option: C
a square
Option C (a square) is correct because the locus of P forms a square with sides parallel to the axes, centered at Q.
167
Easy
167 of 187
Understanding
+4 | -1
The locus of a point which moves so that it is always equidistant from the point and is
View answer and explanation
Correct option: B
Perpendicular bisector of the line segment
Option B (Perpendicular bisector of the line segment ) is correct because the locus of points equidistant from two points is the perpendicular bisector of the line segment joining them.
168
Easy
168 of 187
Application
+4 | -1
A line meets -axis at and -axis at such that incentre of the triangle is . Equation of is
View answer and explanation
Correct option: B
Option B () is correct because the perpendicular distance from the incenter (1,1) to the line equals the inradius, which is 1.
169
Medium
169 of 187
Application
+4 | -1
The coordinates of incentre of , where and are
View answer and explanation
Correct option: B
Option B () is correct because it represents the coordinates of the incenter of the triangle calculated using the given vertices.
170
Medium
170 of 187
Application
+4 | -1
Let and be the vertices of a triangle such that algebraic sum of perpendicular distance from and to the variable line is always '0'. If , then find the value of .
View answer and explanation
Correct option: A
15
Option A (15) is correct because the centroid of the triangle lies on the line, and using the given condition 3a + 2b + c = 0, we can find the value of the sum.
171
Easy
171 of 187
Application
+4 | -1
If the points be collinear and , then
View answer and explanation
Correct option: A
Option A () is correct because it satisfies both the collinearity condition and the distance condition AC = 10.
172
Easy
172 of 187
Application
+4 | -1
The extremities of a diagonal of a parallelogram are the points and . If third vertex is , 1) then the coordinates of the fourth vertex are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the fourth vertex calculated using the parallelogram property doesn't match any of the given options.
173
Easy
173 of 187
Application
+4 | -1
The orthocentre of the triangle formed by the lines and is
View answer and explanation
Correct option: C
Option B () is correct because this is the equation of the line that passes through (3,-2) and is inclined at 60 degrees to the given line.
174
Easy
174 of 187
Application
+4 | -1
If the points are the vertices of a triangle, then this triangle is
View answer and explanation
Correct option: A
Equilateral
Option A (Equilateral) is correct because all three sides of the triangle have the same length, making it equilateral.
175
Easy
175 of 187
Application
+4 | -1
The centroid of a triangle, whose vertices are and , is
View answer and explanation
Correct option: B
Option A () is correct because using the centroid formula, we can solve for the coordinates of the third vertex.
176
Easy
176 of 187
Application
+4 | -1
If and are two fixed points, then the locus of the point on which the line subtends the right angle, is
View answer and explanation
Correct option: D
Option D () is correct because it represents the locus of a point where the line segment AB subtends a right angle.
177
Easy
177 of 187
Application
+4 | -1
If vertices of are , then find its orthocenter
View answer and explanation
Correct option: A
The orthocenter is (8, -2) because the triangle is right-angled, and in a right-angled triangle, the orthocenter is the vertex where the right angle is formed.
178
Easy
178 of 187
Analyzing
+4 | -1
If is the area of the triangle formed by the centroid and two vertices of a triangle, is the area of the triangle formed by the mid-points of the sides of the same triangle, then
View answer and explanation
Correct option: C
Option C () is correct because the ratio of the area of the triangle formed by the centroid and two vertices to the area of the triangle formed by the mid-points of the sides is 4:3.
179
Easy
179 of 187
Application
+4 | -1
If a vertex of a triangle is and the mid-points of two sides through this vertex are and , then the centroid of the triangle, is
View answer and explanation
Correct option: B
Option B (1, 7/3) is correct because by finding the coordinates of the other two vertices using the midpoint formula, we can calculate the centroid.
180
Easy
180 of 187
Application
+4 | -1
The area of the quadrilateral whose vertices are and , is
View answer and explanation
Correct option: B
sq. units
Option B ( sq. units) is correct because it represents the area of the quadrilateral calculated by dividing it into triangles and summing their areas.
Revision notes and key takeaways
Review 2D GEOMETRY SYSTEM in your textbook or class notes, then use this set to check your understanding.
Showing 151-180 of 187 questions
