Skip to main content
Back to 2D GEOMETRY SYSTEM

Your Progress

0 / 187 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • 2D GEOMETRY SYSTEM

2d geometry system practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 151-180 of 187 questions

Preparing practice and saved progress…

151 Easy
151 of 187
Application +4 | -1
The area of a triangle, two of whose vertices are (2,1)(2,1) and (3,−2)(3,-2) is 5 . The coordinates of the third vertex can not be
View answer and explanation
Correct option: D
(2,9)(2,9)
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 (3,2)(3,2) 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 PP moves so that the sum of squares of its distances from the points (1,2)(1,2) and (−2,1)(-2,1) is 14 . Let f(x,y)=0f(x, y)=0 be the locus of PP, which intersects the xx-axis at the points A,BA, B and the yy-axis at the points C,DC, D. Then the area of the quadrilateral ACBDA C B D is equal to
View answer and explanation
Correct option: B
3172\frac{3 \sqrt{17}}{2}
Option B (3172\frac{3 \sqrt{17}}{2}) 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 AA is (2,5),B(2,5), B is (4,−11)(4,-11) and CC lies on 9x+7y+4=09 x+7 y+4=0, then the locus of the centroid of the ΔABC\Delta A B C is a straight line parallel to the straight line is
View answer and explanation
Correct option: C
9x+7y+4=09 x+7 y+4=0
Option C (9x+7y+4=09 x+7 y+4=0) 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 (−1,0)(-1,0) is always three times its distance from the point (0,2)(0,2). 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 8(x2+y2)−36y−2x+35=08(x^{2}+y^{2})-36 y-2 x+35=0, 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 (a2+1,a2+1)\left(a^2+1, a^2+1\right) and (2a,−2a)(2 a,-2 a), then the orthocentre satisfies the equation
View answer and explanation
Correct option: D
(a−1)2x−(a+1)2y=0(a-1)^2 x-(a+1)^2 y=0
Option D (a−1)2x−(a+1)2y=0(a-1)^2 x-(a+1)^2 y=0 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 (h,k)(h, k). The locus of the foot of perpendicular on it drawn from the origin is
View answer and explanation
Correct option: A
x2+y2−hx−ky=0x^2+y^2-h x-k y=0
Option A (x2+y2−hx−ky=0x^2+y^2-hx-ky=0) 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 A,B,CA, B, C, be (4,4),(3,−2)(4,4),(3,-2) and (3,−16)(3,-16) respectively, then the area of the triangle ABCA B C 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 (a,b)(a, b), is
View answer and explanation
Correct option: A
x2+y2−ax−by=0x^2+y^2-a x-b y=0
Option A (x2+y2−ax−by=0x^2+y^2-a x-b y=0) 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 (a,a)(a, a) fall between the lines ∣x+y∣=2|x+y|=2, then
View answer and explanation
Correct option: C
∣a∣<1|a| < 1
Option C (∣a∣<1|a| < 1) is correct because substituting (a, a) into the inequality derived from ∣x+y∣<2|x+y|<2 gives ∣2a∣<2|2a|<2, which simplifies to ∣a∣<1|a|<1.
161 Easy
161 of 187
Application +4 | -1
The orthocentre of the triangle formed by the lines xy=0x y=0 and x+y=1x+y=1 is
View answer and explanation
Correct option: A
(0,0)(0,0)
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 A(1,1),B(−1,−1)A(1,1), B(-1,-1) and C(−3,3)C(-\sqrt{3}, \sqrt{3}) are the vertices of aa 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 (4,2)(4,2) and the xx-axis, is
View answer and explanation
Correct option: B
x2−8x−4y+20=0x^{2}-8 x-4 y+20=0
Option B (x2−8x−4y+20=0x^{2}-8 x-4 y+20=0) 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 (4,0)(4,0) is half that of its distance from the line x=16x=16. The locus of this point is
View answer and explanation
Correct option: A
3x2+4y2=1923 x^2+4 y^2=192
Option A (3x2+4y2=1923 x^2+4 y^2=192) 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 ax±by±c=0a x \pm b y \pm c=0, is
View answer and explanation
Correct option: B
2c2ab\frac{2 c^2}{a b}
Option D ((d1−c1)(d2−c2)a1b2−a2b1\frac{{(d_1-c_1)(d_2-c_2)}}{{a_1 b_2-a_2 b_1}}) is correct because it represents the area of a parallelogram formed by two pairs of parallel lines.
166 Medium
166 of 187
Application +4 | -1
The distance of any point P(x,y)P(x, y) from the point Q(x1,y1)Q\left(x_1, y_1\right) is given by d(P,Q)=Max{∣x−x1∣,∣y−y1∣}d(P, Q)=M a x\{| x-x_1|,| y-y_1 \mid\}. If Q(1,2)Q(1,2) and d(P,Q)=3d(P, Q)=3 then locus of PP 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 A(a,0)A(a, 0) and B(−a,0)B(-a, 0) is
View answer and explanation
Correct option: B
Perpendicular bisector of the line segment ABA B
Option B (Perpendicular bisector of the line segment ABA B) 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 xx-axis at AA and yy-axis at BB such that incentre of the triangle OABO A B is (1,1)(1,1). Equation of ABA B is
View answer and explanation
Correct option: B
3x+4y=123 x+4 y=12
Option B (3x+4y=123x + 4y = 12) 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 △ABC\triangle A B C, where A(−3,−2),B(5,0)A(-3,-2), B(5,0) and C(−13,133)C\left(-\frac{1}{3}, \frac{13}{3}\right) are
View answer and explanation
Correct option: B
(12,1)\left(\frac{1}{2}, 1\right)
Option B ((12,1)\left(\frac{{1}}{{2}}, 1\right)) 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 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) be the vertices of a triangle such that algebraic sum of perpendicular distance from A,BA, B and CC to the variable line ax+by+c=0a x+b y+c=0 is always &#39;0&#39;. If 3a+2b+c=03 a+2 b+c=0, then find the value of ∑i=13(xi+yi)\sum\limits_{{i=1}}^3\left(x_i+y_i\right).
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 A(3,4),B(7,7),C(a,b)A(3,4), B(7,7), C(a, b) be collinear and AC=10A C=10, then (a,b)=(a, b)=
View answer and explanation
Correct option: A
(11,10)(11,10)
Option A ((11,10)(11,10)) 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 (3,−4)(3,-4) and (−6,5)(-6,5). If third vertex is (−2(-2, 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 xy=0x y=0 and 2x+3y−5=02 x+3 y-5=0 is
View answer and explanation
Correct option: C
(0,0)(0,0)
Option B (y−3x+2+33=0y-\sqrt{3} x+2+3 \sqrt{3}=0) 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 (1,1)(−1,−1)(−3,3)(1,1)(-1,-1)(-\sqrt{3}, \sqrt{3}) 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 (2,1),(5,2)(2,1),(5,2) and (3,4)(3,4), is
View answer and explanation
Correct option: B
(103,73)\left(\frac{10}{3}, \frac{7}{3}\right)
Option A ((4,8)(4,8)) 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 A(−a,0)A(-a, 0) and B(a,0)B(a, 0) are two fixed points, then the locus of the point on which the line ABA B subtends the right angle, is
View answer and explanation
Correct option: D
x2+y2=a2x^{2}+y^{2}=a^{2}
Option D (x2+y2=a2x^{2}+y^{2}=a^{2}) 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 Δ\Delta are (8,−2),(2,−2) & (8,6)(8,-2),(2,-2)\, \&\,(8,6), then find its orthocenter
View answer and explanation
Correct option: A
(8,−2)(8,-2)
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 Δ1\Delta_1 is the area of the triangle formed by the centroid and two vertices of a triangle, Δ2\Delta_2 is the area of the triangle formed by the mid-points of the sides of the same triangle, then Δ1:Δ2=\Delta_1: \Delta_2=
View answer and explanation
Correct option: C
4:34: 3
Option C (4:34: 3) 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 (1,1)(1,1) and the mid-points of two sides through this vertex are (−1,2)(-1,2) and (3,2)(3,2), then the centroid of the triangle, is
View answer and explanation
Correct option: B
(1,73)\left(1, \frac{7}{3}\right)
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 (1,2),(6,2),(5,3)(1,2),(6,2),(5,3) and (3,4)(3,4), is
View answer and explanation
Correct option: B
112\frac{11}{2} sq. units
Option B (112\frac{11}{2} 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

Your practice

0 / 187 answered · 187 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.