Values of $\alpha$ if $(\alpha, 2 \alpha)$ lies inside the $\Delta A B C$ if $A(0,2), B(2,0)$ and $C(4,4)$

$\alpha \in\left(\frac{1}{3}, \frac{2}{3}\right)$

$\alpha \in\left(\frac{2}{3}, 1\right)$

$\alpha \in\left(\frac{1}{3}, 1\right)$

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JEE • MATHEMATICS • 2D GEOMETRY SYSTEM

2d geometry system practice · Medium

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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 1-30 of 57 questions

1 Medium

1 of 57

Application +4 | -1

Values of

\alpha

(\alpha, 2 \alpha)

lies inside the

\Delta A B C

A(0,2), B(2,0)

and

C(4,4)

View answer and explanation

Correct option: C

\alpha \in\left(\frac{2}{3}, \frac{4}{3}\right)

Option C (

\alpha \in\left(\frac{{2}}{{3}}, \frac{{4}}{{3}}\right)

) is correct because the point

(\alpha, 2\alpha)

must satisfy the inequalities derived from the equations of the lines forming the triangle.

2 Medium

2 of 57

Application +4 | -1

The locus of the moving point

P

, such that

2 P A=3 P B

where

A

(0,0)

and

B

(4,-3)

, is

View answer and explanation

Correct option: D

5 x^{2}+5 y^{2}-72 x+54 y+225=0

Option D (

5 x^{2}+5 y^{2}-72 x+54 y+225=0

) is correct because it represents the locus of point P satisfying the given distance condition.

3 Medium

3 of 57

Application +4 | -1

The ratio in which the line segment joining the points

(3,-4)

and

(-5,6)

is divided by the

X

-axis, is

View answer and explanation

Correct option: A

2: 3

Option A (2:3) is correct because the X-axis divides the line segment in the ratio 2:3, which is found by setting the y-coordinate of the section formula to zero.

4 Medium

4 of 57

Application +4 | -1

Locus of the points which are at equal distance from

3 x+4 y-11=0

and

12 x+5 y+2=0

and which is near the origin is

View answer and explanation

Correct option: B

99 x+77 y-133=0

Option B (

99 x+77 y-133=0

) is correct because it represents the locus of points equidistant from the two given lines and near the origin.

5 Medium

5 of 57

Application +4 | -1

If the coordinates of a point be given by the equation

x=a(1-\cos \theta), y=a \sin \theta

, then the locus of the point will be

View answer and explanation

Correct option: B

A circle

Option B (A circle) is correct because eliminating

\theta

from the given parametric equations results in the equation

x^{2}+y^{2}-2 a x=0

, which represents a circle.

6 Medium

6 of 57

Understanding +4 | -1

A point moves in such a way that its distance from origin is always 4 . Then the locus of the point is

View answer and explanation

Correct option: B

x^{2}+y^{2}=16

Option B (

x^{2}+y^{2}=16

) is correct because the distance from the origin is 4, which translates to the equation of a circle centered at the origin with radius 4.

7 Medium

7 of 57

Application +4 | -1

Without changing the direction of coordinate axes, origin is transferred to

(h, k)

, so that the linear (one degree) terms in the equation

x^{2}+y^{2}-4 x+6 y-7=0

are eliminated. Then the point

(h, k)

View answer and explanation

Correct option: C

(2,-3)

Option C (

(2,-3)

) is correct because substituting

x=x'+h

and

y=y'+k

into the equation and setting the coefficients of

x'

and

y'

to zero eliminates the linear terms, leading to

h=2

and

k=-3

8 Medium

8 of 57

Application +4 | -1

If the coordinates of a point be given by the equations

x=b \sec \phi, \quad y=a \tan \phi

, then its locus is

View answer and explanation

Correct option: D

A hyperbola

Option D (A hyperbola) is correct because the given parametric equations,

x = b \sec \phi

and

y = a \tan \phi

, satisfy the equation of a hyperbola.

9 Medium

9 of 57

Application +4 | -1

In an equilateral triangle of side 

2\sqrt 3

cm, the circum radius is

View answer and explanation

Correct option: C

2 cm

Option C (2 cm) is correct because the circumradius R of an equilateral triangle is given by

R = \frac{a}{{2\sin A}}

, where a is the side length and A is the angle, which simplifies to 2 cm in this case.

10 Medium

10 of 57

Application +4 | -1

Let

A(a, b)

be a fixed point and

O

be the origin of coordinates. If

A_1

is the mid-point of

O A, A_2

is the mid-point of

A A_1, A_3

is the mid-point of

A A_2

and so on. Then the coordinates of

A_n

are

View answer and explanation

Correct option: A

\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.

11 Medium

11 of 57

Application +4 | -1

Let A be the set of all points

\left( {\alpha ,\beta } \right)

such that the area of triangle formed by the points (5, 6), (3, 2) and

\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

\frac{8}{{\sqrt 5 }}

Option C (

\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.

12 Medium

12 of 57

Application +4 | -1

If the vertices of a triangle be

\left(a m_{1}^{2}, 2 a m_{1}\right),\left(a m_{2}^{2}, 2 a m_{2}\right)

and

\left(a m_{3}^{2}, 2 a m_{3}\right)

, then the area of the triangle is

View answer and explanation

Correct option: C

a^{2}\left(m_{2}-m_{3}\right)\left(m_{3}-m_{1}\right)\left(m_{1}-m_{2}\right)

Option C (

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.

13 Medium

13 of 57

Application +4 | -1

The polar coordinates of a point having cartesian coordinates

(-1,-1)

are

View answer and explanation

Correct option: C

\left(\sqrt{2}, \frac{5 \pi}{4}\right)

Option C (

\left(\sqrt{{2}}, \frac{{5 \pi}}{{4}}\right)

) is correct because it represents the polar coordinates corresponding to the given Cartesian coordinates.

14 Medium

14 of 57

Application +4 | -1

The medians

A D

and

B E

of a triangle with vertices at

A(0, b), B(0,0)

and

C(a, 0)

are perpendicular to each other if

View answer and explanation

Correct option: D

\mathrm{a}= \pm \sqrt{2} \mathrm{~b}

Option D (

\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

a^2 = 2b^2

15 Medium

15 of 57

Application +4 | -1

The locus of the mid-point of the distance between the axes of the variable line

x \cos \alpha+y \sin \alpha=p

, where

p

is constant, is

View answer and explanation

Correct option: B

\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{p^{2}}

Option B (

\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.

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16 Medium

16 of 57

Application +4 | -1

H\left(\frac{13}{2}, \frac{11}{2}\right)

is the orthocentre and

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

\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

\left(\frac{13+2 \sqrt{7}}{6}, \frac{11+2 \sqrt{5}}{6}\right)

17 Medium

17 of 57

Application +4 | -1

If the vertices of a triangle

A B C

are

A(-4,-1)

B(1,2)

and

C(4,-3)

, then the coordinates of the circumcentre of the triangle are,

View answer and explanation

Correct option: C

(0,-2)

Option C ((0,-2)) is correct because the triangle is right-angled at B, making the midpoint of AC the circumcenter.

18 Medium

18 of 57

Application +4 | -1

The foot of the coordinates drawn from

(2,4)

to the line

x+y=1

View answer and explanation

Correct option: B

\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.

19 Medium

19 of 57

Application +4 | -1

Area of the parallelogram formed by the lines

a_1 x+b_1 y+c_1=0, a_1 x+b_1 y+d_1=0

and

a_2 x+b_2 y+c_2=0, a_2 x+b_2 y+d_2=0

View answer and explanation

Correct option: D

\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.

20 Medium

20 of 57

Analyzing +4 | -1

t_1, t_2

and

t_3

are distinct and the points

\left(t_1, 2 a t_1+a t_1^3\right),\left(t_2, 2 a t_2+a t_2^3\right)

\left(t_3, 2 a t_3+a t_3^3\right)

are collinear, then

View answer and explanation

Correct option: C

t_1+t_2+t_3=0

Option C (

t_1+t_2+t_3=0

) is correct because the collinearity condition of the three points leads to this relationship between the parameters.

21 Medium

21 of 57

Application +4 | -1

If the locus of the point, whose distances from the point

(2,1)

and

(1,3)

are in the ratio

5: 4

, is

a x^2+b y^2+c x y+d x+e y+170=0

, then the value of

a^2+2 b+3 c+4 d+e

is equal to:

View answer and explanation

Correct option: C

Option C (37) is correct because it is the result of substituting the coefficients of the locus equation into the given expression after correctly deriving the locus equation.

22 Medium

22 of 57

Application +4 | -1

If the vertices

A, B

and

C

of a triangle are

(2,1),(5,2)

and

(3,4)

, respectively. Then, the circumcentre is

View answer and explanation

Correct option: C

\left(\frac{13}{4}, \frac{9}{4}\right)

Option C ((13/4, 9/4)) is correct because the circumcenter is equidistant from all vertices, and solving the equations PA²=PB² and PB²=PC² gives the correct coordinates.

23 Medium

23 of 57

Application +4 | -1

In a parallelogram

A B C D,

A(1,2),

B(3,4),

C(2,5)

then length of the side

A D

is equal to

View answer and explanation

Correct option: B

\sqrt{2}

Option B (

\sqrt{2}

) is correct because finding the coordinates of point D using the parallelogram property and then calculating the distance AD yields

\sqrt{2}

24 Medium

24 of 57

Application +4 | -1

The area of the triangle formed by the lines

y=m_{1} x+c_{1}

y=m_{2} x+c_{2}

and

x=0

View answer and explanation

Correct option: C

\frac{1}{2} \frac{\left(c_{1}-c_{2}\right)^{2}}{\left(m_{1}-m_{2}\right)}

Option C (

\frac{1}{2} \frac{\left(c_{1}-c_{2}\right)^{2}}{\left(m_{1}-m_{2}\right)}

) is correct because it represents the area of the triangle formed by the given lines, calculated by finding the vertices and using the area formula.

25 Medium

25 of 57

Application +4 | -1

The area of the parallelogram formed by the lines

4 y-3 x=1,

4 y-3 x-3=0,

3 y-4 x+1=0,

3 y-4 x+2=0

View answer and explanation

Correct option: B

02-Jul

Option B (02-Jul) is correct because the area of the parallelogram is calculated to be 2/7.

26 Medium

26 of 57

Application +4 | -1

The circumcentre of the triangle formed by the lines

x+y=0, x y=0

and

x \,\,\,7=0

View answer and explanation

Correct option: D

(7,0)

Option D is correct. Since the lines

x+y=0

and

x-y=0

are perpendicular, the triangle is a right triangle, and the circumcenter is the midpoint of the hypotenuse, which is

(7,0)

27 Medium

27 of 57

Application +4 | -1

The distance between the points

P\left(2,60^{\circ}\right)

and

Q\left(5,270^{\circ}\right)

View answer and explanation

Correct option: A

\sqrt{29+10 \sqrt{3}}

Option A (

\sqrt{{29+10 \sqrt{{3}}}}

) is correct because it represents the distance between the points in polar coordinates.

28 Medium

28 of 57

Application +4 | -1

Incentre of triangle, whose vertices are

A(-36,7), B(20,7)

and

C(0,-8)

, is

View answer and explanation

Correct option: C

(-1,0)

Option A (-1,0) is correct because the incenter is calculated using the formula involving side lengths and vertex coordinates.

29 Medium

29 of 57

Application +4 | -1

The co-ordinates of the foot of perpendicular from the point

(2,3)

on the line

x+y-11=0

are

View answer and explanation

Correct option: B

(5,6)

Option D ((6,5)) is correct because it represents the coordinates of the foot of the perpendicular from the given point to the line.

30 Medium

30 of 57

Understanding +4 | -1

If the point

P\left(x_1+t\left(x_2 x_1\right), y_1+t\left(y_2 y_1\right)\right)

divides the line segment joining

A\left(x_1, y_1\right)

and

B\left(x_2, y_2\right)

internally, then

\ldots .

View answer and explanation

Correct option: B

0 < t < 1

Option B (

0 < t < 1

) is correct because for P to divide AB internally, t must be between 0 and 1.

Revision notes and key takeaways

Review 2D GEOMETRY SYSTEM in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 57 single-correct-answer MCQs on 2D GEOMETRY SYSTEM.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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