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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 121-150 of 187 questions

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121 Medium

121 of 187

Application +4 | -1

Two points

A

and

B

have coordinates

(1,0)

and

(-1,0)

respectively and

Q

is a point which satisfies the relation

A Q-B Q= \pm 1

. The locus of

Q

View answer and explanation

Correct option: B

12 x^{2}-4 y^{2}=3

Option B (

12 x^{2}-4 y^{2}=3

) is correct because the given condition

AQ - BQ = \pm 1

defines a hyperbola, and after simplification, the equation becomes

12x^2 - 4y^2 = 3

122 Easy

122 of 187

Application +4 | -1

Area of the triangle formed by the lines

y-x=0

x+y=0

and

y=k

in square units is

View answer and explanation

Correct option: B

k^2

Option D (

k^2 / 2

) is correct because the area of the triangle formed by the given lines is calculated using the coordinates of the vertices.

123 Easy

123 of 187

Application +4 | -1

Area of pentagon formed by

(1,2),(2,-1),(-2,-1),(2,1)

and

(-1,2)

View answer and explanation

Correct option: D

Option D, 10, is the correct area of the pentagon, calculated using the Shoelace formula.

124 Medium

124 of 187

Application +4 | -1

Let two points be

A\left( {1, - 1} \right)

and

B\left( {0,2} \right).

If a point

P\left( {x',y'} \right)

be such that the area of

\Delta PAB = 5\,\,\,sq.\,\,units

and it lies on the line,

3x + y - 4\lambda = 0,

then a value of

\lambda

View answer and explanation

Correct option: C

Option C (3) is correct because it is the value of

\lambda

that satisfies the area condition and the line equation.

125 Medium

125 of 187

Application +4 | -1

The points

A(-4,-1), B(-2,-4), C(4,0)

and

D(2,3)

are the vertices of

View answer and explanation

Correct option: B

Rectangle

Option B (Rectangle) is correct. The midpoints of the diagonals coincide, and the product of the slopes of adjacent sides is -1, indicating perpendicularity, thus forming a rectangle.

126 Easy

126 of 187

Application +4 | -1

If the coordinates of the mid-points of side

A B

and

A C

\triangle A B C

are

D(3,5)

and

E(-3,-3)

respectively, the

B C=

View answer and explanation

Correct option: C

Option C (20) is correct because the length of BC is twice the length of the line segment joining the midpoints of AB and AC.

127 Medium

127 of 187

Application +4 | -1

The area of the triangle formed by the lines

7 x-2 y+10=0,7 x+2 y-10=0

and

y+2=0

View answer and explanation

Correct option: C

14 sq. unit

Option C (14 sq. unit) is correct because it is the area of the triangle formed by the intersection of the given lines, calculated using the determinant formula.

128 Easy

128 of 187

Application +4 | -1

A B C D

is a rectangle in the clockwise direction. The coordinates of

A

are

(1,3)

and of

C

are

(5,1)

, vertices

B

and

D

lie on the line

y=2 x+c

, then the coordinates of

D

are

View answer and explanation

Correct option: A

(2,0)

Option A (2,0) is correct because the midpoint of AC lies on the line y=2x+c, and AD is perpendicular to CD. Since D lies below A, its coordinates are (2,0).

129 Easy

129 of 187

Application +4 | -1

(x+3,4-y)=(1,7)

, then the value of

4+y

View answer and explanation

Correct option: A

Option A (1) is correct because solving for y gives y = -3, and thus 4 + y = 1.

130 Medium

130 of 187

Application +4 | -1

The orthocentre of the triangle with vertices

(-2,-6),(-2,4)

and

(1,3)

View answer and explanation

Correct option: C

(1,3)

Option C (

(1,3)

) is correct because it is the intersection point of the altitudes of the triangle.

131 Medium

131 of 187

Analyzing +4 | -1

The points

A(2 a, 4 a), B(2 a, 6 a)

and

C(2 a+\sqrt{3} a, 5 a)(

when

a>0)

are vertices of

View answer and explanation

Correct option: B

an equilateral triangle

Option B (an equilateral triangle) is correct because all sides of the triangle are equal in length, which is 2a.

132 Easy

132 of 187

Application +4 | -1

O

is the origin and

P\left(x_1, y_1\right), Q\left(x_2, y_2\right)

are two points, then

O P

OQ

\cos \angle P O Q

View answer and explanation

Correct option: A

x_1 x_2+y_1 y_2

Option A (

x_1 x_2+y_1 y_2

) is correct because it represents the dot product of the vectors OP and OQ.

133 Medium

133 of 187

Application +4 | -1

The coordinates of point which divides the line segment joining points

A(0,0)

and

B(9,12)

in the ratio

1: 2

, are

View answer and explanation

Correct option: B

(3,4)

Option B ((3,4)) is correct because it is found by using the section formula to divide the line segment in the given ratio.

134 Easy

134 of 187

Application +4 | -1

Let a variable line of slope

\mathrm{m} > 0

passing through the point

(4,-9)

intersect the coordinate axes at the points

\mathrm{A}

and

\mathrm{B}

. the minimum value of the sum of the distances of

\mathrm{A}

and

\mathrm{B}

from the origin is

View answer and explanation

Correct option: A

Option A (25) is correct because it is the minimum value of the sum of the distances of A and B from the origin, found using the AM-GM inequality.

135 Easy

135 of 187

Application +4 | -1

If the coordinates of two opposite vertices of a square are

(6,-3)

and

(-2,3)

, then area of square is

View answer and explanation

Correct option: A

50 \mathrm{sq}

units

Option A (50 sq units) is correct because the area of a square is half the square of the diagonal, and the diagonal is calculated correctly.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

136 Easy

136 of 187

Application +4 | -1

A(4,-3), B(3,-2)

and

C(2,8)

are the vertices of a triangle, then its centroid will be

View answer and explanation

Correct option: C

(3,1)

Option C (

(3,1)

) is correct because the centroid is found by averaging the x-coordinates and y-coordinates of the vertices.

137 Easy

137 of 187

Application +4 | -1

A triangle with vertices

(4,0),(-1,-1)

(3,5)

, is

View answer and explanation

Correct option: A

isosceles and right angled

Option A (isosceles and right angled) is correct because the triangle satisfies both the conditions for being isosceles and right-angled.

138 Easy

138 of 187

Application +4 | -1

The points of trisection of line segment joining the points

A(2,1)

and

B(5,3)

, are

View answer and explanation

Correct option: C

\left(3, \frac{5}{3}\right),\left(4, \frac{7}{3}\right)

Option C ((3,

\frac{{5}}{{3}}

), (4,

\frac{{7}}{{3}}

)) is correct because these are the coordinates that divide the line segment into three equal parts.

139 Easy

139 of 187

Application +4 | -1

If the point

(x, y)

is equidistant from the points

(a+b, b-a)

and

(a-b, a+b)

, then

View answer and explanation

Correct option: B

b x=a y

Option B (

bx=ay

) is correct because it represents the locus of a point equidistant from two given points.

140 Medium

140 of 187

Application +4 | -1

If the vertices of triangle are

(0,2),(1,0)

and

(3,1)

, then the triangle is

View answer and explanation

Correct option: D

Isosceles right angled

Option D (Isosceles right angled) is correct. Two sides are equal (isosceles) and the square of the longest side equals the sum of the squares of the other two sides (right-angled).

141 Easy

141 of 187

Analyzing +4 | -1

\triangle A B C

, base

B C

and area of triangle '

\triangle

' are fixed. Locus of the centroid of

\triangle A B C

is a straight line that is

View answer and explanation

Correct option: A

parallel to side

B C

Option A (parallel to side

BC

) is correct because with a fixed base and area, the height of the triangle is constant, implying the locus of the centroid is a line parallel to the base.

142 Easy

142 of 187

Application +4 | -1

The centroid of

\Delta A B C

(2,7)

. If

A=(4,8), B=(a, 0)

and

C=(0, b)

, then

View answer and explanation

Correct option: B

a=2, b=13

Option B (

a=2, b=13

) is correct because the centroid coordinates are calculated by averaging the x and y coordinates of the vertices.

143 Easy

143 of 187

Application +4 | -1

(x, 3)

and

(3,5)

are the extremities of a diameter of a circle with centre at

(2, y)

, then the values of

x

and

y

are

View answer and explanation

Correct option: A

x=1, y=4

Option A (

x=1, y=4

) is correct because the center of the circle is the midpoint of the diameter.

144 Easy

144 of 187

Application +4 | -1

The area of the triangle with vertices at

(a, b+c),(b, c+a)

and

(c, a+b)

, is

View answer and explanation

Correct option: A

Option A (0) is correct because the area of the triangle is zero, as the points are collinear. This can be shown by applying properties of determinants.

145 Easy

145 of 187

Application +4 | -1

If the vertices of a triangle are

(1,2),(4,-6)

and

(3,5)

, then its area is

View answer and explanation

Correct option: B

\frac{25}{2}

sq units

Option B (

\frac{{25}}{{2}}

sq units) is correct because it is the area calculated using the determinant formula with the given vertices.

146 Easy

146 of 187

Application +4 | -1

The coordinates of a point on the line

\frac{x-1}{2}=\frac{y+1}{-3}=z

at a distance

4 \sqrt{14}

from the point

(1,-1,0)

nearer the origin are

View answer and explanation

Correct option: D

(-7,11,-4)

Option D ((-7,11,-4)) is correct because it represents a point on the given line at a distance of

4\sqrt{14}

from (1,-1,0) and is nearer to the origin compared to the other possible point.

147 Easy

147 of 187

Application +4 | -1

If the points

(-2,-5),(2,-2),(8, a)

are collinear, then the value of

a

View answer and explanation

Correct option: B

\frac{5}{2}

Option B (

\frac{5}{2}

) is correct because the slope between any two pairs of collinear points must be equal.

148 Easy

148 of 187

Application +4 | -1

The coordinates of the points

A

and

B

are

(a, 0)

and

(-a, 0)

respectively. If a point

P

moves so that

P A^{2}-P B^{2}=2 k^{2}

, when

k

is constant, then the equation to the locus of the point

P

, is

View answer and explanation

Correct option: B

2 a x+k^{2}=0

Option B (

2ax + k^2 = 0

) is correct because the equation represents a vertical line, derived from the given condition

PA^2 - PB^2 = 2k^2

149 Easy

149 of 187

Application +4 | -1

A(6,3), B(-3,5), C(4,-2)

and

D(x, 3 x)

are four points. If the ratio of area of

\triangle D B C

and

\triangle A B C

1: 2

, then the value of

x

, will be

View answer and explanation

Correct option: A

\frac{11}{8}

Option A (

\frac{11}{8}

) is correct because it is the value of x that satisfies the given ratio of areas of the two triangles, calculated using the area formula.

150 Medium

150 of 187

Analyzing +4 | -1

a, b, c

and

d

are the sides of a quadrilateral, then the minimum value of

\frac{a^2+b^2+c^2}{d^2}

View answer and explanation

Correct option: C

\frac{1}{3}

Option C (

\frac{{1}}{{3}}

) is correct because the minimum value of the expression is found by relating the sum of squares to the square of the sum, and using the triangle inequality.

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