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JEE • MATHEMATICS • 2D GEOMETRY SYSTEM

2d geometry system practice · Medium

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Showing 31-57 of 57 questions
31 Medium
31 of 57
Analyzing +4 | -1
Let A(α,1α)B(β,1β)A\left(\alpha, \frac{1}{\alpha}\right) B\left(\beta, \frac{1}{\beta}\right) and C(γ,1γ)C\left(\gamma, \frac{1}{\gamma}\right) be the vertices of a △ABC\triangle A B C, where α,β\alpha, \beta are the roots of the equation x2−6p1x+2=0x^2-6 p_1 x+2=0 and β,γ\beta, \gamma are the roots of the equation x2−6p2x+3=0x^2-6 p_2 x+3=0 and γ,α\gamma, \alpha are the roots of the equation x2−6p3x+6=0,p1,p2,p3x^2-6 p_3 x+6=0, p_1, p_2, p_3 being positive. Then, the coordinates of the centroid of △ABC\triangle A B C are
View answer and explanation
Correct option: C
(2,1118)\left(2, \frac{11}{18}\right)
Option C (2,1118)\left(2, \frac{{11}}{{18}}\right) is correct because it represents the centroid coordinates calculated using the roots of the given quadratic equations.
32 Medium
32 of 57
Application +4 | -1
The coordinates of the foot of the perpendicular from the point (2,3)(2,3) on the line y=3x+4y=3 x+4 are given by
View answer and explanation
Correct option: B
(−110,3710)\left(-\frac{1}{10}, \frac{37}{10}\right)
Option B ((-1/10), (37/10)) is correct because it represents the coordinates of the foot of the perpendicular from the given point to the line.
33 Medium
33 of 57
Application +4 | -1
The locus of PP such that area of △PAB=12\triangle P A B=12 sq. units, where A(2,3)A(2,3) and B(−4,5)B(-4,5) is
View answer and explanation
Correct option: B
(x+3y+1)(x+3y−23)=0(x+3 y+1)(x+3 y-23)=0
Option B ((x+3y+1)(x+3y−23)=0(x+3 y+1)(x+3 y-23)=0) is correct because it represents the locus of point P such that the area of triangle PAB is 12 sq. units.
34 Medium
34 of 57
Application +4 | -1
The values of tt, if the area of the pentagon ABCDEA B C D E is 452sq\frac{45}{2} s q units, where A=(1,3),B=(−2,5),C=(−3,−1),D(0,−2)A=(1,3), B=(-2,5), C=(-3,-1), D(0,-2) and E=(2,t)E=(2, t) and here the points are given in order, are
View answer and explanation
Correct option: A
−1,89-1,89
Option A (-1, 89) is correct because the area of the pentagon is calculated using the determinant formula and solving for t.
35 Medium
35 of 57
Application +4 | -1
The orthocentre of the triangle formed by the lines x+y=1,2x+3y=6x+y=1,2 x+3 y=6 and 4x−y+4=04 x-y+4=0 lies in quadrant
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Correct option: A
First
Option A (First) is correct because the orthocentre, calculated as the intersection of altitudes, lies in the first quadrant with coordinates (37,227)\left(\frac{{3}}{{7}}, \frac{{22}}{{7}}\right).
36 Medium
36 of 57
Analyzing +4 | -1
The orthocentre of the triangle formed by the line 2x2−xy−6y2=02 x^2-x y-6 y^2=0 and the line x+y+1=0x+y+1=0, is
View answer and explanation
Correct option: A
(−43,−43)\left(-\frac{4}{3},-\frac{4}{3}\right)
Option A ((-4/3, -4/3)) is correct because it represents the intersection point of the altitudes of the triangle formed by the given lines.
37 Medium
37 of 57
Application +4 | -1
If the distance of any point PP from the point A(a+b,a−b)A(a+b, a-b) and B(a−b,a+b)B(a-b, a+b) are equal, then the locus of PP is
View answer and explanation
Correct option: A
x−y=0x-y=0
Option A (x−y=0x-y=0) is correct because it represents the locus of point P equidistant from points A and B.
38 Medium
38 of 57
Application +4 | -1
The ratio in which the line 3x+y−9=03 x+y-9=0 divides the segment joining the points (1,3)(1,3) and (2,7)(2,7), is
View answer and explanation
Correct option: C
3:43: 4
Option C (3:4) is correct because the line divides the segment in the ratio 3:4, which is found by substituting the section formula into the line's equation.
39 Medium
39 of 57
Understanding +4 | -1
If (α,β),(xˉ,yˉ)(\alpha, \beta),(\bar{x}, \bar{y}) and (u,v)(u, v) are respectively coordinates of the circumoentre, centroid and orthocentre of a triangle. Then,
View answer and explanation
Correct option: A
3xˉ=2α+u3 \bar{x}=2 \alpha+u and 3yˉ=2β+v3 \bar{y}=2 \beta+v
Option A is correct because the centroid divides the line joining the circumcentre and orthocentre in the ratio 2:1.
40 Medium
40 of 57
Application +4 | -1
Two points AA and BB have coordinates (1,0)(1,0) and (−1,0)(-1,0) respectively and QQ is a point which satisfies the relation AQ−BQ=±1A Q-B Q= \pm 1. The locus of QQ is
View answer and explanation
Correct option: B
12x2−4y2=312 x^{2}-4 y^{2}=3
Option B (12x2−4y2=312 x^{2}-4 y^{2}=3) is correct because the given condition AQ−BQ=±1AQ - BQ = \pm 1 defines a hyperbola, and after simplification, the equation becomes 12x2−4y2=312x^2 - 4y^2 = 3.
41 Medium
41 of 57
Application +4 | -1
Let two points be A(1,−1)A\left( {1, - 1} \right) and B(0,2).B\left( {0,2} \right). If a point P(x′,y′)P\left( {x',y'} \right) be such that the area of ΔPAB=5   sq.  units\Delta PAB = 5\,\,\,sq.\,\,units and it lies on the line, 3x+y−4λ=0,3x + y - 4\lambda = 0, then a value of λ\lambda is
View answer and explanation
Correct option: C
3
Option C (3) is correct because it is the value of λ\lambda that satisfies the area condition and the line equation.
42 Medium
42 of 57
Application +4 | -1
The points A(−4,−1),B(−2,−4),C(4,0)A(-4,-1), B(-2,-4), C(4,0) and D(2,3)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.
43 Medium
43 of 57
Application +4 | -1
The area of the triangle formed by the lines 7x−2y+10=0,7x+2y−10=07 x-2 y+10=0,7 x+2 y-10=0 and y+2=0y+2=0 is
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.
44 Medium
44 of 57
Application +4 | -1
The orthocentre of the triangle with vertices (−2,−6),(−2,4)(-2,-6),(-2,4) and (1,3)(1,3) is
View answer and explanation
Correct option: C
(1,3)(1,3)
Option C ((1,3)(1,3)) is correct because it is the intersection point of the altitudes of the triangle.
45 Medium
45 of 57
Analyzing +4 | -1
The points A(2a,4a),B(2a,6a)A(2 a, 4 a), B(2 a, 6 a) and C(2a+3a,5a)(C(2 a+\sqrt{3} a, 5 a)( when a>0)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.
46 Medium
46 of 57
Application +4 | -1
The coordinates of point which divides the line segment joining points A(0,0)A(0,0) and B(9,12)B(9,12) in the ratio 1:21: 2, are
View answer and explanation
Correct option: B
(3,4)(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.
47 Medium
47 of 57
Application +4 | -1
If the vertices of triangle are (0,2),(1,0)(0,2),(1,0) and (3,1)(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).
48 Medium
48 of 57
Analyzing +4 | -1
If a,b,ca, b, c and dd are the sides of a quadrilateral, then the minimum value of a2+b2+c2d2\frac{a^2+b^2+c^2}{d^2} is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\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.
49 Medium
49 of 57
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.
50 Medium
50 of 57
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.
51 Medium
51 of 57
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).
52 Medium
52 of 57
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.
53 Medium
53 of 57
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.
54 Medium
54 of 57
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.
55 Medium
55 of 57
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 '0'. 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.
56 Medium
56 of 57
Understanding +4 | -1
The locus of a point PP which moves in such a way that the segment OPO P, where OO is the origin, has slope 3\sqrt{3} is
View answer and explanation
Correct option: D
3x−y=0\sqrt{3} x-y=0
Option D (3x−y=0\sqrt{3} x-y=0) is correct because the slope of OP is y/x, and setting that equal to 3\sqrt{3} and rearranging gives the equation of the locus.
57 Medium
57 of 57
Analyzing +4 | -1
If vertices of a quadrilateral are A(0,0),B(3,4),C(7,7)A(0,0), B(3,4), C(7,7) and D(4,3)D(4,3), then quadrilateral ABCDA B C D is
View answer and explanation
Correct option: D
rhombus
Option D (rhombus) is correct because all sides of the quadrilateral are equal, but the diagonals are not equal.
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