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JEE • MATHEMATICS • 2D GEOMETRY SYSTEM

2d geometry system practice · Mixed difficulty

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Showing 61-90 of 187 questions

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61 Medium
61 of 187
Application +4 | -1
If the locus of the point, whose distances from the point (2,1)(2,1) and (1,3)(1,3) are in the ratio 5:45: 4, is ax2+by2+cxy+dx+ey+170=0a x^2+b y^2+c x y+d x+e y+170=0, then the value of a2+2b+3c+4d+ea^2+2 b+3 c+4 d+e is equal to:
View answer and explanation
Correct option: C
37
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.
62 Medium
62 of 187
Application +4 | -1
If the vertices A,BA, B and CC of a triangle are (2,1),(5,2)(2,1),(5,2) and (3,4)(3,4), respectively. Then, the circumcentre is
View answer and explanation
Correct option: C
(134,94)\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.
63 Medium
63 of 187
Application +4 | -1
In a parallelogram ABCD,A B C D, A(1,2),A(1,2), B(3,4),B(3,4), C(2,5)C(2,5) then length of the side ADA D is equal to
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because finding the coordinates of point D using the parallelogram property and then calculating the distance AD yields 2\sqrt{2}.
64 Easy
64 of 187
Analyzing +4 | -1
If each of the vertices of a triangle has integral coordinates, then the triangle will not be
View answer and explanation
Correct option: B
equilateral
Option B (equilateral) is correct because a triangle with integral coordinates cannot be equilateral due to area considerations involving 3\sqrt{3}.
65 Medium
65 of 187
Application +4 | -1
The area of the triangle formed by the lines y=m1x+c1y=m_{1} x+c_{1}, y=m2x+c2y=m_{2} x+c_{2} and x=0x=0 is
View answer and explanation
Correct option: C
12(c1−c2)2(m1−m2)\frac{1}{2} \frac{\left(c_{1}-c_{2}\right)^{2}}{\left(m_{1}-m_{2}\right)}
Option C (12(c1−c2)2(m1−m2)\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.
66 Easy
66 of 187
Analyzing +4 | -1
Let BB and CC be the two points on the line y+x=0y+x=0 such that BB and CC are symmetric with respect to the origin. Suppose AA is a point on y−2x=2y-2 x=2 such that △ABC\triangle A B C is an equilateral triangle. Then, the area of the △ABC\triangle A B C is
View answer and explanation
Correct option: D
83\frac{8}{\sqrt{3}}
Option D (83\frac{8}{\sqrt{3}}) is correct because it represents the area of the equilateral triangle calculated using the distance from point A to the origin and the properties of equilateral triangles.
67 Easy
67 of 187
Analyzing +4 | -1
If the points O(0,0),A(cos⁡α,sin⁡α)O(0,0), A(\cos \alpha, \sin \alpha), B(cos⁡β,sin⁡β)B(\cos \beta, \sin \beta) are the vertices of a right-angled triangle, then ∣sin⁡(α−β2)∣=\left|\sin \left(\frac{\alpha-\beta}{2}\right)\right|=
View answer and explanation
Correct option: B
12\frac{1}{\sqrt{2}}
Option B (12\frac{1}{\sqrt{2}}) is correct because the given points form a right-angled triangle, and using the distance formula and trigonometric identities, we can find the value of ∣sin⁡(α−β2)∣\left|\sin \left(\frac{\alpha-\beta}{2}\right)\right|.
68 Easy
68 of 187
Application +4 | -1
If OO is the origin and the coordinates of AA and BB are (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) respectively then OA×OBcos⁡∠AOBO A \times O B \cos \angle A O B is equal to
View answer and explanation
Correct option: B
x1x2+y1y2x_1 x_2+y_1 y_2
Option B (x1x2+y1y2x_1 x_2+y_1 y_2) is correct because it represents the dot product of the vectors OA and OB, which is equal to |OA||OB|cos(∠AOB).
69 Easy
69 of 187
Application +4 | -1
If the points (a,b),(a′,b′)(a, b),\left(a^{\prime}, b^{\prime}\right) and (a−a′,b−b′)\left(a-a^{\prime}, b-b^{\prime}\right) are collinear, then
View answer and explanation
Correct option: A
ab′=a′ba b^{\prime}=a^{\prime} b
Option A (ab′=a′ba b^{\prime}=a^{\prime} b) is correct because the condition for collinearity leads to the proportionality of the coordinates.
70 Medium
70 of 187
Application +4 | -1
The area of the parallelogram formed by the lines 4y−3x=1,4 y-3 x=1, 4y−3x−3=0,4 y-3 x-3=0, 3y−4x+1=0,3 y-4 x+1=0, 3y−4x+2=03 y-4 x+2=0 is
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.
71 Easy
71 of 187
Application +4 | -1
For what value of 'aa' the lines x=3,y=4x=3, y=4 and 4x−3y+a=04 x-3 y+a=0 are concurrent
View answer and explanation
Correct option: A
0
Option A (0) is correct because substituting x=3 and y=4 into the third equation and solving for 'a' yields a=0.
72 Easy
72 of 187
Analyzing +4 | -1
If two vertices of an equilateral triangle have integral coordinates, then the third vertex will have
View answer and explanation
Correct option: C
at least one coordinate irrational
Option C is correct. If two vertices have integral coordinates, the third vertex of an equilateral triangle must have at least one irrational coordinate due to the area involving sqrt(3).
73 Easy
73 of 187
Application +4 | -1
Three points (p+1,1),(2p+1,3)(p+1,1),(2 p+1,3) and (2p+2,2p)(2 p+2,2 p) are collinear, if p=p=
View answer and explanation
Correct option: C
2
Option C (2) is correct because the value of p is found by setting the determinant of the matrix formed by the points to zero, indicating collinearity.
74 Easy
74 of 187
Application +4 | -1
If (−6,−4),(3,5),(−2,1)(-6,-4),(3,5),(-2,1) are the vertices of a parallelogram, then remaining vertex cannot be
View answer and explanation
Correct option: A
(0,−1)(0,-1)
Option A ((0,−1)(0,-1)) is correct because the possible fourth vertices of the parallelogram are (-11, -8), (7, 10), and (-1, 0), making (0, -1) the only impossible vertex.
75 Medium
75 of 187
Application +4 | -1
The circumcentre of the triangle formed by the lines x+y=0,xy=0x+y=0, x y=0 and x   7=0x \,\,\,7=0 is
View answer and explanation
Correct option: D
(7,0)(7,0)
Option D is correct. Since the lines x+y=0x+y=0 and x−y=0x-y=0 are perpendicular, the triangle is a right triangle, and the circumcenter is the midpoint of the hypotenuse, which is (7,0)(7,0).
76 Easy
76 of 187
Application +4 | -1
If points O(0,0),A(3,3)O(0,0), A(3, \sqrt{3}) and B(3,a)B(3, a) are the vertices of an equilateral triangle, then a=a=
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the calculated value of 'a' does not match any of the provided options.
77 Medium
77 of 187
Application +4 | -1
The distance between the points P(2,60∘)P\left(2,60^{\circ}\right) and Q(5,270∘)Q\left(5,270^{\circ}\right) is
View answer and explanation
Correct option: A
29+103\sqrt{29+10 \sqrt{3}}
Option A (29+103\sqrt{{29+10 \sqrt{{3}}}}) is correct because it represents the distance between the points in polar coordinates.
78 Medium
78 of 187
Application +4 | -1
Incentre of triangle, whose vertices are A(−36,7),B(20,7)A(-36,7), B(20,7) and C(0,−8)C(0,-8), is
View answer and explanation
Correct option: C
(−1,0)(-1,0)
Option A (-1,0) is correct because the incenter is calculated using the formula involving side lengths and vertex coordinates.
79 Medium
79 of 187
Application +4 | -1
The co-ordinates of the foot of perpendicular from the point (2,3)(2,3) on the line x+y−11=0x+y-11=0 are
View answer and explanation
Correct option: B
(5,6)(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.
80 Easy
80 of 187
Analyzing +4 | -1
If O′\mathrm{O}^{\prime} is the orthocentre of triangle ABCA B C whose vertices are at A(at12,2at1),B(at22,2at2)A\left(a t_1{ }^2, 2 a t_1\right), B\left(a t_2{ }^2, 2 a t_2\right) and C(at32,2at3)C\left(a t_3{ }^2, 2 a t_3\right), then the coordinates of the orthocentre of △O′BC\triangle O^{\prime} B C are
View answer and explanation
Correct option: C
(at12,2at1)\left(a t_1^2, 2 a t_1\right)
Option C ((at12,2at1)\left(a t_1^2, 2 a t_1\right)) is correct because the orthocenter of △O′BC\triangle O'BC is simply point A, given that O' is the orthocenter of triangle ABC.
81 Easy
81 of 187
Application +4 | -1
The points (0,0),(a,0)(0,0),(a, 0) and (a2,a32)\left(\frac{a}{2}, \frac{a \sqrt{3}}{2}\right) are vertices of
View answer and explanation
Correct option: B
Equilateral triangle
Option B (Equilateral triangle) is correct because the distances between all three points are equal, indicating an equilateral triangle.
82 Easy
82 of 187
Application +4 | -1
The equation to the locus of a point which moves so that its distance from xx-axis is always one half its distance from the origin, is
View answer and explanation
Correct option: B
x2−3y2=0x^{2}-3 y^{2}=0
Option B (x2−3y2=0x^{2}-3 y^{2}=0) is correct because the distance from the point to the x-axis (∣y∣|y|) is half its distance from the origin (x2+y2\sqrt{{x^2+y^2}}), leading to the equation ∣y∣=12x2+y2|y| = \frac{{1}}{{2}} \sqrt{{x^2+y^2}}, which simplifies to x2−3y2=0x^{2}-3 y^{2}=0.
83 Easy
83 of 187
Application +4 | -1
The incentre of the triangle formed by the lines xy=0\mathrm{xy}=0 and x3+y4=1\frac{x}{3}+\frac{y}{4}=1 is (h,k)(\mathrm{h}, \mathrm{k}), then
View answer and explanation
Correct option: D
(h,k)(\mathrm{h}, \mathrm{k}) lies on the circle (x−4)2+(y−5)2=25(x-4)^2+(y-5)^2=25
The correct answer is D ((x−4)2+(y−5)2=25x-4)^2+(y-5)^2=25) because the incenter (1,1) satisfies this equation.
84 Easy
84 of 187
Application +4 | -1
The coordinates of two points AA and BB are (3,4)(3,4) and (5,−2)(5,-2) respectively. If PP is a point not lying on any of the coordinate axes such that PA=PBP A=P B and Area of △PAB=10\triangle P A B=10, then the coordinates of PP, are
View answer and explanation
Correct option: B
(7,2)(7,2)
Option B ((7,2)) is correct because it satisfies both the conditions PA=PB and Area of triangle PAB=10.
85 Easy
85 of 187
Application +4 | -1
If points (5,5),(10,k)(5,5),(10, k) and (−5,1)(-5,1) are collinear, then k=k=
View answer and explanation
Correct option: C
7
Option C (7) is correct because substituting k=7 into the determinant equation for collinearity makes the determinant equal to zero.
86 Easy
86 of 187
Application +4 | -1
The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm160 \mathrm{~cm}, then what can you say about breadth?
View answer and explanation
Correct option: C
Breadth ≥20\geq 20
Option C (Breadth ≥20\geq 20) is correct because the perimeter is 2(l+b)2(l+b) where l=3bl=3b, and the minimum perimeter is 160 cm, leading to b≥20b \geq 20.
87 Medium
87 of 187
Understanding +4 | -1
If the point P(x1+t(x2x1),y1+t(y2y1))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(x1,y1)A\left(x_1, y_1\right) and B(x2,y2)B\left(x_2, y_2\right) internally, then ….\ldots ..
View answer and explanation
Correct option: B
0<t<10 < t < 1
Option B (0<t<10 < t < 1) is correct because for P to divide AB internally, t must be between 0 and 1.
88 Easy
88 of 187
Application +4 | -1
A point PP moves so that its distance from the point (a,0)(a, 0) is always equal to its distance from the line x+a=0x+a=0. The locus of the point is
View answer and explanation
Correct option: A
y2=4axy^{2}=4 a x
Option A (y2=4axy^{2}=4 a x) is correct because the distance from the point (x,y)(x,y) to (a,0)(a,0) being equal to its distance from the line x+a=0x+a=0 leads to the equation (x−a)2+y2=(x+a)2(x-a)^2 + y^2 = (x+a)^2, which simplifies to y2=4axy^{2}=4 a x.
89 Easy
89 of 187
Application +4 | -1
If the vertices of a triangle are (5,2),(2/3,2)(5,2),(2 / 3,2) and (−4,3)(-4,3), then the area of the triangle is
View answer and explanation
Correct option: D
13/613 / 6
Option D (13/613 / 6) is correct because the area of the triangle is calculated using the determinant formula and absolute value.
90 Medium
90 of 187
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.
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