Coordinate Geometry Test

Result

Coordinate Geometry Test - 48

Score
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Weekly Quiz Competition

Question 1
1 / -0

If three lines $$x-3y=p,ax+2y=q$$ and $$ax+y=r$$ form a right angled triangle, then

A
$${ a }^{ 2 }-9a+18=0$$

B
$${ a }^{ 2 }-6a-12=0$$

C
$${ a }^{ 2 }-6a-18=0$$

D
$${ a }^{ 2 }-9a+12=0$$
Question 2
1 / -0

The line segment joining the points A(4,8) and B(6,10) has a third point C on it such that time passing though C is equidistant fro A and B The length of BC is equal to

A
$$\sqrt { 8 } $$ Units

B
$$\sqrt { 7 } $$ Units

C
$$\sqrt { 2 } $$ Units

D
$$\sqrt { 5 } $$ Units

Solution
Question 3
1 / -0

If the tangent at $$\theta =\frac { \pi }{ 4 } $$ to the curve $$x=a\cos ^{ 3 }{ \theta } ,y=a\sin ^{ 3 }{ \theta } $$ meets the x and y axes in A and B then the area of the triangle OAB is

A
$$\frac { { a }^{ 2 } }{ 4 } sq.units$$

B
$$\frac { { a }^{ 2 } }{ 2 } sq.units$$

C
$$\frac { { 3a }^{ 2 } }{ 4 } sq.units$$

D
$$\frac { { 5a }^{ 2 } }{ 4 } sq.units$$

Solution
Question 4
1 / -0

If $$P , Q$$ are two points on the line $$3 x + 4 y + 15 = 0$$ such that $$O P = O Q = 9$$ then the area of $$\Delta O P Q$$ is

A
6$$\sqrt { 2 }$$

B
9$$\sqrt { 2 }$$

C
12$$\sqrt { 2 }$$

D
18$$\sqrt { 2 }$$

Solution
Question 5
1 / -0

Area of a triangle whose vertices are $$(a\cos \theta ,b\sin \theta ),(-a\sin \theta ,b\cos \theta)$$ and $$(-a\cos \theta ,-b\sin \theta )$$ is-

A
$$ab\sin \theta \cos \theta $$

B
$$a\cos \theta \sin \theta$$

C
$$\frac 12ab$$

D
ab

Solution
Question 6
1 / -0

Mid point of $$A(0,0)$$ and $$B(1024,2048)$$ is $${ A }_{ 1 }$$, midpoint of $${ A }_{ 1 }$$ and B is $${ A }_{ 2 }$$ and so on. Coordinates of $${ A }_{ 10 }$$ are

A
(1022, 2044)

B
(1025, 2050)

C
(1023, 2046)

D
(1, 2)

Solution
Question 7
1 / -0

Let A(-4, 0) & B(4,0). Then the number of points C=(x,y) on the circle $${x^2} + {y^2} = 16$$ lying in first quadrant $$(x,y \geqslant 0)$$ such that the area of the triangle whose vertices are A,B,C is a integer is

A
14

B
15

C
16

D
None of these
Question 8
1 / -0

Area of the triangle formed by the tangents at the points $$\left( {4,6} \right),\left( {10,8} \right)$$ and $$\left( {2,4} \right)$$ on the parabola $${y^2} - 2x = 8y - 20,$$is (in sq. units)

A
4

B
2

C
1

D
8

Solution
Question 9
1 / -0

The area of the triangle inscribed in the parabola $$y^2$$ = $$4x$$ , the ordinates of whose vertices are 1 , 2 and 4 is :

A
7/2

B
5/2

C
3/2

D
3/4
Question 10
1 / -0

If the area of triangle formed by the points $$(2a, b) (a + b, 2b + a)$$ and $$(2b, 2a)$$ be $$\lambda$$, then the area of the triangle whose vertices are $$(a + b, a - b), (3b - a, b + 3a)$$ and $$(3a - b, 3b - a)$$ will be

A
$$\frac{3}{2}\lambda$$

B
$$3\lambda$$

C
$$4\lambda$$

D
none of these

Solution