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Geometry Paper

1) The point (-3,-4) lies in which quadrant?

a) First Quadrant b) Second Quadrant

c) Third Quadrant d) Fourth Quadrant

2) The perpendicular distance between the point (-4,0) from X-axis is ___________units.

a) 0 units b) 4 units

c) Cannot be determined d) 8 units

3) The point (5,-1/2) lies in which Quadrant?

a) First Quadrant b) Second Quadrant

c) Third Quadrant d) Fourth Quadrant

4) The distance between the points (3,4) and (-3,4) is __________?

a) 3 units b) 9 units

c) 6 units d) none of these

5) The distance between X-axis and (0,-3) is ___________units?

a) 9 b) 3

c) Square root of 3 d) none of these

6) The distance between Origin (0,0) and the point (3,4) is _________units?

a) 3 b) 4

c) 5 d) none of these

7) One of the points of tri-section of the line joining the points A(3,4) and B(-5,3) is________?

a) (2/3,1/3) b) (1/3,-7/3)

c) (-7/3, 11/3) d) (1/3,11/3)

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8) The mid pint of the line joining the points (5,-1) and (-3,5) is ________?

a) (2,1) b) (-1,2)

c) (1,-2) d) (1,2)

9) The mid-point of the line joining (3,k) and (-1,0) is (1,2), then the value of k = ?

a) -1 b) 0

c) 4 d) 7

10) The angle between X-axis (in the Positive direction) and the line joining the points Origin

(0,0) & (2,2) is__________?

a) 30 degrees b) 45 degrees

c) 90 degrees d) 60 degrees

11) A line makes an angle of 60 degrees with X-axis in the positive direction. Then the slope of

the line Is __________?

a) (Sqrt3)/2 b) 1/(sqrt2)

c) 1/(sqrt3) d) None of these

12) The area of the triangle formed by three points (0,0), (1,2) & (-1,0) is ______sq. units?

a) 2 b) 3

c) 1 d) 4

13) The area of the triangle formed by three points (1,k), (-1,0) & (0,0) is Zero units then the

The value of k =?

a) -1 b) 1

c) -2 d) 0

14) The distance between the points (0,K) & (-1, 0) is sqrt(2) then the value of k = ?

a) 2 b) 1

c) -1 d) both (b) & (c)

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15) The point which divides the line joining (-1,0) and (2,0) in the ratio 1 : 2 is ______?

a) (0,1) b) (0,0)

c) (1,0) d) (-1,0)

16) The points of tri section of the line joining the points A(-1,0), B(-4,0) are

a) (-2,0) & (-3,0) b) (0,0) & (-2,0)

c) (0,0) & (-1,0) d) None of these

17) The centroid of the triangle whose vertices are (0,0), (2,2) & (4,0) is _______?

a) (2/3, 2) b) (1/3, 2/3)

c) (2, 2/3) d) ( 2/3, 1/3)

18) The distance between the points (0,1) from X- axis is ________units?

a) 1 2) 2

c) 0 4) None of these

19) The distance between the points (k,0) from Y-axis is 4 units. Then the value of k is___?

a) 0 b) 1

c) 4 d) 3

20) The point (2,0) divides the line joining the points (-1,0) & (1,0) in the ratio____?

a) 3:1, internally b) 3:1,externally

c) 1:3, internally c) 1:3, external

21) The points (0,k), (3,5) & (1,1) are collinear, then the value of K = ?

a) 1 b) -1

c) 0 d) 2

22) The slope of the line joining the points (0,0) & (3,4) is __________?

a) 3/4 b) -3/4

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c) -4/3 d) 4/3

23) The slope of X-axis & Y-axis respectively are

a) infinity, 0 b) 0, infinity

c) -1, 0 d) 0,-1

24) The slope of the line which is parallel to X- axis is_____?

a) 0 b) infinity

c) 1 d) None of these

25)The center of the circle, whose ends of the diameter are (3,4), (5,6) is ____?

a) (3,2) b) (-4,5)

c) (-4,-5) d) (4,5)

Answers

1) C Solution: The point both of whose co-ordinates have negative signs lies in the Third quadrant.

2) A Solution: The point (-4,0) lies on the X-axis . Hence the perpendicular distance between (-4,0) is

Zero (0)

3) D. Solution: X- co-ordinate is +ve and Y- co-ordinate is –ve. Hence the point is in Fourth Quadrant.

4) C Solution: The distance between two points (x1,y1), (x2 ,y2) is equal to

Sqrt((x2-x1)^2+(y2-y1)^2) = Sqrt((-3-3)^2+(4-4)^2) = 6 units

5) B

6) C Solution: The formula for calculating the distance between Origin(0,0) and (x1,y1)

= Sqrt(x1^2+y1^2)

Using the above formula the distance between (0,0) and (3,4) is Sqrt(3^2+4^2) = Sqrt(25) =5

7) D Solution: The point which divides the line joining the points (3,4), (-5,3) in the ratio 1:2 or 2: 1 is

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called point of Tri-section. Here (x1,y1), (x2,y2) are the points and m:n = 1:2 is the ratio

Using the formula for point of tri-section ((mx2+nx1)/(m+n), (my2+ny1)/(m+n))

=((1x(-5)+(2X3))/(1+2) , (1x3+2x4)/(1+2))= (1/3,11/3)

8) D Solution:The mid-point of a line joining two pints (x1,y1), (x2,y2) is ( (x1+x2)/2, (y1+y2)/2 )

Using the above formula the mid-poins is ( (5-3)/2, (-1+5)/2 ) = (1,2)

9) C Solution: Using the above formula in Q.No.8 we can find the value of K=4

10) B

11) D Solution: Slope = tan(theta)= tan 60 degrees=Sqrt(3)

12) C Solution: The area of the traingle formed by three points (x1,y1),(x2,y2),(x3,y3) is

= 1/2x (( x1-x2)x(y2-y3))-(x2-x3)x(y1-y2))

Using the Above Fromula we can find Area= 1 sq.units

13) D Solution: Using the above formula in Q.No. 12, we can find the value of k = 0

14) D Solution: Using the distance formula stated in Q.No.1 , we can find the value of K=±1

15) B Solution: Using the formula for point which divided two points in the given ratio stated in Q.No. 7

we can find the required point.

Here m:n=1:2 & (x1,y1)=(-1,0), (x2,y2) =(2,0)

16) A

17) C Solution: The centriod of a traingle whose vertices are (x1,y1), (x2,y2),(x3,y3) is given by

( (X1+x2+x3)/3, (y1+y2+y3)/3 )

You can find the centriod now.

18) A Solution: The distance between any point from X-axis is its Y-Co-ordinate

and the distance between any point from Y-axis is its X-Co-ordinate.

19) C Solution: (note the explanation given in the above answer)

20) B Solution: You can find the ratio using the formula given in Answer No.7. If you get the ratio m:n

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as negative, then the division is external, otherwise it is internal division.

21) B Solution: When three pointa are collinear, then area formed by those three points is Zero.

Use the formula stated in Answer No.12, then you get the value of K=-1

22) D Solution: The formula to find out the slopeof a line joining the points ( x1,y1),(x2,y2)

Slope = m = (y2-y2)/(x2-x1) =4/3

23) B

24) A

25) D Solution: The mid-point of the diametre is the centre of the circle. Use the mid-point formula

stated in Answer No.8 to find out the centre of the circle.