Coordinate Geometry Test

Result

Coordinate Geometry Test - 4

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Weekly Quiz Competition

Question 1
3 / -1

The point on the x – axis which is equidistant from the points (2, – 5) and ( – 2, 9) is

A
(7, 0)

B
( – 7, 0)

C
(0, – 7)

D
(0, 7)

Solution
Question 2
3 / -1

The distance of the point ( – 5, 12) from the y – axis is

A
5 units

B
12 units

C
- 5 units

D
13 units

Solution

The distance of any point from y−axis is its abscissa. Therefore, the required distance is 5 units.
Question 3
3 / -1

The point where the medians of a triangle meet is called the ________ of the triangle

A
none of these

B
centroid

C
circumcentre

D
orthocentre

Solution

The point where three medians of a triangle meet is called the centroid of the triangle.it is the centre of gravity of the triangle. it divides the median in the ratio 2 :1
Question 4
3 / -1

If the co – ordinates of a point are ( – 5, 11), then its abscissa is

A
– 11

B
5

C
– 5

D
11

Solution

Since x−coordinate of a point is called abscissa.Therefore, abscissa is −5.
Question 5
3 / -1

If the end points of a diameter of a circle are ( – 4, – 3) and (2, 7), then the co – ordinates of the centre are

A
( – 1, 2)

B
(2, – 1)

C
(0, 0)

D
none of these

Solution

Let the coordinates of centre O be (x,y). The end points of diameter of the circle are A (- 4, - 3) and B(2, 7).

Since centre is the midpoint of diameter.

Therefore, the coordinates of the centre O is (−1,2).
Question 6
3 / -1

The centroid of a triangle divides the median in the ratio

A
3 : 1

B
1 : 3

C
1:2

D
2 : 1

Solution

The centroid of a triangle is the centre of the triangle which is the point of intersection of all the three medians of the triangle and divides the median in the ratio 2 : 1The median is a line drawn from the mid-point of a side to the opposite vertix
Question 7
3 / -1

The abscissa of any point on the y – axis is

A
0

B
y

C
– 1

D
1

Solution

Since coordinates of any point on y−axis is (0,y).Therefore, abscissa is 0.
Question 8
3 / -1

The point on the y – axis which is equidistant from the points (6, 5) and ( – 4, 3) is

A
( – 9, 0)

B
(0, – 9)

C
(9, 0)

D
(0, 9)

Solution
Question 9
3 / -1

If one end of a diameter of a circle is (4, 6) and the centre is ( – 4, 7), then the other end is

A
(8, – 12)

B
(8, – 6)

C
( – 12, 8)

D
(8, 10)

Solution

one end of a diameter is A(4 , 6) and the centre is O ( - 4 , 7) .... ( Given)

Let the other end be B (x,y)
Since centre is the mid-point of diameter of the circle.

therefore coordinates of centre O are x = (4 + x) / 2

Therefore, the required coordinates of other end of the diameter are (−12,8).
Question 10
3 / -1

The co – ordinates of the mid – point of the line segment joining the points ( – 2, 3) and (4, – 5) are

A
( – 1, 1)

B
(1, – 1)

C
( – 2, 4)

D
(0, 0)

Solution

Let the coordinates of midpoint C(x,y)(x,y) of the line segment joining the points A(−2,3) and B(4,−5).
Question 11
3 / -1

If the mid – point of the line segment joining the points (a, b – 2) and ( – 2, 4) is (2, – 3), then the values of ‘a’ and ‘b’ are

A
6, 8

B
4, – 5

C
6, – 8

D
– 6, 8

Solution

Let the coordinates of midpoint O(2,−3) is equidistance from the points A(a,b−2) and B(−2,4).
Question 12
3 / -1

The abscissa of any point on the x – axis is

A
– 1

B
0

C
x

D
1

Solution

Since coordinates of any point on x−axis is (x,0).Therefore, abscissa is x.
Question 13
3 / -1

Find the value of ‘k’, if the point (0, 2) is equidistant from the points (3, k) and (k, 5)

A
1

B
– 1

C
0

D
2

Solution

Let point C (0, 2) is equidistant fromthe points A(3,k) and B(k,5).
Question 14
3 / -1

The points A( – 1, 0), B(3, 1), C(2, 2) and D( – 2, 1) are the vertices of a

A
Square

B
Parallelogram

C
Rhombus

D
Rectangle

Solution

Given: The points A( – 1, 0), B(3, 1), C(2, 2) and D( – 2, 1)

Therefore diagonals AC and BD are not equal
Since, opposite sides of the given fig are equal and both diagonals are not equal.
Therefore, the given figure (Quadrilateral) is a parallelogram.
Question 15
3 / -1

If one end of a diameter of a circle is (2, 3) and the centre is ( – 2, 5), then the other end is

A
( – 6, 7)

B
(0, 8)

C
(0, 4)

D
(6, – 7)

Solution

Let the coordinates of the other end be B(x2,y2).

One end of the diameter is A (2, 3) and the centre is O(−2,5).
Since the centre is midpoint of the diameter of the circle.

Therefore, the coordinates of other end of the diameter are (−6,7).