Coordinate Geometry Test

Result

Coordinate Geometry Test - 3

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Weekly Quiz Competition

Question 1
3 / -1

The distance of the point ( – 3, 4) from the origin is

A
5 units

B
1 unit

C
25 units

D
7 units

Solution
Question 2
3 / -1

If A and B are the points ( – 6, 7) and ( – 1, – 5) respectively, then the distance 2AB is equal to

A
26 units

B
13 units

C
15 units

D
20 units

Solution
Question 3
3 / -1

Three consecutive vertices of a parallelogram ABCD are A(1, 2), B(1, 0) and C(4, 0). The co – ordinates of the fourth vertex D are

A
( – 4, – 2)

B
(4, 2)

C
(4, – 2)

D
( – 4, 2)

Solution

Explanation:
Coordinates are given for A(1 , 2) , B(1 , 0) and C(4 , 0)
Let coordinates of D be (x,y).
Since diagonals of a parallelogram bisect each other. at point O
Therefore O is the mid point of diagonal AC
Question 4
3 / -1

The distance between the points ( – 1, – 5) and ( – 6, 7) is

A
12 units

B
13 units

C
169 units

D
144 units

Solution
Question 5
3 / -1

The length of the median through A of ΔABC with vertices A(7, – 3), B(5, 3) and C(3, – 1) is

A
7 units

B
25 units

C
5 units

D
3 units

Solution

Explanation:
ABC is a triangle with A(7 ,- 3), B(5 , 3) and C(3 , - 1)
Let median on BC bisects BC at D. (AD is given as the median)
Question 6
3 / -1

If (a, 0) , (0, b) and (x, y) are collinear, then

A
ay – bx = 1

B
ay + bx = ab

C
ax – by = ab

D
ax + by = 1

Solution

Explanation:
If given points are collinear, the area of triangle formed by these three points is 0.
Question 7
3 / -1

If the distance between the points (p, – 5) and (2, 7) is 13 units, then the value of ‘p’ is

A
– 3, 7

B
3, – 7

C
– 3, – 7

D
3, 7

Solution

Explanation:
Let point A be (p,−5) and point B (2, 7) and distance between A and B = 13 units
Question 8
3 / -1

The vertices of a quadrilateral are (1, 7), (4, 2), ( – 1, – 1) and ( – 4, 4). The quadrilateral is a

A
square

B
parallelogram

C
rhombus

D
rectangle

Solution

Explanation:
Let A (1, 7), B (4, 2), C(−1,−1) and D(−4,4) are the vertices of a quadrilateral ABCD.
Question 9
3 / -1

The point ( – 3, 5) lies in the ___________ quadrant

A
III

B
II

C
I

D
IV

Solution

Explanation:
Since x−coordinate is negative and y−coordinate is positive.Therefore, the point (−3,5) lies in II quadrant.
Question 10
3 / -1

A
0 units

B
p²units

C
1 units

D
p units

Solution

The sum is according to distance formula
= √(x₂-x₁)²+(y₂-y₁)²
= √(0-p cos 25)² + (p cos 65-0)
= √(p cos 25)² + (p cos 65)²
= √(p cos 25)² + (p sin 25)²
= √(p² cos²25) + (a² sin²25)
= √p²(cos²25 + sin²25) we know that sin²Θ + cos²Θ = 1
= √p²(1)
= √p²
= p Units
Question 11
3 / -1

The base of an equilateral triangle ABC lies on the y – axis. The co – ordinates of the point C is (0, – 3). If origin is the midpoint of BC, then the co – ordinates of B are

A
(0, 3)

B
(3, 0)

C
( – 3, 0)

D
(0, – 3)

Solution

Explanation:
Let the coordinate of B be (0,a).
It is given that (0, 0) is the mid-point of BC.
Therefore 0 = (0 + 0) /2 , 0 =(a - 3) /2 a - 3 = 0 , a = 3
Therefore, the coordinates of B are (0, 3).
Question 12
3 / -1

If the co – ordinates of a point are (3, – 7), then its ordinate is

A
– 7

B
3

C
– 3

D
7

Solution

Explanation:
Since y−coordinate of a point is called ordinate.its distance from the x-axis measured parralel to the y - axisTherefore, ordinate is −7.
Question 13
3 / -1

The values of ‘y’ for which the distance between the points (2, – 3) and (10, y) is 10 units is

A
– 3, – 9

B
3, – 9

C
3, 9

D
– 3, 9

Solution
Question 14
3 / -1

If A is point on the x – axis whose abscissa is 5 and B is the point (1, – 3), then the distance AB is

A
25 units

B
5 units

C
8 units

D
9 units

Solution

Explanation:
Since A is a point on the x - axis, therefore coordinates of A are (5 , 0)
Question 15
3 / -1

The point where the perpendicular bisector of the line segment joining the points A(2, 5) and B(4, 7) cuts is:

A
(0, 0)

B
(3, 6)

C
(6, 3)

D
(2, 5)

Solution

Explanation:
Since, the point, where the perpendicular bisector of a line segment joining the points A(2 , 5) and B(4 , 7) cuts, is the mid-point of that line segment.