Coordinate Geometry Test

Result

Coordinate Geometry Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

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

A
169 units

B
12 units

C
13 units

D
144 units

Solution
Question 2
1 / -0

The triangle whose vertices are ( – 3, 0), (1, – 3) and (4, 1) is _____________ triangle.

A
equilateral

B
right angled isosceles

C
scalene

D
Obtuse triangle

Solution
Question 3
1 / -0

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
Question 4
1 / -0

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

A
25 units

B
5 units

C
1 unit

D
7 units

Solution
Question 5
1 / -0

A circle has its centre at the origin and a point P(5, 0) lies on it. Then the point Q(8, 6) lies ________ the circle.

A
out side

B
on

C
inside

D
none of these

Solution
Question 6
1 / -0

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 7
1 / -0

The distance between the points (a, b) and ( – a, b) is

A
2a units

B

C

D
2b

Solution
Question 8
1 / -0

If the point P(2, 4) lies on a circle, whose centre is C(5, 8), then the radius of the circle is

A
5 units

B
4 units

C
8 units

D
25 units

Solution
Question 9
1 / -0

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

A
2

B
1

C
– 1

D
0

Solution

Let point C (0, 2) is equidistant from the points A(3,k) and B(k,5).

i.e. AC = BC
∴AC²= BC²

⇒(3−0)²+(k−2)²= (k−0)²+ (5−2)²
• ⇒9 + k²+ 4 − 4k = k²+9

⇒4k = 4

⇒k = 1
Question 10
1 / -0

The vertices of a square are (0, – 1), (2, 1), (0, 3) and ( – 2, 1). The side of the square is

A
2√2 units

B
√2 units

C
2 units

D
2√3 units

Solution
Question 11
1 / -0

The distance between the points (x₁, y₁) and (x₂, y₂) is given by

A
√[(x₂ - x₁)² + (y2 - y1)²] units

B
√[(x₂ + x₁)² + (y₂ + y₁)²] units

C
√[(x₂ + x₁)² - (y₂ + y₁)²] units

D
√[(x₂ - x₁)² - (y₂ - y₁)²] units

Solution
Question 12
1 / -0

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

A
( – 7, 0)

B
(0, 7)

C
(0, – 7)

D
(7, 0)

Solution

Let A(2, −5) and B(−2, 9).
Since point is on x−axis C(x,0).
∴AC2 = BC2
⇒(2−x) 2 + (−5−0) 2 = (−2−x)2 + (9−0)²
⇒4 + x2 − 4x + 25 = 4 + x2 + 4x + 81
⇒−8x = 56
⇒x = −7
Therefore, the point on x−axis is (−7,0).
Question 13
1 / -0

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

A
(0, 9)

B
(0, – 9)

C
(9, 0)

D
( – 9, 0)

Solution

Let one point is A(6, 5) and 2nd point is B(−4, 3).
3rd point on y-axis is C (0, y).

since C is equidistant from A and B , i.e. AC = BC
∴AC2 = BC2
⇒(6 − 0)²+(5−y)²= (−4 − 0)²+(3 − y)²
⇒36 + 25 + y² − 10y = 16 + 9 + y2 − 6y
⇒−4y = −36
⇒y = 9
Therefore, the point on y−axis is (0, 9).
Question 14
1 / -0

If the point (x, y) is equidistant from the points (5, 1) and ( – 1, 5), then the relation between ‘x’ and ‘y’ is given by

A
– 2x = 3y

B
3x = – 2y

C
3x = 2y

D
2x = 3y

Solution

Let the point C(x,y) is equidistant from the points A (5, 1) and B(−1,5).

i.e. AC = BC
∴AC² = BC²

⇒(5−x)²+ (1−y)²= (−1−x)²+ (5−y)²
⇒25 + x2 − 10x + 1 + y2 − 2y = 1+ x2 + 2x + 25 + y2 − 10y
⇒−10x − 2y − 2x + 10 y = 26 − 26
⇒−12x + 8y = 0
⇒−12x = −8y
⇒3x = 2y
Question 15
1 / -0

The distance of the point P(x, y) from the origin O(0, 0) is given by

A
√[(x - y)²] units

B
√[(x² - y²)²] units

C
√[(x + y)²] units

D
√[(x² + y²)] Units

Solution