Coordinate Geometry - 1 - Free MCQ Practice Test with solutions,

2. I Quadrant

Explanation:

The point A(3, 4) lies in the I Quadrant because:

• In the I Quadrant, both the x-coordinate (abscissa) and the y-coordinate (ordinate) are positive.
• Since 3 (x-coordinate) is positive and 4 (y-coordinate) is also positive, the point lies in the I Quadrant.

We can see in diagram that distance of point P is 2 from y – axis in positive x – axis direction and 3 from x – axis but in negative y – axis direction.
Therefore, x – coordinate and y – coordinate of point p will be 2 and -3 respectively.
Coordinates of any points are shown in (x – coordinate, y – coordinate) form.
Therefore, coordinates of point P are (2, -3).

The point (7,0) lies on the positive direction of the x-axis.

Explanation:

• The x-coordinate of the point is 7, and the y-coordinate is 0.
• When the y-coordinate is 0, the point lies on the x-axis.
• Since the x-coordinate is positive, the point is on the positive direction of the x-axis.

So, the correct answer is:

4. on the positive direction of x-axis

• A point on the y-axis always has an x-coordinate of 0.
• Being 6 units in the positive direction means the y-coordinate is +6.
• Therefore, the correct coordinates are (0, 6).

• In a coordinate plane, the quadrants are defined as follows:
  
  
  • I Quadrant: (x > 0, y > 0)
  
  
  • II Quadrant: (x < 0, y > 0)
  
  
  • III Quadrant: (x < 0, y < 0)
  
  
  • IV Quadrant: (x > 0, y < 0)
  
   
• The point B(-3, 4) has x-coordinate -3 and y-coordinate 4.
• Since x is negative and y is positive, it lies in the II Quadrant.

A point of the form (0,b) lies on the y-axis because the x-coordinate is 0. This means the point is located directly on the y-axis at the value bbb.

- The coordinate point (0, -4) is given.
- The x-coordinate is 0, which means the point is not to the left or right of the origin.
- The y-coordinate is -4, indicating the point is 4 units below the origin.
- Points with x-coordinate 0 lie on the y-axis.
- Thus, the point (0, -4) lies on the y-axis.

The correct answer is B: on y-axis.

To understand this, let's break down the information given:

1. x-axis: The x-axis is the horizontal line in the coordinate plane. Any point on the x-axis will have a y-coordinate of 0 because it does not move up or down from the axis.

2. Distance of 4 units in the negative direction of x-axis:

   • The x-axis has a positive direction (towards the right) and a negative direction (towards the left).
   • Moving 4 units in the negative direction means moving 4 units to the left from the origin (0,0).

3. Coordinates:

   • When you move 4 units to the left on the x-axis, the x-coordinate becomes -4, and since the point is on the x-axis, the y-coordinate remains 0.

Therefore, the point that lies on the x-axis 4 units in the negative direction is (−4,0)(-4, 0)(−4,0).

This corresponds to option (b) (-4, 0).

The point D(3, -6) lies in the IV Quadrant.
Explanation: In the Cartesian plane, the X-axis and Y-axis divide the plane into four quadrants. The I Quadrant is where both x and y are positive. The II Quadrant is where x is negative and y is positive. The III Quadrant is where both x and y are negative. The IV Quadrant is where x is positive and y is negative.
The point D(3, -6) has a positive x-coordinate and a negative y-coordinate. Therefore, it lies in the IV Quadrant.

The abscissa of a point refers to its x-coordinate in the Cartesian coordinate system.

• On the x-axis, the y-coordinate (ordinate) is always 0, while the x-coordinate (abscissa) can be any real number.
• This means that any point on the x-axis is of the form (x, 0), where x can be any number.

Thus, the correct answer is:

c) any number.

The given three points are A(1,2) B(3,-4) and C(5,-6).

Let P (x, y) be the point equidistant from these three points.

So, PA = PB = PC

⇒ x² + 1– 2x + y² + 4 – 4y = x² + 9 -6x + y² + 16 + 8y = x 2 + 25– 10x + y² + 36 + 12y

⇒  – 2x– 4y + 5  = -6x + 8y +25=  – 10x + 12y+61

– 2x– 4y + 5  = -6x + 8y +25

⇒  – 2x– 4y + 5 +6x - 8y -25=0

⇒  4x– 12y -20=0

⇒  x– 3y - 5 =0....(i)

- 2x– 4y + 5  =  – 10x + 12y+61

⇒- 2x– 4y + 5 +10x - 12y-61=0

⇒8x– 16y -56=0

⇒x– 2y -7=0....(ii)

Solving (i) and (ii)

x = 11, y = 2

Thus, the required point is (11, 2)