Test: Coordinate Geometry - 2 - Class 9 MCQ

The point at which X and Y axes intersect is known as the origin.
The origin is shown in the diagram below:

The origin is the point where the axes intersect. At this point the x-coordinate and y-coordinate both are zero.

The distance between the point (2, 3) and x-axis can be determined by assuming a point (2, 0) on x-axis.
The distance between 
Therefore, the distance between (2, 3) and (2, 0)​​​​​​​ 

Hence, the distance of the point (2, 3) from x-axis is 3 units.

The point (–3, 2) is located in the second quadrant. This is because the x coordinate is negative and the y coordinate is positive, which are the characteristics of points in Quadrant II.

The correct answer is: B: Q₂

The x value is POSITIVE which means that the point is to the right of the origin.

The y value is NEGATIVE which means that the point is below the origin.

The 4th quadrant is to the right and below the origin,(The bottom right quadrant)

All points in the 4th quadrant have the signs as (+,−)

The point (3, 2) is located in the first quadrant. This is because both the x coordinate and the y coordinate are positive, which are the characteristics of points in Quadrant I.
The correct answer is: A: Q₁

The point (-2, -3) is located in the third quadrant. This is because both the x coordinate and the y coordinate are negative, which are the characteristics of points in Quadrant III.
The correct answer is: C: Q₃

The point (−2, 0) lies on the negative x-axis because the x coordinate is negative and the y coordinate is zero, indicating a location along the x-axis to the left of the origin.

The correct answer is: C: –ve x-axis

The point (0, −2) lies on the negative y-axis because the x coordinate is zero and the y coordinate is negative, indicating a location along the y-axis below the origin.

The correct answer is: D: –ve y-axis

The point (3, 0) lies on the positive x-axis because the x coordinate is positive and the y coordinate is zero, indicating a location along the x-axis to the right of the origin.

The correct answer is: A: +ve x-axis

The point (0, 3) lies on the positive y-axis because the x coordinate is zero and the y coordinate is positive, indicating a location along the y-axis above the origin.

The correct answer is: B: +ve y-axis

Use the distance formula to determine the distance between two points.

√{1 - (-4)}² + {(-5) - 7}²

= √25 + 144

= √169

= 13

Given, the distance between points (p, -5) and (2, 7) is 13 units.
∵ The distance between two points (x₁, y₁) and (x₂, y₂) is 

⇒ (2 − p)² + (7 + 5)² = 132

⇒ (2 − p)² = 169 − 144

⇒ (2 − p)² = 25

⇒ 2 − p = ±5

case 1:
2 − p = 5 ⇒ p = −3

case 2:
2 − p = −5 ⇒ p = 7

Hence, p can be -3 or 7

The given points are let

⇒ ΔABC is an equilateral triangle.
Hence proved.

In rhombus, all sides are equal and diagonals are not equal.

Distance between two points = [(x2−x1)2 + (y2−y1)2]1/2

AB =  [(3−2)2 + (4+1)2]1/2 =(26)1/2

BC = [(3+2)2 + (3−4)2]1/2 = (26)1/2

CD = [(-3+2)2 + (-2-3)2]1/2 = (26)1/2

DA = [(-3-2)2 + (-2+1)2]1/2 = (26)1/2

AC = [(2+2)2 + (4)2]1/2 = 4(2)1/2

BD = [(-3-3)2 + (4+2)2]1/2 = 6(2)1/2

AB=BC=CD=DA All sides are equal

AC and BD Diagonals are not equal

The points (0, 0), (–2, 0) and (3, 0) all have their Y-coordinates as zero and different x-coordinates, indicating they are positioned along the x-axis. Additionally, because they are on the same line (the x-axis), they are collinear.

The correct answer is: D: Both A and B

The point on the Y-axis that is equidistant from the points (–3, 4) and (7, 6) can be found using the distance formula.

The distance between the two points is √(7- (-3))² + (6-4)² = √(10² + 2²) = √(100 + 4) = √104 = 10.2.

The point on the Y-axis that is equidistant from the two points is (0, 10.2).

The closest answer to this is (0, 15).

Let P(x, 0) be the point on the x-axis. Then as per the question we have

AP = 10

Hence, the points on the x-axis are (5, 0) and (17,0) .

Given three points (1,2),(3,−4)and(5,−6)

Let the coordinates o centre be O(h,k)

The distance from the centre to a point on the circumference is equal to radius.

∴ OA=OB=OC

From distance formula 

OA²=OB²

Let the given points be A(-5, 6), B(3, 0) and C(9, 8).

Therefore,  AB =  BC = units 

This show that  Δ ABC  is right angled at B. Therefore, the points A(-5,6), B(3,0) and C(9,8). are the vertices of an isosceles rightangled triangle
Also, area of a triangle = 1/2 x base x height
If AB is the height and BC is the base,
Area = 1/2 x 10 x 10
= 50 square units

The vertices of the given triangles are A(4, 4), B (3, 5) and C (- 1, - 1).

It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by m = (y₂ - y₁)/(x₂ - x₁), x2 ≠ x₁

Therefore, slope of AB (m₁) = (5 - 4)/(3 - 4) = - 1

Slope of BC (m₂) = (- 1 - 5)/(- 1 - 3) = - 6/- 4 = 3/2

Slope of CA (m₃) = (4 + 1)/(4 + 1) = 5/5 = 1

It is observed that m₁m₃ = - 1

This shows that line segments AB and CA are perpendicular to each other i.e., the given triangle is right-angled at A(4, 4).

Thus, the points (4, 4), (3, 5) and (- 1, - 1) are the vertices of a right-angled triangle