Test: Coordinate Geometry - 2 - Grade 9 MCQ

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 (+,−)

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

√{1-(-4)}² +{(-5)-7}²

=√25+144

=√169

=13

 The correct answer is a.

Distance between two points (x1,y1) and (x2,y2) can be calculated using the formula −

Distance between the points (p,−5) and (2,7)=

(2−p)2  +(7−(−5)) 2 =13 

(2-p)2 + 144 = 13

(2-p)2 + 144 = 169

(2-p)2  = 25

2-p = 5  or 2-p = -5

p = -3 or p = 7

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 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