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Test: Coordinate Geometry - 2


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25 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Coordinate Geometry - 2

Test: Coordinate Geometry - 2 for Class 9 2022 is part of Mathematics (Maths) Class 9 preparation. The Test: Coordinate Geometry - 2 questions and answers have been prepared according to the Class 9 exam syllabus.The Test: Coordinate Geometry - 2 MCQs are made for Class 9 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Coordinate Geometry - 2 below.
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Test: Coordinate Geometry - 2 - Question 1

The point of intersection of X and Y axes is called :

Test: Coordinate Geometry - 2 - Question 2

The point (2, 3) is at a distance of _______________ units from x-axis :

Test: Coordinate Geometry - 2 - Question 3

The point (3, 2) is at a distance of _______________ units from y-axis :

Test: Coordinate Geometry - 2 - Question 4

The point (–3, 2) belongs to Quadrant _______________ :

Test: Coordinate Geometry - 2 - Question 5

The point (2, –3) belongs to quadrant _______________ :

Detailed Solution for Test: Coordinate Geometry - 2 - Question 5

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 (+,−)

Test: Coordinate Geometry - 2 - Question 6

The point (3, 2) belongs to quadrant _______________ :

Test: Coordinate Geometry - 2 - Question 7

The point (–2, –3) belongs to Quadrant :

Test: Coordinate Geometry - 2 - Question 8

The point (–2, 0) lies on :

Test: Coordinate Geometry - 2 - Question 9

The point (0, –2) lies on :

Test: Coordinate Geometry - 2 - Question 10

The point (3, 0) lies on :

Test: Coordinate Geometry - 2 - Question 11

The point (0, 3) lies on :

Detailed Solution for Test: Coordinate Geometry - 2 - Question 11 This is because y is also called X=0 and x=0,y=3 so it will lie on positive y axis(+ve y axis)
Test: Coordinate Geometry - 2 - Question 12

The distance between the points (–4, 7) and (1, –5) is :

Detailed Solution for Test: Coordinate Geometry - 2 - Question 12

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

√{1-(-4)}2 +{(-5)-7}2

=√25+144

=√169

=13

Test: Coordinate Geometry - 2 - Question 13

The distance of the point (–2, –2) from the origin is :

Detailed Solution for Test: Coordinate Geometry - 2 - Question 13 Solving it by Pythagoras theorem the perpendicular will be 2 and the base will also be 2 the distance will be the hypotenuse =>. h² = b² + p² h²= 2² + 2² h² = 4+4 h²= 8 h=√8 √8 can be written as √2 x √2 x √2 which is equal to 2√2units.
Test: Coordinate Geometry - 2 - Question 14

If the distance between points (p, –5), (2, 7) is 13 units, then p is _____________ :

Detailed Solution for Test: Coordinate Geometry - 2 - Question 14

 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

Test: Coordinate Geometry - 2 - Question 15

The points (a, a) (–a, a) and (– (√3) a, (√3)a) form the vertices of an :

Test: Coordinate Geometry - 2 - Question 16

Find the ratio in which the line joining the points (6, 4) and (1, –7) is divided by x-axis.

Detailed Solution for Test: Coordinate Geometry - 2 - Question 16

Test: Coordinate Geometry - 2 - Question 17

The points (2, –1), (3, 4), (–2, 3) and (–3, –2) taken in the order from the vertices of _________________:

Detailed Solution for Test: Coordinate Geometry - 2 - Question 17

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

Test: Coordinate Geometry - 2 - Question 18

The points (0, 0), (–2, 0) and (3, 0) _________________:

Test: Coordinate Geometry - 2 - Question 19

The point on Y-axis equidistant from (–3, 4) and (7, 6) is _________________:

Test: Coordinate Geometry - 2 - Question 20

The point on X-axis equidistant from (5, 4) and (–2, 3) is _________________:

Test: Coordinate Geometry - 2 - Question 21

The points on X-axis at a distance of 10 units from (11, –8) are :

Detailed Solution for Test: Coordinate Geometry - 2 - Question 21

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) .

Test: Coordinate Geometry - 2 - Question 22

The points on Y-axis at a distance of 13 units from point (–5, 7) are :

Detailed Solution for Test: Coordinate Geometry - 2 - Question 22
Let  P(0,y) be the required point on y−axis.
Now, distance between the point P(0,y) and A(−5,7) is 13 units.
Now, PA = 13 units
⇒√(0+5)^2 + (y−7)^2 = 13
⇒√25 + y2 + 49 − 14y = 13
⇒√y^2 + 74 − 14y = 13
⇒y^2 − 14y + 74 = 169
⇒y^2 − 14y − 95 = 0
⇒y^2 − 19y + 5y − 95 = 0
⇒y(y−19) + 5(y−19) = 0
⇒(y+5)(y−19) = 0⇒y+5 = 0   or  y−19 = 0
⇒y = −5   or  y = 19
So, the required points are :(0,−5) and (0,19)
Test: Coordinate Geometry - 2 - Question 23

The coordinates of the centre of a circle passing through (1, 2), (3, –4) and (5, –6) is _________________:

Detailed Solution for Test: Coordinate Geometry - 2 - Question 23

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

 

Test: Coordinate Geometry - 2 - Question 24

The points (–5, 6), (3, 0) and (9, 8) form the vertices of :

Test: Coordinate Geometry - 2 - Question 25

The points (4, 4), (3, 5) and (–1, –1) from the vertices of :

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