Mathematics Class IX Questions Bank

Question 1 of 15

1. Question
1 point(s)
Where will you find all points with positive abssissa and negative ordinate ?
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Hint

We will find all points with positive abssissa and negative ordinate in quadrant IV.
Question 2 of 15

2. Question
1 point(s)
Write whether the following statements are true or false. Justify your answer.
(i) Point (5, 0) lies in the first quadrant.
(ii) The co-ordinates of the point whose ordinate is $-\dfrac{1}{2} $ and abscissa is 1 are $\left(-\dfrac{1}{2},1\right) $.
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Hint

(i) No, point $(5, 0)$ lies on the x-axis
(ii) No, the coordinates will be $\left(1,\dfrac{-1}{2}\right)$ as the coordinates are in the form $(x,y)$
Question 3 of 15

3. Question
1 point(s)
What is the perpendicular distance of the points A(7, – 4) from (i) x-axis (ii) y-axis ?
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Hint

$A(7,-4)$
(i) From x-axis it is 7 units in the +ve direction
(ii) From y-axis it is 4 units
Question 4 of 15

4. Question
1 point(s)
In figure, ABDC is a square. Find the co-ordinates of points A and D.
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Hint

$C(-1,-2),\ B(3,2)\\ \\ A(-1,2),\ D(3,-2)$
If we plot the given points in the graph and take the reflection of the points with respect to x-axis and y-axis we get A and D coordinates.
Question 5 of 15

5. Question
1 point(s)
In the figure, if $\triangle $ ABC and $\triangle $ ABD are equilateral, then find the co-ordinates of points C and D.
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Hint

$C(0,a),D(0,-a)$
As ABC and ABD are equilateral triangle. So all sides will be $a$
Question 6 of 15

6. Question
1 point(s)
In the figure, ABCD is a square of side 2a. Find the co-ordinates of the vertices of the square :
(i) Taking AB and AD as axes.
(ii) Taking the centre of the square as origin and axes parallel to the sides AB and AD.
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Hint

$A(0,0),B(2a,0)\\ \\ C(2a,2a)\ D(0,2a)$
(ii) Side $=2a$
Centre $=(0,0)$
$y$ Coordinate of $AB=\dfrac{(0-2a)}{2}=-a$
$y$ Coordinate of $AD=\dfrac{0+2a}{2}=a\\ \\ A=(-a,-a)\\ \\ B=(a,-a)\\ \\ C=(a,a)\\ \\ D=(-a,a)$
Question 7 of 15

7. Question
1 point(s)
Plot the following points and write the name of the figure thus obtained.
A (–3, 2), B (–7,– 3), C (6, – 3), D (2, 2)
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Hint

Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point A(-3,2) lies in II quadrant, B(-7,-3) lies in III quadrant, C(6, -3) lies in IV quadrant and D(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium ABCD.
Question 8 of 15

8. Question
1 point(s)
Plot the following points and check whether they are collinear or not.
(i) (0, 0), (2, 2), (5, 5)
(ii) (1, 3), (– 1, – 1), (–2, – 3)
(iii) (1, 1), (2, – 3), (–1, – 2)
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Hint

(i) From the graph, we find that all the three points lie on the same straight line. Hence, the given points are collinear.
(ii) From the graph, we find that all the three points lie on the same straight line. Hence, the given points are collinear.
(iii) From the graph, we find that all the three points do not lie on the same straight line. Hence, the given points are not collinear.
Question 9 of 15

9. Question
1 point(s)
In the figure, $\triangle $ PQR is an equilateral triangle with coordinates of vertices Q and R as (–2, 0) and (2, 0). Find the coordinates of the vertex P.
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Hint

vertices of $Q=(-2,0)$
$R=(2,0)$
$QR=2+2=4$
So $PQ=QR=RP=4$
$OP^2=PR^2-OR^2\\ \\ OP^2=4^2-2^2=12\\ \\ OP=2\sqrt3$
Coordinates of $P=(0,2\sqrt3)$
Question 10 of 15

10. Question
1 point(s)
The following table gives the number of pairs of shoes and their corresponding price. Plot these as ordered pairs and join them. What type of graph do you get ?
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Hint

We get bargraph
Question 11 of 15

11. Question
1 point(s)
In the figure, AOB is a triangle with co-ordinates of A and O as (4,0) and (0,0) respectively. AB = 5. Find the co-ordinates of B.
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Hint

$A=(4,0),\ O(0,0)\\ \\ AB=5$
In $\triangle AOB$
$AB^2=OA^2+OB^2\\ \\ 5^2=4^2+OB^2\\ \\ OB^2=9\\ \\ OB=3$
Coordinates of $B(0,3)$
Question 12 of 15

12. Question
1 point(s)
In the figure, $\triangle $ ABC is an equilateral triangle with co-ordinate of B and C as B (1, 0) and C (5, 0). Find the co-ordinate of vertex A.
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Hint

$B(1,0),\ C(5,0)\\ \\ BC=\sqrt{(5-1)^2+(0-0)^2}\\ \\ =\sqrt{4^2}=\sqrt{6}=4\\ \\ BC=4\\ \\ BD=3-1=2\\ \\ AB=4$
Question 13 of 15

13. Question
1 point(s)
Three vertices of a rectangle are (– 4, 5), (– 4, 2) and (3, 2). Plot these points and find the co-ordinates of the fourth vertex.
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Hint

$A(-4, 5),B(-4,2)$ and $C(3,2)$
Question 14 of 15

14. Question
1 point(s)
Points A (5, 3), B (–2, 3) and D (5, –4) are three vertices of a square ABCD. Plot these points on a graph paper and hence find the coordinates of the vertex C.
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Hint

Coordinates of C are $(-2,-4)$
Question 15 of 15

15. Question
1 point(s)
Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, one vertex at the origin, the longer side lies on the x-axis and one of the vertices lies in the third quadrant.
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Hint

Let the rectangle be OABC with the vertex O at the origin.
Let OA be the longer side.
$\therefore$ The co-ordinate of O is O(0,0).
Since one of the vertex is in the third quadrant and the longer side of length 5 units lie on the x-axis, the co-ordinate of A is A(−5,0).
Also, the rectangle lie in the III quadrant.
$\therefore$ The side OC will lie on the y-axis to the negative direction.
Then, the length of OC is 3 units.
$\therefore$ The co-ordinates of C will be C(0,−3).
Then obviously, the co-ordinates of B will be (−5,−3).
Here, OC || AB and $\angle$ AOC = $90^o$ , hence is a rectangle.
Therefore, the vertices of the rectangle are O(0,0), A(−5, 0), B(−5, −3), C(0, −3).