Rd Sharma 2022 for Class 9 Maths Chapter 8 - Coordinate Geometry

Answer:

The following points are given below. 

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Letand be the coordinate axes.

(i) Here for the given point the abscissa is 2 units and ordinate is 5 units.

The point is in the first quadrant. So it will look like as shown in the following figure.

(ii) Here for the given point the abscissa is 4 units and ordinate is -3 units.

The point is in the fourth quadrant. So it will look like as shown in the following figure.

(iii) Here for the given point the abscissa is -5 units and ordinate is -7 units.

The point is in the third quadrant. So it will look like as shown in the following figure.

(iv) Here for the given point the abscissa is 7 units and ordinate is -4 units.

The point is in the fourth quadrant. So it will look like as shown in the following figure.

(v) Here for the given point the abscissa is -3 units and ordinate is 2 units.

The point is in the second quadrant. So it will look like as shown in the following figure.

(vi) Here for the given point the abscissa is 7 units and ordinate is 0 units.

The point is on the x-axis. So it will look like as shown in the following figure.

(vii) Here for the given point the abscissa is -4 units and ordinate is 0 units.

The point is on the x-axis. So it will look like as shown in the following figure. 

(viii) Here for the given point the abscissa is 0 units and ordinate is 7 units.

The point is on the y-axis. So it will look like as shown in the following figure.

(ix) Here for the given point the abscissa is 0 units and ordinate is -4 units.

The point is on the y-axis. So it will look like as shown in the following figure.

(x) Here for the given point the abscissa is 0 units and ordinate is 0 units.

The point is basically intersection of the coordinate axes. So it will look like as shown in the following figure.

Answer:

The following graph is given in the question with the marked points and we are asked to write down their coordinates.

The distance of point A from y-axis is 3 units and that of from x-axis is 1 units. Since A lies in the first quadrant, so its coordinates are.

The distance of point B from y-axis is 6 units and that of from x-axis is 0 units. Since B lies on x-axis, so its coordinates are. 

The distance of point C from y-axis is 0 units and that of from x-axis is 6 units. Since C lies on y-axis, so its coordinates are.

The distance of point D from y-axis is -3 units and that of from x-axis is 0 units. Since D lies on x-axis, so its coordinates are.

The distance of point E from y-axis is -4 units and that of from x-axis is 3 units. Since E lies in the second quadrant, so its coordinates are.

The distance of point F from y-axis is -2 units and that of from x-axis is -4 units. Since F lies in the third quadrant, so its coordinates are.

The distance of point G from y-axis is 0 units and that of from x-axis is -5 units. Since G lies on y-axis, so its coordinates are.

The distance of point H from y-axis is 3 units and that of from x-axis is -6 units. Since H lies in the fourth quadrant, so its coordinates are.

The distance of point P from y-axis is 7 units and that of from x-axis is -3 units. Since P lies in the fourth quadrant, so its coordinates are.

The distance of point Q from y-axis is 7 units and that of from x-axis is 6 units. Since Q lies in the first quadrant, so its coordinates are.

Answer:

Given that, P(–3, 2), Q(–7, –3), R(6, –3), S(2, 2).
On plotting the above points, we have


From the figure,
$\mathrm{PS}\parallel \mathrm{QR}$
Hence, PQRS is a trapezium.

Answer:

(i) Given points: (1, 3), (–1, –1), (–2, –3)

The points (1, 3), (–1, –1), (–2, –3) lie in a straight line.

Hence, the points are collinear.

(ii) Given points: (1, 1), (2, –3), (–1, –2)

The points (1, 1), (2, –3), (–1, –2) do not lie in a straight line.

Hence, the points are not collinear.

(iii) Given points: (0, 0), (2, 2), (5, 5)

The points (0, 0), (2, 2), (5, 5)lie in a straight line.

Hence, the points are collinear.

Answer:

(i) Given point is $\left(-3,5\right)$
Thus, the point lies in the II quadrant.

(ii) Given point is $\left(-5,-3\right)$
Thus, the point lies in the III quadrant.

(iii) Given point is $\left(-5,3\right)$
Thus, the point lies in the II quadrant.

(iv) Given point is $\left(3,5\right)$
Thus, the point lies in the I quadrant.

Answer:

We have,
(–3, 5) : abscissa is – 3 and ordinate is 5
Thus the given point lies in the II quadrant.

(4, –1) : abscissa is 3 and ordinate is –1
Thus the given point lies in the IV quadrant.

(2, 0) : abscissa is 2 and ordinate is 0
Thus the given point lies in the X-axis

(–3, –6) : abscissa is –3 and ordinate is –6
Thus the given point lies in the III quadrant.






 

Answer:

We have,
A(1, 1), B(1, 0), C(0, 1), D(0, 0), E(0, –1), F(–1, 0), G(0, 5), H(–7, 0), I(3, 3)

Now, a point will lie on the y-axis if its abscissa is zero.

Thus, points C(0, 1), D(0, 0), E(0, –1), G(0, 5) will lie on y-axis.
 

Answer:

The given point lies in the positive direction of the x-axis.
So its y- coordinate will be zero and it is at a  distance of 7 units from the y - axis, so its coordinates are (7, 0).

If it lies in the negative direction of the y-axis, then its x-coordinate will be zero and its distance from the x-axis is 7 units, so its coordinate is (0, −7).

Answer:

Since ABCD is a square, i.e., AB = BC = CD = DA.
Therefore, the abscissa of C should be equal to the abscissa of B that is, −2, and the ordinate of C should be equal to the ordinate of D which is −4.
Hence, the coordinates of C are (−2,–4).

Answer:

Since PQRS is a square, i.e., PQ = QR = RS = SP.

Now, the abscissa of R should be equal to the abscissa of Q that is, 4, and the ordinate of R should be equal to the ordinate of S which is 3.
Hence, the coordinates of R are (4, 3).

Answer:

Based on the information given in the question, we have



Thus, the coordinates of the vertices will be O(0, 0), A(−5, 0), B(−5, −3), C(0, −3).

Answer:

Answer:

Given Points: A(1, 3), B(–3, –1), C(1, –4), D(–2, 3), E(0, –8), F (1, 0)

Answer:

If we take any point on the y-axis, then the distance of this point from the y-axis is 0. Therefore, the abscissa of this point is 0.

The co-ordinate of a point on the y-axis are of the form (0, y). Thus, the abscissa of all points on the y-axis is 0.

Abscissa of all the points on y-axis is _____0_____.

Answer:

If we take any point on the x-axis, then the distance of this point from the x-axis is 0. Therefore, the ordinate of this point is 0. 

​The co-ordinate of a point on the x-axis are of the form (x, 0). Thus, the ordinate of all points on the x-axis is 0.

Ordinate of all the points on x-axis is _____0_____.

Answer:

The given point is (–7, 0).

We know that, the co-ordinates of a point on the x-axis are of the form (x, 0).

If x > 0, then the point (x, 0) lies on the positive direction of the x-axis.

If x < 0, then the point (x, 0) lies on the negative direction of the x-axis.

The given point (–7, 0) is of the form (x, 0), where x < 0. Thus, the point (–7, 0) lies on the negative direction of the x-axis.


Point (–7, 0) lies on the __negative__ direction of  __x-__ axis.

Answer:

The given point is (0, –3).

We know that, the co-ordinates of a point on the y-axis are of the form (0, y).

If y > 0, then the point (0, y) lies on the positive direction of the y-axis.

If y < 0, then the point (0, y) lies on the negative direction of the y-axis.

The given point (0, –3) is of the form (0, y), where y < 0. Thus, the point (0, –3) lies on the negative direction of the y-axis.


Point (0, –3) lies on the __negative__ direction of  __y-__ axis.

Answer:

The point of intersection of the coordinate axes is called the origin.

The point at which the two coordinate axes meet is called the __origin__.

Answer:

In the third quadrant, x < 0, y < 0. Thus, the point whose abscissa and ordinate both are negative lies in third quadrant.

A point whose abscissa and ordinate both are negative lies in __third quadrant__.

Answer:

The y-coordinate of every point on x-axis is 0. So, the co-ordinates of any point on x-axis are of the form (x, 0).

If y-coordinate of a point is zero, then it always lies on __x-__ axis.

Answer:

The given points are (–3, 2) and (2, –3).

In II quadrant: x < 0, y > 0

So, the point (–3, 2) lie in the second quadrant.

In IV quadrant: x > 0, y < 0

So, the point (2, –3) lie in the fourth quadrant.

The points (–3, 2) and (2, –3) lie in __second__ and __fourth__ quadrants respectively.

Answer:

We know that, the co-ordinates of a point on the y-axis are of the form (0, y). The point (0, y) lies on the negative direction of the y-axis if y is negative.

The point lies on the y-axis so its x-coordinate is 0. Also, the point is at a distance of 5 units in the negative direction of y-axis so its y-coordinate is −5. Thus, the point is (0, −5).

The point which lies on y-axis at a distance of 5 units in the negative direction of y-axis has the coordinates __(0, −5)__.

Answer:

The given points are P(5, 1), Q(8, 0), R(0, 4), S(0, 5) and O(0, 0).  These points can be plotted on the graph paper as shown below.



It can be seen that, the points Q(8, 0) and O(0, 0) lie on the x-axis.

If P(5, 1), Q(8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the point(s) on the x-axis are __Q(8, 0) and O(0, 0)__.

Answer:

The point which lies on both x and y-axes is the point of intersection of x-axis and y-axis. The point of intersection of x-axis and y-axis is the origin. The coordinates of the origin are (0, 0).

The coordinates of the point which lies on x and y-axes both are ___(0, 0)___.

Answer:

The ordinate of the point is –4.

So, y = –4

Now, the coordinates of a point on y-axis are of the form (0, y).

Thus, the coordinates of the point whose ordinate is –4 and which lies on y-axis are (0, –4).

The coordinates of the point whose ordinate is –4 and which lies on y-axis, are ___(0, –4)___.

Answer:

The abscissa of the point is 5.

So, x = 5

Now, the coordinates of a point on x-axis are of the form (x, 0).

Thus, the coordinates of the point whose abscissa is 5 and which lies on x-axis are (5, 0).

The coordinates of point whose abscissa is 5 and which lies on x-axis, are ___(5, 0)___.

Answer:

Under reflection of a point in the x-axis, the abscissa of the point remains unchanged while the sign of the ordinate is changed. So, the image of the point (x, y) in the x-axis is (x, −y).

Thus, the image of the point (–3, –2) in the x-axis is (–3, 2).

In II quadrant: x < 0, y > 0

In the point (–3, 2), abscissa is negative and ordinate is positive. So, this point lies in the second quadrant.

The image of the point (–3, –2) in x-axis lies in __second__ quadrant.

Answer:

Under reflection of a point in the y-axis, the ordinate of the point remains unchanged while the sign of the abscissa is changed. So, the image of the point (x, y) in the y-axis is (−x, y).

If a > 0 and b < 0, then the point (a, b) lies in the fourth quadrant.

The coordinates of the image of point (a, b), where a > 0 and b < 0, in the y-axis are of the form (a, b), where a < 0 and b < 0.

Now, for a < 0 and b < 0, the point (a, b) lies in the third quadrant.

If a > 0 and b < 0, then the image of (a, b) in y-axis lies in __third__ quadrant.

Answer:

The given points are O(0, 0), A(5, 0) and B(0, 5).  These points can be plotted on the graph paper as shown below.



Here, OA = 5 units and OB = 5 units

∴ OA = OB

Hence, the points O(0, 0), A(5, 0) and B(0, 5) form an isosceles right ∆OAB right angled at O.

The points O(0, 0), A(5, 0) and B(0, 5) are joined in order to form a/an __isosceles right__ triangle.

Answer:

The given points are O(0, 0), A(7, 0), B(7, 4) and C(0, 4).  These points can be plotted on the graph paper as shown below.



From the figure, we have

OA = 7 units, AB = 4 units, BC = 7 units and OC = 4 units

In quadrilateral OABC,

OA = BC = 7 units and AB = OC = 4 cm

Thus, the quadrilateral OABC is a rectangle.

The points O(0, 0), A(7, 0), B(7, 4) and C(0, 4) form a ___rectangle___.

Answer:

The given points are O(0, 0), A(6, 0) and B(0, 4). These points can be plotted on the graph paper as shown below.



So, the points O(0, 0), A(6, 0) and B(0, 4) form right ∆OAB right angled at O.

From the figure, we have

OA = 6 units and OB = 4 units

∴ Area of ∆OAB = $\frac{1}{2}$ × OA × OB = $\frac{1}{2}$ × 6 × 4 = 12 square units


The points O(0, 0), A(6, 0) and B(0, 4) form a __right__ triangle of area ___12___ sq. units.