## Selina Concise Mathematics Class 9 ICSE Solutions Co-ordinate Geometry

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Selina ICSE Solutions for Class 9 Maths Chapter 26 Co-ordinate Geometry

Exercise 26(A)

Solution 1:

Solution 2:

On the graph paper, let us draw the co-ordinate axes XOX’ and YOY’ intersecting at the origin O. With proper scale, mark the numbers on the two co-ordinate axes.
Now for the point A(8,7)
Step I
Starting from origin O, move 8 units along the positive direction of X axis, to the right of the origin O
Step II
Now from there, move 7 units up and place a dot at the point reached. Label this point as A(8,7)
Similarly plotting the other points

Solution 3:

Solution 4:
(i) The abscissa is 2
Now using the given graph the co-ordinate of the given point A is given by (2,2)
(ii) The ordinate is 0
Now using the given graph the co-ordinate of the given point B is given by (5,0)
(iii) The ordinate is 3
Now using the given graph the co-ordinate of the given point C and E is given by (-4,3)& (6,3)
(iv) The ordinate is -4
Now using the given graph the co-ordinate of the given point D is given by (2,-4)
(v) The abscissa is 5
Now using the given graph the co-ordinate of the given point H, B and G is given by (5,5) ,(5,0) & (5,-3)
(vi)The abscissa is equal to the ordinate.
Now using the given graph the co-ordinate of the given point I,A & H is given by (4,4),(2,2) & (5,5)
(vii)The ordinate is half of the abscissa
Now using the given graph the co-ordinate of the given point E is given by (6,3)

Solution 5:
(i)The ordinate of a point is its x-co-ordinate.
False.
(ii)The origin is in the first quadrant.
False.
(iii)The y-axis is the vertical number line.
True.
(iv)Every point is located in one of the four quadrants.
True.
(v)If the ordinate of a point is equal to its abscissa; the point lies either in the first quadrant or in the second quadrant.
False.
(vi)The origin (0,0) lies on the x-axis.
True.
(vii)The point (a,b) lies on the y-axis if b=0.
False

Solution 6:

Solution 7:

After plotting the given points A(2,0), B(8,0) and C(8,4) on a graph paper; joining A with B and B with C. From the graph it is clear that the vertical distance between the points B(8,0) and C(8,4) is 4 units, therefore the vertical distance between the points A(2,0) and D must be 4 units. Now complete the rectangle ABCD
As is clear from the graph D(2,4)
(ii)A(4,2), B(-2,-2) and D(4,-2)

After plotting the given points A(4,2), B(-2,2) and D(4,-2) on a graph paper; joining A with B and A with D. From the graph it is clear that the vertical distance between the points A(4,2) and D(4,-2) is 4 units and the horizontal distance between the points A(4,2) and B(-2,2) is 6 units , therefore the vertical distance between the points B(-2,2)and C must be 4 units and the horizontal distance between the points B(-2,2) and C must be 6 units. Now complete the rectangle ABCD
As is clear from the graph C(-2,2)

Solution 8:

After plotting the given points A(2,-2), B(8,2) and C(4,-4) on a graph paper; joining B with C and B with A . Now complete the parallelogram ABCD.
As is clear from the graph D(-6,4)
Now from the graph we can find the mid points of the sides AB and CD.
Therefore the co-ordinates of the mid-point of AB is E(3,2) and the co-ordinates of the mid-point of CD is F(-1,-4)

Solution 9:

Solution 10:

Solution 11:

Solution 12:

Solution 13:

Exercise 26(B)

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Solution 7:

Solution 8:

Solution 9:

Solution 10:

Solution 11:

Solution 12:

Solution 13:

Exercise 26(C)

Solution 1:
The angle which a straight line makes with the positive direction of x-axis (measured in anticlockwise direction) is called inclination o the line.
The inclination of a line is usually denoted by θ
(i)The inclination is θ = 45°
(ii) The inclination is θ = 135°
(iii) The inclination is θ = 30°

Solution 2:
(i)The inclination of a line parallel to x-axis is θ = 0°
(ii)The inclination of a line perpendicular to x-axis is θ = 90°
(iii) The inclination of a line parallel to y-axis is θ = 90°
(iv) The inclination of a line perpendicular to y-axis is θ = 0°

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Solution 7:

Solution 8:
Given line is 3x + 4y = 12
The graph of the given line is shown below.

Clearly from the graph we can find the y-intercept.
The required y-intercept is 3.

Solution 9:
Given line is
2x – 3y – 18 = 0
The graph of the given line is shown below.

Clearly from the graph we can find the y-intercept.
The required y-intercept is -6

Solution 10:
Given line is
x + y = 5
The graph of the given line is shown below.

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## What Is The Cartesian Coordinate System

In Cartesian co-ordinates the position of a point P is determined by knowing the distances from two perpendicular lines passing through the fixed point. Let O be the fixed point called the origin and XOX’ and YOY’, the two perpendicular lines through O, called Cartesian or Rectangular co-ordinates axes.
Draw PM and PN perpendiculars on OX and OY respectively. OM (or NP) and ON (or MP) are called the x-coordinate (or abscissa) and y-coordinate (or ordinate) of the point P respectively.

1. Axes of Co-ordinates
In the figure OX and OY are called as x-axis and y-axis respectively and both together are known as axes of co-ordinates.
2. Origin
It is point O of intersection of the axes of co-ordinates.
3. Co-ordinates of the Origin
It has zero distance from both the axes so that its abscissa and ordinate are both zero. Therefore, the coordinates of origin are (0, 0).
4. Abscissa
The distance of the point P from y-axis is called its abscissa. In the figure OM = PN is the Abscissa.
5. Ordinate
The distance of the point P from x-axis is called its ordinate. ON = PM is the ordinate in the figure.
The axes divide the plane into four parts. These four parts are called quadrants. So, the plane consists of axes and quadrants. The plane is called the cartesian plane or the coordinate plane or the xy-plane. These axes are called the co-ordinate axes.
A quadrant is 1/4 part of a plane divided by co-ordinate axes.

(i) XOY is called the first quadrant
(ii) YOX’ the second.
(iii) X’OY’ the third.
(iv) Y’OX the fourth
as marked in the figure.

RULES OF SIGNS OF CO-ORDINATES

1. In the first quadrant, both co-ordiantes i.e., abscissa and ordinate of a point are positive.
2. In the second quadrant, for a point, abscissa is negative and ordinate is positive.
3. In the third quadrant, for a point, both abscissa and ordinate are negative.
4. In the fourth quadrant, for a point, the abscissa is positive and the ordinate is negative.

## Cartesian Coordinate System Example Problems With Solutions

Example 1:    From the adjoining figure find

(i) Abscissa
(ii) Ordinate
(iii) Co-ordinates of a point P
Solution:    (i) Abscissa = PN = OM = 3 units
(ii) Ordinate = PM = ON = 4 units
(iii) Co-ordinates of the point P = (Abscissa, ordinate) = (3, 4)

Example 2:    Determine

(i) Abscissa (ii) ordinate (iii) Co-ordinates of point P given in the following figure.
Solution:    (i) Abscissa of the point P = – NP = –OM = – a
(ii) Ordinate of the point P = MP = ON = b
(iii) Co-ordinates of point P = (abscissa, ordinate)
= (–a, b)

Example 3:    Write down the (i) abscissa (ii) ordinate (iii) Co-ordinates of P, Q, R and S as given in the figure.

Solution:    Point P
Abscissa of P = 2; Ordinate of P = 3
Co-ordinates of P = (2, 3)
Point Q
Abscissa of Q = – 2; Ordinate of Q = 4
Co-ordinate of Q = (–2, 4)
Point R
Abscissa of R = – 5; Ordinate of R = – 3
Co-ordinates of R = (–5, –3)
Point S
Abscissa of S = 5; Ordinate of S = – 1
Co-ordinates of S = (5, – 1)

Example 4:    Draw a triangle ABC where vertices A, B and C are (0, 2), (2, – 2), and (–2, 2) respectively.
Solution:    Plot the point A by taking its abscissa O and ordinate = 2.
Similarly, plot points B and C taking abscissa 2 and –2 and ordinates – 2 and 2 respectively. Join A, B and C. This is the required triangle.

Example 5:    Draw a rectangle PQRS in which vertices P, Q, R and S are (1, 4), (–5, 4), (–5, –3) and
(1, – 3) respectively.
Solution:    Plot the point P by taking its abscissa 1 and ordinate – 4.
Similarly, plot the points Q, R and S taking abscissa as –5, –5 and 1 and ordinates as 4, – 3 and –3 respectively.
Join the points PQR and S. PQRS is the required rectangle.

Example 6:    Draw a trapezium ABCD in which vertices A, B, C and D are (4, 6), (–2, 3), (–2, –5) and
(4, –7) respectively.
Solution:    Plot the point A taking its abscissa as 4 and ordinate as 6.
Similarly plot the point B, C and D taking abscissa as – 2, –2 and 4 and ordinates as 3, – 5, and –7 respectively. Join A, B, C and D ABCD is the required trapezium.

## Mirror Image Of Coordinates Of A Point

With respect to x axis:
In image of a point P(x, y) with respect to x axis, the change in only sign of ordinate of point so the image is Q(x, –y).

P1 (–3, 1) & P2 (5, –2) are two points in II & IV quadrant respectively then their images are
Q1 (–3, –1) & Q2 (5, 2) in III & I quadrant respectively.

With respect to y axis:
In image of a point P(x, y) the change in sign of its abscissa so the image is Q(–x, y).

## Mirror Image Of Coordinates Of A Point Example Problems With Solutions

Example 1:    Find the images of the following points with respect to x axis,
(1, 2), (3/8, 4/3), (-2/3, 3),  (2, 5), (5, 0), (0, 7), (– 3, – 4)
Solution:

Example 2:    Find the images, of points (0, 0), (3, 0), (0, 2), (5, 1), (–2, 3), (–3, –3), (6, –7) with respect to y axis.
Solution:    The images are (0, 0), (–3, 0), (0, 2), (–5, 1),
(2, 3), (3, –3), (–6, –7) respectively.