Selina Solutions for Class 9 Maths Chapter 26 Coordinate Geometry are provided here. Since these concepts are continued in Class 10, it is highly important to understand the concepts taught in Class 9 in depth. To score good marks in Class 9 Mathematics examination, students are advised to solve questions provided in each exercise in the book by Selina publication. This Selina solutions for Class 9 Maths helps students in understanding the concepts better. Download pdf of Class 9 Maths Chapter 26 Selina Solutions from the link given below.
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Exercise 26A page: 315
1. For each equation given below; name the dependent and independent variables.
Solution:
(i)
y is the dependent variable
x is the independent variable
(ii) x = 9y + 4
x is the dependent variable
y is the independent variable
(iii)
x is the dependent variable
y is the independent variable
(iv)
y is the dependent variable
x is the independent variable
2. Plot the following points on the same graph paper:
(i) (8, 7)
(ii) (3, 6)
(iii) (0, 4)
(iv) (0, 4)
(v) (3, 2)
(vi) (2, 5)
(vii) (3, 0)
(viii) (5, 0)
(ix) (4, 3)
Solution:
Consider the points as
(i) (8, 7) = A
(ii) (3, 6) = B
(iii) (0, 4) = C
(iv) (0, 4) = D
(v) (3, 2) = E
(vi) (2, 5) = F
(vii) (3, 0) = G
(viii) (5, 0) = H
(ix) (4, 3) = I
3. Find the values of x and y if:
(i) (x – 1, y + 3) = (4, 4)
(ii) (3x + 1, 2y – 7) = (9, 9)
(iii) (5x – 3y, y – 3x) = (4, 4)
Solution:
We know that two ordered pairs are equal.
(i) (x – 1, y + 3) = (4, 4)
It can be written as
x – 1 = 4 and y + 3 = 4
x = 5 and y = 1
(ii) (3x + 1, 2y – 7) = (9, 9)
It can be written as
3x + 1 = 9 and 2y – 7 = 9
3x = 8 and 2y = 2
x = 8/3 and y = 1
(iii) (5x – 3y, y – 3x) = (4, 4)
It can be written as
5x – 3y = 4 ….. (1)
y – 3x = 4 ….. (2)
By multiplying equation (2) by 3
3y – 9x = – 12 …… (3)
Now add equations (1) and (3)
(5x – 3y) + (3y – 9x) = 4 + (12)
– 4x = 8
x = 2
Substituting the value of x in equation (2)
y – 3x = 4
y = 3x – 4
y = 3 (2) – 4
y = 2
Therefore, x = 2 and y = 2.
4. Use the graph given alongside, to find the coordinates of point (s) satisfying the given condition:
(i) The abscissa is 2.
(ii) The ordinate is 0.
(iii) The ordinate is 3.
(iv) The ordinate is 4.
(v) The abscissa is 5.
(vi) The abscissa is equal to the ordinate.
(vii) The ordinate is half of the abscissa.
Solution:
(i) The abscissa is 2.
Based on the graph,
The coordinate of the point A is given by (2, 2).
(ii) The ordinate is 0.
Based on the graph,
The coordinate of the point B is given by (5, 0).
(iii) The ordinate is 3.
Based on the graph,
The coordinates of the points C and E are given by (4, 3) and (6, 3).
(iv) The ordinate is 4.
Based on the graph,
The coordinate of the point D is given by (2, 4).
(v) The abscissa is 5.
Based on the graph,
The coordinates of the points H, B and G are given by (5, 5), (5, 0) and (5, 3).
(vi) The abscissa is equal to the ordinate.
Based on the graph,
The coordinates of the points I, A and H are given by (4, 4), (2, 2) and (5, 5).
(vii) The ordinate is half of the abscissa.
Based on the graph,
The coordinate of the point E is given by (6, 3).
5. State true or false:
(i) The ordinate of a point is its x coordinate.
(ii) The origin is in the first quadrant.
(iii) The yaxis is the vertical number line.
(iv) Every point is located in one of the four quadrants.
(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.
(vi) The origin (0, 0) lies on the xaxis.
(vii) The point (a, b) lies on the yaxis if b = 0.
Solution:
(i) False
(ii) False
(iii) True
(iv) True
(v) False
(vi) True
(vii) False
6. In each of the following, find the coordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation:
Solution:
(i)
We know that
3 – 2x = 7
3 – 7 = 2x
– 4 = 2x
x = – 2
Similarly
2y + 1 – 10 – 2 ½ y
2y + 1 = 10 – 5/2 y
By cross multiplication
4y + 2 = 20 – 5y
4y + 5y = 20 – 2
9y = 18
y = 2
Hence, the coordinates of the point are (2, 2).
(ii)
We know that
2a/3 – 1 = a/2
2a/3 – a/2 = 1
Taking LCM
(4a – 3a)/ 6 = 1
a = 6
Similarly
By taking LCM
45 – 12b = 14b – 7
45 + 7 = 14b + 12b
52 = 26b
b = 2
Hence, the coordinates of the point are (6, 2)
(iii)
We know that
5x – (5 – x) = ½ (3 – x)
It can be written as
(5x + x) – 5 = ½ (3 – x)
By cross multiplication
12x – 10 = 3 – x
12x + x = 3 + 10
13x = 13
x = 1
Similarly
By cross multiplication
12 – 9y = 4 + y
12 – 4 = y + 9y
8 = 10y
y = 8/10
y = 4/5
Hence, the coordinates of the point are (1, 4/5).
7. In each of the following, the coordinates of the three vertices of a rectangle ABCD are given. By plotting the given points; find, in each case, the coordinates of the fourth vertex:
(i) A (2, 0), B (8, 0) and C (8, 4).
(ii) A (4, 2), B (2, 2) and D (4, 2).
(iii) A (4,6), C (6, 0) and D (4, 0).
(iv) B (10, 4), C (0, 4) and D (0, 2).
Solution:
(i) A (2, 0), B (8, 0) and C (8, 4)
From the graph the coordinates of the fourth vertex is D (2, 4).
(ii) A (4, 2), B (2, 2) and D (4, 2).
From the graph the coordinates of the fourth vertex is C (2, 2).
(iii) A (4,6), C (6, 0) and D (4, 0).
From the graph the coordinates of the fourth vertex is B (6, 6).
(iv) B (10, 4), C (0, 4) and D (0, 2)
From the graph the coordinates of the fourth vertex is A (10, 2).
8. A (2, 2), B (8, 2) and C (4, 4) are the vertices of a parallelogram ABCD. By plotting the given points on a graph paper; find the coordinates of the fourth vertex D.
Also, form the same graph, state the coordinates of the midpoints of the sides AB and CD.
Solution:
It is given that
A (2, 2), B (8, 2) and C (4, 4) are the vertices of the parallelogram ABCD
By joining A, B, C and D we get the parallelogram ABCD.
From the graph, we get D (6, 4)
Using the graph,
The coordinates of the midpoint of AB is E (3, 2)
The coordinates of the midpoint of CD is F (1, 4)
9. A (2, 4), C (4, 10) and D (2, 10) are the vertices of a square ABCD. Use the graphical method to find the coordinates of the fourth vertex B. Also, find:
(i) The coordinates of the midpoint of BC;
(ii) The coordinates of the midpoint of CD and
(iii) The coordinates of the point of intersection of the diagonals of the square ABCD.
Solution:
It is given that
A (2, 4), C (4, 10) and D (2, 10) are the vertices of a square ABCD.
From the graph, we get B (4, 4)
Using the graph,
The coordinates of the midpoint of BC is E (4, 7)
The coordinates of the midpoint of CD is F (1, 10)
The coordinates of the diagonals of the square is G (1, 7)
10. By plotting the following points on the same graph paper. Check whether they are collinear or not:
(i) (3, 5), (1, 1) and (0, 1)
(ii) (2, 1), (1, 4) and (4, 1)
Solution:
After plotting the points, we clearly see from the graph that
(i) A (3, 5), B (1, 1) and C (0, 1) are collinear
(ii) P (2, 1), Q (1, 4) and R (4, 1) are noncollinear.
Exercise 26B page: 320
1. Draw the graph for each linear equation given below:
(i) x = 3
(ii) x + 3 = 0
(iii) x – 5 = 0
(iv) 2x – 7 = 0
(v) y = 4
(vi) y + 6 = 0
(vii) y – 2 = 0
(viii) 3y + 5 = 0
(ix) 2y – 5 = 0
(x) y = 0
(xi) x = 0
Solution:
(i)
x 
3 
3 
3 
y 
1 
0 
1 
(ii)
x 
3 
3 
3 
y 
1 
0 
1 
(iii)
x 
5 
5 
5 
y 
1 
0 
1 
(iv)
x 
7/2 
7/2 
7/2 
y 
1 
0 
1 
(v)
x 
1 
0 
1 
y 
4 
4 
4 
(vi)
x 
1 
0 
1 
y 
6 
6 
6 
(vii)
x 
1 
0 
1 
y 
2 
2 
2 
(viii)
x 
1 
0 
1 
y 
6 
6 
6 
(ix)
x 
1 
0 
1 
y 
5/2 
5/2 
5/2 
(x)
x 
1 
0 
1 
y 
0 
0 
0 
(xi)
x 
0 
0 
0 
y 
1 
0 
1 
2. Draw the graph for each linear equation given below:
(i) y = 3x
(ii) y = – x
(iii) y = – 2x
(iv) y = x
(v) 5x + y = 0
(vi) x + 2y = 0
(vii) 4x – y = 0
(viii) 3x + 2y = 0
(ix) x = 2y
Solution:
(i)
x 
1 
0 
1 
y 
3 
0 
3 
(ii)
x 
1 
0 
1 
y 
1 
0 
1 
(iii)
x 
1 
0 
1 
y 
2 
0 
2 
(iv)
x 
1 
0 
1 
y 
1 
0 
1 
(v)
x 
1 
0 
1 
y 
5 
0 
5 
(vi)
x 
1 
0 
1 
y 
1/2 
0 

(v)
x 
1 
0 
1 
y 
4 
0 
4 
(viii)
x 
1 
0 
1 
y 
3/2 
0 
3/2 
(ix)
x 
1 
0 
1 
y 
½ 
0 

3. Draw the graph for each linear equation given below:
(i) y = 2x + 3
Solution:
(i)
x 
1 
0 
1 
y 
5/3 
3 
5 
(ii)
x 
1 
0 
1 
y 
5/3 
1 
1/3 
(iii)
x 
1 
0 
1 
y 
5 
4 
3 
(iv)
x 
1 
0 
1 
y 
13/2 
5/2 
3/2 
(v)
x 
1 
0 
1 
y 
5/6 
2/3 
13/6 
(vi)
x 
1 
0 
1 
y 
2 
4/3 
2/3 
(vii) We can write the equation as
2x – 3y = 8
x 
1 
0 
1 
y 
10/3 
8/3 
2 
(viii) We can write the equation as
5x – 2y = 17
x 
1 
0 
1 
y 
11 
17/2 
6 
(ix)
x 
1 
0 
1 
y 
1/5 
2/5 
3/5 
4. Draw the graph for each equation given below:
(i) 3x + 2y = 6
(ii) 2x – 5y = 10
In each case, find the coordinates of the points where the graph (line) drawn meets the coordinates axes.
Solution:
(i)
x 
2 
0 
2 
y 
6 
3 
0 
From the graph, the line intersects xaxis at (2, 0) and yaxis at (0, 3).
(ii)
x 
1 
0 
1 
y 
12/5 
2 
8/5 
From the graph, the line intersects xaxis at (5, 0) and yaxis at (0, 2).
(iii)
x 
1 
0 
1 
y 
5.25 
4.5 
3.75 
From the graph, the line intersects xaxis at (10, 0) and yaxis at (0, 7.5).
(iv)
x 
1 
0 
1 
y 
3 
1/3 
11/3 
From the graph, the line intersects xaxis at (1/10, 0) and yaxis at (0, 4.5).
5. For each linear equation, given above, draw the graph and then use the graph drawn (in each case) to find the area of a triangle enclosed by the graph and the coordinates axes:
(i) 3x – (5 – y) = 7
(ii) 7 – 3 (1 – y) = – 5 + 2x
Solution:
(i)
We know that
Area of the right triangle obtained = ½ × base × altitude
= ½ × 4 × 12
= 24 sq. units
(ii)
We know that
Area of the right triangle obtained = ½ × base × altitude
= ½ × 9/2 × 3
= 27/4
= 6.75 sq. units
Exercise 26C page: 323
1. In each of the following, find the inclination of line AB:
Solution:
The angle which is a straight line makes with the positive direction of xaxis (measured in anticlockwise direction) is called as inclination of the line.
(i) The inclination of line AB is θ = 45^{0}
(ii) The inclination of line AB is θ = 135^{0}
(iii) The inclination of line AB is θ = 30^{0}
2. Write the inclination of a line which is:
(i) Parallel to xaxis.
(ii) Perpendicular to xaxis.
(iii) Parallel to yaxis.
(iv) Perpendicular to yaxis.
Solution:
(i) The inclination of a line which is parallel to xaxis is θ = 0^{0}.
(ii) The inclination of a line which is perpendicular to xaxis is θ = 90^{0}.
(iii) The inclination of a line which is parallel to yaxis is θ = 90^{0}.
(iv) The inclination of a line which is perpendicular to yaxis θ = 0^{0}.
3. Write the slope of the line whose inclination is:
(i) 0^{0}
(ii) 30^{0}
(iii) 45^{0}
(iv) 60^{0}
Solution:
The slope of the line is tan θ if θ is the inclination of a line.
Here slope is usually denoted by the letter m.
(i) The inclination of a line is 0^{0} then θ = 0^{0}.
Therefore, the slope of the line is m = tan 0^{0} = 0
(ii) The inclination of a line is 30^{0} then θ = 30^{0}.
Therefore, the slope of the line is m = tan θ = tan 30^{0} = 1/ √3
(iii) The inclination of a line is 45^{0} then θ = 45^{0}.
Therefore, the slope of the line is m = tan θ = tan 45^{0} = 1
(iv) The inclination of a line is 60^{0} then θ = 60^{0}.
Therefore, the slope of the line is m = tan θ = tan 60^{0} = √3
4. Find the inclination of the line whose slope is:
(i) 0
(ii) 1
(iii) √3
(iv) 1/√3
Solution:
If tan θ is the slope of a line; then the inclination of the line is θ
(i) If the slope of the line is 0; then tan θ =0
tan θ = 0
tan θ = tan 0^{0}
θ = 0^{0}
Hence, the inclination of the given line is θ = 0^{0}.
(ii) If the slope of the line is 1; then tan θ = 1
tan θ = 1
tan θ = tan 45^{0}
θ = 45^{0}
Hence, the inclination of the given line is θ = 45^{0}.
(iii) If the slope of the line is √3; then tan θ = √3
tan θ = √3
tan θ = tan 60^{0}
θ = 60^{0}
Hence, the inclination of the given line is θ = 60^{0}.
(iv) If the slope of the line is 1/√3; then tan θ = 1/√3
tan θ = 1/√3
tan θ = tan 30^{0}
θ = 30^{0}
Hence, the inclination of the given line is θ = 30^{0}.
5. Write the slope of the line which is:
(i) Parallel to xaxis.
(ii) Perpendicular to xaxis.
(iii) Parallel to yaxis.
(iv) Perpendicular to yaxis.
Solution:
(i) We know that the inclination of line parallel to xaxis θ = 0^{0}
So the slope (m) = tan θ = tan 0^{0} = 0
(ii) We know that the inclination of line perpendicular to xaxis θ = 90^{0}
So the slope (m) = tan θ = tan 90^{0} = ∞ (not defined)
(iii) We know that the inclination of line parallel to yaxis θ = 90^{0}
So the slope (m) = tan θ = tan 90^{0} = ∞ (not defined)
(iv) We know that the inclination of line perpendicular to yaxis θ = 0^{0}
So the slope (m) = tan θ = tan 0^{0} = 0
Selina Solutions for Class 9 Maths Chapter 26Coordinate Geometry
The Chapter 26, Coordinate Geometry, contains 3 exercises and the Selina Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.
26.1 Introduction
26.2 Dependent and Independent Variable
26.3 Ordered Pair
26.4 Cartesian Plane
26.5 Coordinates of Points
26.6 Quadrants and Sign Convention
26.7 Plotting of Points
26.8 Graphing a linear equation
26.10 Inclination And Slope
26.11 YIntercept
Selina Solutions for Class 9 Maths Chapter 26 Coordinate Geometry
The branch of Mathematics that deals with questions related to shape, size, relative position of figures, and the properties of space is known as Geometry. Further, the branch of geometry in which the position of the points on the plane is defined by coordinates are known as Coordinate Geometry. Read and learn the Chapter 26 of Selina textbook to learn more about Coordinate Geometry along with the concepts covered in it. Solve the Selina Solutions for Class 9 effectively to score high in the examination.