Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry Exercise 26A Solutions
EXERCISE – 26A
(Q1) For each equation given below; name the dependent and independent variables.
(i) y = 4/3 x – 7
Solution:
If we have to put the value of x then we have to find the value of y.
E.x. y = 4/3 x – 7
Put x = 2
y = 4/3 × 2 – 7
y = 8/3 – 7
y = 8-21/3
y = -13/3
So, y is dependent and x is independent variable.
(ii) x = 9y + 4
Solution:
If we have to put the value of y then we have to find the value of x
E.x. x = 9y + 4
Put y = 2
x = 9×2+4
x = 18 + 4
x = 22
So, x is dependent and y is independent variable.
(iii) x = 5y + 3/2
Solution:
If we have to put the value of y then we have to find the value of x.
E.x. x = 5y+3/2
Put y = 2
x = 5×2+3/2
= 10+3/2
x = 13/2
So, x is dependent and y is independent variable.
(iv) y = 1/7 (6x + 5)
Solution:
If we have to put the value of x then we have to find the value of y.
Ex. y = 1/7 (6x + 5)
Put x = 2
y = 1/7 (6×2 + 5)
= 1/7 (12+5)
= 1/7 (17)
y = 17/7
So, y is dependent and x is independent variable.
(Q2) 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:
(Q3) 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:
(i) (x-1, y+3) = (4, 4)
x = x – 1
∴ x – 1 = 4
x = 4 + 1
x = 5
y = y + 3
y + 3 = 4 {(x, y) = (4, 4)}
y = 4 – 3
y = 1
(ii) (3x+1, 2y-7) = (9, -9)
∴ 3x+1 = 9
3x = 9-1
3x = 8
x = 8/3
2y – 7 = -9
2y = -9+7
2y = -2
2y = -2
y = -2/2
y = -1
(iii) (5x – 3y, y – 3x) = (4, -4)
=> 5x – 3y = 4 —– (i)
5x – 3 (-4 + 3x) = 4
5x + 12 – 9x = 4
-4x + 12 = 4
-4x = 4-12
-4x = -8
x = -8/-4
x = 2
y – 3a = -4
y = -4 + 3x —- (ii)
Put in equation (i)
Put x = 2 in equation (ii),
y = -4 + 3 (2)
= -4 + 6
y = 2
(Q4) Use the graph given alongside, to find the co-ordinates 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:
(v) The abscissa is 5
=> According to graph
The co-ordinates of the point H, B and G are given by (5, 5), (5, 0) and (5, -3)
(vi) The abscissa is equal to the ordinate
=> According to graph
The co-ordinates 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
=> According to graph
The co-ordinate of the point E is given by (6, 3)
=> (i) The abscissa is 2.
=> According to graph. {Value of x is abscissa value of y is ordinate}
The co-ordinate of the point A is given by (2, 2)
(ii) The ordinate is 0
=> according to graph
The co-ordinate of the point B is given by (5, 0)
(iii) The ordinate is 3.
=> According to graph
The co-ordinate of the points C and E are given by (-4, 3) and (6, 3)
(iv) The ordinate is -4
= According to graph
The co-ordinate of the point D is given by (2, -4)
(Q5) State true or false:
(i) The ordinate of a point is it’s x co-ordinate.
Solution:
The ordinate of a point is it’ y co-ordinate. So, the statement is False.
(ii) The origin is in the first quadrant.
Solution:
The origin is not in the first quadrant and not a second quadrant.
The origin is on the first and second quadrant.
So, this statement is False.
(iii) The y-axis is the vertical number line.
Solution:
Yes, The y-axis is the vertical number line.
So, this statement is True.
(iv) Every point is located in one of the four quadrants.
Solution:
Every point is located in one of the four quadrants is not necessary. Some points is located in first quadrant, some points in second quadrant.
So, This statement is False.
(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.
Solution:
This is False statement.
(vi) The origin (0, 0) lies on the x-axis.
Solution:
Yes, the origin (0, 0) lies on the x-axis.
So, This is True Statement.
(vii) The point (a, b) lies on the y-axis if b = 0.
Solution:
If a = 0 then the point (a, b) lies on the y-axis.
So, this statement is False.
(Q6) In each of the following, find the co-ordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation:
(i) 3 – 2x = 7; 2y + 1 = 10 – 2 1/2 y
Solution:
3 – 2x = 7
-2x = 7 – 3
-2x = 4
x = 4/-2
x = -2
2y + 1 = 10 – 2 1/2 y
2y + 1 = 10 – 4+1/2 y
2y + 1 = 10 – 5/2 y
2y + 5/2 y = 10-1
4y +5y/2 = 9
9y/2 = 9
y = 9×2/9
y = 2
∴ The co-ordinates of the point are (-2, 2)
(ii) 2a/3 – 1 = a/2; 15-4b/7 = 2b-1/3
Solution:
2a-1/3 = a/2
2 (2a – 3) = 3a
4a – 6 = 3a
-6 = 3a – 4a
-6 = -1a
-6/-1 = a
6 = a
15-4b/7 = 2b-1/3
3 (15 – 4b) = 7 (2b – 1)
45 – 12b = 14b – 7
45 + 7 = 14b + 12b
52 = 26b
52/26 = b
2 = b`
∴ The co-ordinates of the point are (6, 2)
(iii) 5x – (5-x) = 1/2 (3 – x); 4 – 3y = 4+y/3
Solution:
5x – (5-x) = 1/2 = (3 – x)
5x – 5+x = 1/2 (3 – x)
6x – 5 = 1/2 (3 – x)
2 (6x – 5) = 3 – x
12x – 10 = 3 – x
12x + x = 3 + 10
13x = 13
x = 13/13
x = 1
4 – 3y = 4+y/3
3 (4 – 3y) = 4 + y
12 – 9y = 4 + y
12 – 4 = y + 9y
8 = 10y
8/10 = y
4/5 = y
∴ The co-ordinates of the point are (1, 4/5)
(Q7) In each of the following, the co-ordinates of the three vertices of a rectangle ABCD are given. By plotting the given points; find, in each case, the co-ordinates of the fourth vertex:
(i) A (2, 0), B (8, 0) and C (8, 4)
Solution:
Now we have to plot this three points on a graph and we have to find fourth vertex of rectangle ABCD.
Fourth vertex is D (2, 4).
(ii) A (4, 2), B (-2, 2) and D (4, -2)
Solution:
Now, we have to plot this three points on a graph and we have to find fourth vertex of rectangle ABCD.
Fourth vertex is C (-2, -2)
(iii) A (-4, -6), C (6, 0) and D (-4, 0)
Solution:
Now, we have to plot this three points on a graph and we have to find fourth vertex of rectangle ABCD.
Fourth vertex is B (6, -6)
(iv) B (10, 4), 0 (0, 4) and D (0, -2)
Solution:
Now we have to plot this three points on graph and we have to find fourth vertex of rectangle ABCD.
Fourth vertex is A (10, -2)
(Q8) 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 co-ordinates of the fourth vertex D.
Also, form the same graph, state the co-ordinates of the mid-points of the sides AB and CD.
Solution:
We have to plot the given three vertices and we have to find the fourth vertex of a parallelogram ABCD.
∴ The fourth vertex is D (-6, -4).
∴ The co-ordinates of the mid-points of the sides AB and CD are,
P (3, 2) and Q (-1, -4)
The mid-point of AB is P (3, 2) and The mid-point of CD is Q (-11-4)
(Q9) A (-2, 4), C (4, 10) and D (-2, 10) are the vertices of a square ABCD. Use the graphical method to find the co-ordinates of the fourth vertex B. Also, find:
(i) The co-ordinates of the mid-point of BC
(ii) The co-ordinates of the mid-point of CD and
(iii) The co-ordinates of the point of intersection of the diagonals of the square ABCD.
Solution:
(i) The co-ordinates of the mid-point of BC is P (4, 7)
(ii) The co-ordinates of the mid-point of CD is Q (1, 10)
(iii) The co-ordinates of the point of intersection of the diagonals of the square ABCD is O (1, 7)
BD and AC are diagonals of square ABCD.
∴ The fourth vertex of square is B (4, 4)
(Q10) 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:
(i) According to graph it is clear that points P, Q and R are collinear points.
(ii) According to graph it is clear that points A, B and C are non-collinear points.
Here is your solution of Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry Exercise 26A
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