ICSE Solutions for Selina Concise Chapter 26 Co-ordinate Geometry Class 9 Maths
Exercise 26(A)
1. For the equation given below; name the dependent and independent variables(i) y = (4/3)x - 7
(ii) x = 9y + 4
(iii) x = (5y+3)/2
(iv) y = (1/7)(6x + 5)
Answer
(i) y = (4/3)x - 7
Dependent variable is y
Independent variable is x
(ii) x = 9y + 4
Dependent variable is x
Independent variable is y
(iii) x = (5y + 3)/2
Dependent variable is x
Independent variable is y
(iv) y = (1/7) (6x + 5)
Dependent variable is y
Independent variable is x
(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)
Answer
Let us take the point as
A(8, 7), B(3, 6), C(0, 4), D(0, -4), E(3, -2), F(-2, 5), G(-3, 0), H(5,0), I(-4, -3)
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,
B(3,6), C(0,4), D(0, - 4), E(3, - 2), F(-2, 5), G(-3,0), H(5,0), I(-4, -3)
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)
Answer
Two ordered pairs are equal.
⇒ Therefore their first components are equal and their second components too are separately equal.
(i) (x-1, y+3) = (4, 4)
(x-1, y+3) = (4, 4)
⇒ x -1 = 4 and y + 3 = 4
⇒ x =5 and y = 1
(ii) (3x + 1, 2y -7) = (9, -9)
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)
5x - 3y = 4 ...(A) and y - 3x = -4 ...(B)
Now multiplying the equation (B) by 3, we get
3y -9x = -12 ...(C)
Now adding both the equations (A) and (C), we get
(5x - 3y) + (3y - 9x) = ( 4 + (-12))
⇒ - 4x = -8
⇒ x = 2
Putting the value of x in the equation (B), we get
y - 3x = -4
⇒ y = 3x - 4
⇒ y = 3(2) - 4
⇒ y = 2
Therefore we get,
x = 2, y = 2
(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.
Answer
(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)
(i) The ordinate of a point is its x-co-ordinate.
(ii) The origin is in the first quadrant.
(iii) The y-axis 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 x-axis.
(vii) The point (a,b) lies on the y-axis if b=0.
Answer
(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
(i) 3 – 2x = 7; 2y + 1 = 10 – 2 1/2y.
(ii) (2a/3) -1 = (a/2), (15- 4b)/7 = (2b - 1)/3
(iii) 5x -(5-x) = (1/2) (3-x); 4 - 3y = (4+y)/3
7. In the following, the coordinates of the three vertices of a rectangle ABCD are given. By plotting the given points; find, in 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).
Answer
(i) A(2, 0) , B(8, 0) and C(8, 4)
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)
As is clear from the graph B(6, -6)
(iv) B(10, 4), C(0, 4) and D(0, -2)
After plotting the given points B(10, 4), C(0, 4) and D(0, -2) on a graph paper ; joining C with B and C with D. From the graph it is clear that the vertical distance between the points C(0, 4) and D(0, -2) is 6 units and the horizontal distance between the points C (0, 4) and B(10, 4) is 10 units, therefore the vertical distance between the points B(10, 4) and A must be 6 units and the horizontal distance between the points D(0, -2) and A must be 10 units. Now complete the rectangle ABCD.
As is clear from the graph 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 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
Answer
Given A(2, -2) , B(8, 2) and C(4 , -4) are the vertices of the parallelogram ABCD
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).
(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.
Answer
Given A(-2 , 4) , C(4, 10) and D(-2, 10) are the vertices of a square ABCD
After plotting the given points A(-2, 4) , C(4, 10) and D(-2, 10) on a graph paper; joining D with A and D with C. Now complete the square ABCD
As is clear from the graph B(4, 4)
Now from the graph we can find the mid points of the sides BC and CD and the co - ordinates of the diagonals of the square.
Therefore the co - ordinates of the mid - point of BC is E (4, 7) and the co- ordinates of the mid- point of CD is F(1, 10) and the co - ordinates 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 col linear or not:
(i) (3, 5), (1, 1) and (0, -1)
(ii) (-2, -1), (-1, -4) and (-4, 1)
Answer
After plotting the given points, we have clearly seen from the graph that
(i) A(3, 5), B(1, 1) and C(0, -1) are collinear.
11. Plot point A(5, -7). From point A, draw AM perpendicular to the x-axis and AN perpendicular to the y-axis. Write the coordinates of points M and N
Answer
Given A (5, -7 )
After plotting the given point A(5, -7) on a graph paper. Now let us draw a perpendicular AM form the point A(5, -7 )on the x - axis and a perpendicular AN from the point A(5, -7) on the y-axis.
As from the graph clearly we get the co - ordinates of the points M and N
Co-ordinate of the point M is (5, 0)
Co-ordinate of the point N is (0, -7)
12. In square ABCD; A = (3, 4), B = (-2, 4) and C = (-2, -1). By plotting these points on a graph paper, find the co-ordinates of vertex D. Also, find the area of the square
Answer
Given that in square ABCD ; A(3,4 ) , B(-2 , 4) and C(-2 , -1)
13. In rectangle OABC; point O is the origin, OA = 10 units along x-axis and AB = 8 units. Find the co-ordinates of vertices A, B and C
Answer
After plotting the given points A(3, 4) , B(-2, 4) and C(-2, -1 ) on a graph paper ; joining B with C and B with A. From the graph it is clear that the vertical distance between the points B(-2 , 4) and C(-2, -1) is 5 units and the horizontal distance between the points B(-2 , 4) and A(3, 4) is 5 units, therefore the vertical distance between the points A(3, 4) and D must be 5 units and the horizontal distance between the points C(-2, -1) and D must be 5 units. Now complete the square ABCD
As is clear from the graph D(3, -1)
Now the area of the square ABCD is given by
area of ABCD = (side)2 = (5)2 = 25 units
Given that in rectangle OABC : point O is origin and OA = 10 units along x - axis therefore we get O(0, 0) and A(10, 0). Also it is given that AB = B units . Therefore we get B(10, 8) and C(0, 8)
After plotting the points O(0, 0) , A (10, 0) , B(10, 8) and C(0, 8) on a graph paper ; we get the above rectangle OABC and the required co -ordinates of the vertices are A(10, 0) , A(10, 0) , B(10, 8) and C(0, 8) on a graph paper ; we get the above rectangle OABC and the required co - ordinates of the vertices are A(10, 0) , B(10, 8) and C(0, 8)
Exercise 26(B)
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 + = 0
(ix) 2y – 5 = 0
(x) y = 0
(xi) x = 0
Answer
(i) Since x = 3, therefore the value of y can be taken as any real no.
First prepare a table as follows:
Thus the graph can be drawn as follows :
(ii) First prepare a table as follows :
(iii) First prepare a table as follows :
Thus the graph can be drawn as follows:
(iv) The equation can be written as :
x =7/2
First prepare a table as follows :
Thus the graph can be drawn as follows :
(v) First prepare a table as follows :
Thus the graph can be drawn as follows:
(vi) First prepare a table as follows :
(vii) First prepare a table as follows :
Thus the graph can be drawn as follows :
(viii) First prepare a table as follows :
Thus the graph can be drawn as follows :
(ix) First prepare a table as follows :
Thus the graph can be drawn as follows :
Thus the graph can be drawn as follows :
(xi) First prepare a table as follows :
Thus the graph can be drawn as follows :
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
Answer
(i) First prepare a table as follows :
Thus the graph can be drawn as follows :
(ii) First prepare a table as follows :
Thus the graph can be drawn as follows:
(iii) First prepare a table as follows :
Thus the graph can be drawn as follows :
(iv) First prepare a table as follows :
Thus the graph can be drawn as follows :
(v) First prepare a table as follows :
Thus the graph can be drawn as follows :
(vi) First prepare a table as follows:
Thus the graph can be drawn as follows :
(viii) First prepare a table as follows:
Thus the graph can be drawn as follows :
(ix) First prepare a table as follows :
Thus the graph can be drawn as follows :
(i) y = 2x + 3
(ii) y = 2x /3 -1
(iii) y = – x + 4
(iv) y = 4x – 5/2
(v) y = 3x/2 + 2/3
(vi) 2x – 3y = 4
(vii) (x-1)/3 - (y+2)/2 = 0
(viii) x - 3 = 2/5 (y+1)
(ix) x + 5y + 2 = 0
Answer 3
(i) First prepare a table as follows:
Thus the graph can be drawn as follows:
Thus the graph can be drawn as follows :
Thus the graph can be drawn as follows :
(iv) First prepare a table as follows :
Thus the graph can be drawn as follows :
Thus the graph can be drawn as follows :
(vi) First prepare a table as follows :
Thus the graph can be drawn as follows :
2x - 3y = 8
First prepare a table as follows :
Thus the graph can be drawn as follows :
5x - 2y = 17
First prepare a table as follows :
(i) 3x + 2y =6
(ii) 2x – 5y =10
(iii) x/2 + 2y/3 = 5
(iv) (2x - 1)/3 - (y-2)/5 = 0
Answer
(i) To draw the graph of 3x + 2y = 6 follows the steps :
First prepare a table as below :
Now sketch the graph as shown :
From the graph it can verify that the line intersect x axis at(2, 0) and y at (0, 3).
Now sketch the graph as shown :
(i) 3x – (5 – y) = 7
(ii) 7 – 3 (1 – y) = – 5 + 2x.
Answer
(i) First draw the graph as follows :
Thus the area of the triangle will be :
= (1/2) × base × altitude
= 24 sq. units
= (1/2)× (9/2) ×3
= (27/4) = 6.75 sq. units
(i) y = 3x – 1 , y = 3x + 2
(ii) y = x – 3 , y = – x + 5
(iii) 2x – 3y = 6 , x/2 + y/3 =1
(iv) 3x + 4y = 24, x/4 + y/3 = 1
Answer
(i) To draw the graph of y = 3x - 1 and y = 3x +2 follows the steps :
First prepare a table as below : 
Now sketch the graph as shown :
From the graph it can verify that the lines are parallel.
First prepare a table as below :
Now sketch the graph as shown :
First prepare a table as below :
Now sketch the graph as shown :
From the graph it can verify that the lines are perpendicular.
First prepare a table as below :
Now sketch the graph as shown :
From the graph it can verify that the lines are parallel.
Answer
First prepare a table as follows :
Now the graph can be drawn as follows :
8: On the same graph paper, plot the graphs of y = 2x – 1, y = 2x and y = 2x + 1 from x = – 2 to x = 4. Are the graphs (lines) drawn parallel to each other?
Answer
First prepare a table as follows :
Now the graph can be drawn as follows :
The lines are parallel to each other.
Answer
To draw the graph of 3x + 2y = 6 follows the steps :
First prepare a table as below :
Now sketch the graph as shown :
From the graph it can verify that the line intersect x axis at (2,0) and y at (0,3), therefore the co ordinates of P(x - axis) and Q(y - axis) are (2,0) and (0, 3) respectively.
(i) x1, the value of x, when y = 3
(ii) x2, the value of x, when y = – 2.
Answer
First prepare a table as follows :
Thus the graph can be drawn as shown :
(ii) For y = -2 we have x = 7
Use the graph drawn to find:
(i) y1, the value of y, when x = 4.
(ii) y2, the value of y, when x = 0.
Answer
First prepare a table as follows :
If x = 4 the value of y = 0
If x = 0 the value of y = -3.
(i) x1, the value of x, when y = 10
(ii) y1, the value of y, when x = 8.
Answer
First prepare a table as follows:
The graph of the equation can be drawn as follows :
for y = 10, the value of x = -4 .
for x = 8 the value of y = -5 .
Answer
The equations can be written as follows :
y = 2 -x
⇒ y = (1/2)(x-5)
⇒ y = -x/3
First prepare a table as follows :
From the graph it is clear that the equation of lines are passes through the same point.
Exercise 26(C)
1. In each of the following, find the inclination of line AB(i)
(ii)
(iii)
Answer
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°
(i) Parallel to the x-axis
(ii) Perpendicular to the x-axis
(iii)Parallel to the y-axis.
(iv) Perpendicular to the y-axis.
Answer
(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°
(i) 0
(ii) 30
(iii) 45
(iv) 60
Answer
If θ is the inclination of a line; the slope of the line is tanθ and is usually denoted by letter m.
(i) Here the inclination of a line is 0, then θ = 0°.
Therefore, the slope of the line is m = tan0° = 0
(ii) Here the inclination of a line is 30°, then θ = 30°.
Therefore the slope of the line is m = tanθ = 30° = 1/√3
(iii) Here the inclination of a line is 45° , then θ = 45°.
Therefore, the slope of the line is m = tan 45° = 1
(iv) Here the inclination of a line is 60°, then θ = 60°.
Therefore the slope of the line is m = tan 60° =√3
(i) 0
(ii) 1
(iii) √3
(iv) 1/√3
Answer
If tan θ is the slope of a line; then inclination of the line is tan θ
(i) Here the slope of line is 0; then tan θ = 0
Now,
tan θ = 0
⇒ tan θ = tan 0°
⇒ θ = 0°
Therefore, the inclination of the given line is θ = 0°
(ii) Here the slope of line is 1 ; then tan θ = 1
Now,
tan θ = 1
⇒ tan θ = tan 45°
⇒ θ = 45°
Therefore, the inclination of the given line is θ = 45°
(iii) Here the slope of line is √3; then tan θ = √3
Now,
tan θ = √3
⇒ tan θ = tan 60°
⇒ θ = 60°
Therefore, the inclination of the given line is θ = 60°
(iv) Here the slope of line is 1/√3 ; then tan θ = 1/√3
Now,
tan θ = 1/√3
⇒ tan θ = tan 30°
⇒ θ = 30°
Therefore, the inclination of the given line is θ = 30°
(i) Parallel to the x-axis
(ii) Perpendicular to the x-axis.
(iii) Parallel to the y-axis.
(iv) Perpendicular to the y-axis.
Answer
(i) For any line which is parallel to x - axis, the inclination if θ = 0°
Therefore, Slope (m) = tan θ = tan0° = 0
(ii) For any line which is perpendicular to x - axis, the inclination is θ = 90°
Therefore, Slope (m) = tan θ = tan 90° = ∞ (not defined)
(iii) For any line which is parallel to y - axis, the inclination is θ = 90°
Therefore, Slope (m) = tan θ = tan 90° = ∞ (not defined)
(iv) For any line which is perpendicular to y - axis, the inclination is θ = 0°
Therefore, Slope (m) = tan θ = tan 0° = 0
(i) x + 3y + 5 = 0
(ii) 3x – y – 8 = 0
(iii) 5x = 4y + 7
(iv) x= 5y – 4.
(v) y = 7x – 2
(vi) 3y = 7
(vii) 4y + 9 = 0
Answer
Equation of any straight line in the form y = mx + c, where slope = m (co-efficient of x) and y - intercept = c (constant term)
(i) x + 3y + 5 = 0
⇒ 3y = -x - 5
y - intercept = constant term = 7/3
(i) Slope = 2 and y-intercept = 3
(ii) Slope = 5 and y-intercept = – 8
(iii) slope = – 4 and y-intercept = 2
(iv) slope = – 3 and y-intercept = – 1
(v) slope = 0 and y-intercept = – 5
(vi) slope = 0 and y-intercept = 0
Answer
(i) Given,
Slope is 2, therefore m = 2
y - intercept is 3, therefore c = 3
Therefore,
y = mx + c
⇒ y = 2x + 3
Therefore, the equation of the required line is y = 2x + 3
(ii) Given
Slope is 5, therefore m = 5
Y - intercept is - 8, therefore c = -8
Therefore,
y = mx + c
⇒ y = 5x + (-8)
Therefore, the equation of the required line is y = 5x + (-8)
(iii) Given,
Slope is - 4, therefore m = -4
Y - intercept is 2, therefore c = 2
Therefore,
y = mx + c
⇒ y = -4x + 2
Therefore the equation of the required line is y = -4x + 2
(iv) Given
Slope is -3 , therefore m = -3
Y - intercept is -1, therefore c = -1
Therefore,
y = mx + c
⇒ y = -3x - 1
Therefore the equation of the required line is y = -3x - 1
(v) Given
Slope is 0, therefore m = 0
Y - intercept is - 5, therefore c = -5
Therefore,
y = mx + c
⇒ y = 0.x + (-5)
⇒ y = -5
Therefore the equation of the required line is y = -5
(vi) Given
Slope is 0, therefore m = 0
Y - intercept is 0 , therefore c = 0
Therefore,
y = mx +c
⇒ y =0.x + 0
⇒ y = 0
Therefore the equation of the required line is y = 0
Answer
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.
9.Draw the line 2x – 3y – 18 = 0 on a graph paper. From the graph paper, read the y-intercept of the line.
Answer
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
Answer 10
Given line is
x + y = 5
The graph of the given line is shown below.
Again, we know that equation of any straight line in the form y = mx +c , where m is the gradient and c is the intercept. Again we have if slope of a line is tanθ then inclination of the line is θ.