Class 9 SELINA Solutions Maths Chapter 26  Coordinate Geometry
Coordinate Geometry Exercise Ex. 26(A)
Solution 1
(i)_{}
Dependent variable is _{}
Independent variable is _{}
(ii)_{}
Dependent variable is _{}
Independent variable is _{}
(iii)_{}
Dependent variable is _{}
Independent variable is _{}
(iv)_{}
Dependent variable is _{}
Independent variable is _{}
Solution 2
Let us take the point as
_{},_{}, _{},_{},_{},_{},_{},_{},_{}
On the graph paper, let us draw the coordinate axes XOX' and YOY' intersecting at the origin O. With proper scale, mark the numbers on the two coordinate 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
Two ordered pairs are equal.
_{}Therefore their first components are equal and their second components too are separately equal.
(i)_{}
_{}
(ii)_{}
_{}
(iii)_{}
_{}
Now multiplying the equation (B) by 3_{}, we get
_{}
Now adding both the equations (A) and (C) , we get
Putting the value of x in the equation (B), we get
Therefore we get,
x = 2, y = 2
Solution 4
(i) The abscissa is 2
Now using the given graph the coordinate of the given point A is given by (2,2)
(ii) The ordinate is 0
Now using the given graph the coordinate of the given point B is given by (5,0)
(iii) The ordinate is 3
Now using the given graph the coordinate 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 coordinate of the given point D is given by (2,4)
(v) The abscissa is 5
Now using the given graph the coordinate 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 coordinate 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 coordinate of the given point E is given by (6,3)
Solution 5
(i)The ordinate of a point is its xcoordinate.
False.
(ii)The origin is in the first quadrant.
False.
(iii)The yaxis 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 xaxis.
True.
(vii)The point (a,b) lies on the yaxis if b=0.
False
Solution 6
(i)_{}
Now
_{}
Again
_{}
_{}
_{}The coordinates of the point _{}
(ii)_{}
Now
_{}
Again
_{}
_{}
_{}The coordinates of the point _{}
(iii)_{}
Now
_{}
_{}
Again
_{}
_{}
_{}The coordinates of the point _{}
Solution 7
(i)_{},_{} and _{}
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)
(iii)_{}, _{}and _{}
After plotting the given points_{},_{} and _{}on a graph paper; joining _{}with _{}and _{}with_{}. From the graph it is clear that the vertical distance between the points _{}and _{}is _{}units and the horizontal distance between the points _{}and _{}is _{}units , therefore the vertical distance between the points _{}and _{}must be _{}units and the horizontal distance between the points _{}and _{}must be _{}units . Now complete the rectangle _{}
As is clear from the graph _{}
(iv)_{}, _{}and _{}
After plotting the given points_{},_{} and _{}on a graph paper; joining _{}with _{}and _{}with_{}. From the graph it is clear that the vertical distance between the points _{}and _{}is _{}units and the horizontal distance between the points _{}and _{}is _{}units , therefore the vertical distance between the points _{}and _{}must be _{}units and the horizontal distance between the points _{}and _{}must be _{}units. Now complete the rectangle _{}
As is clear from the graph _{}
Solution 8
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 coordinates of the midpoint of AB is E(3,2) and the coordinates of the midpoint of CD is F(1,4)
Solution 9
Given _{}, _{}and _{}are the vertices of a square _{}
After plotting the given points_{},_{} and _{}on a graph paper; joining _{}with _{}and _{}with_{}. Now complete the square _{}
As is clear from the graph _{}
Now from the graph we can find the mid points of the sides _{}and _{}and the coordinates of the diagonals of the square.
Therefore the coordinates of the midpoint of _{}is _{}and the coordinates of the midpoint of _{}is _{}and the coordinates of the diagonals of the square is _{}
Solution 10
After plotting the given points, we have clearly seen from the graph that
(i) _{}, _{} and _{} are collinear.
(ii)_{}, _{} and _{} are noncollinear.
Solution 11
Given _{}
After plotting the given point _{} on a graph paper. Now let us draw a perpendicular _{} from the point _{} on the xaxis and a perpendicular _{} from the point _{} on the yaxis.
As from the graph clearly we get the coordinates of the points _{} and _{}
Coordinate of the point _{} is _{}
Coordinate of the point _{} is _{}
Solution 12
Given that in square _{}; _{}, _{} and _{}
After plotting the given points_{},_{} and _{} on a graph paper; joining _{} with _{} and _{} with_{}. From the graph it is clear that the vertical distance between the points _{} and _{} is _{}units and the horizontal distance between the points _{} and _{} is _{} units , therefore the vertical distance between the points _{} and _{} must be _{}units and the horizontal distance between the points _{} and _{} must be _{} units. Now complete the square _{}
As is clear from the graph _{}
_{ }Now the area of the square _{} is given by
_{}
Solution 13
Given that in rectangle _{} ; point _{} is origin and _{} units along xaxis therefore we get _{} and _{}. Also it is given that _{} units. Therefore we get _{} and _{}
After plotting the points _{}, _{}, _{} and _{} on a graph paper; we get the above rectangle _{} and the required coordinates of the vertices are _{},_{} and _{}
Coordinate Geometry Exercise Ex. 26(B)
Solution 1
(i) Since x = 3, therefore the value of y can be taken as any real no.
First prepare a table as follows:
x 
3 
3 
3 
y 
1 
0 
1 
Thus the graph can be drawn as follows:
(ii)
First prepare a table as follows:
x 
3 
3 
3 
y 
1 
0 
1 
Thus the graph can be drawn as follows:
(iii)
First prepare a table as follows:
x 
5 
5 
5 
y 
1 
0 
1 
Thus the graph can be drawn as follows:
(iv)
The equation can be written as:
_{}
First prepare a table as follows:
x 
_{} 
_{} 
_{} 
y 
1 
0 
1 
Thus the graph can be drawn as follows:
(v)
First prepare a table as follows:
x 
1 
0 
1 
y 
4 
4 
4 
Thus the graph can be drawn as follows:
(vi)
First prepare a table as follows:
x 
1 
0 
1 
y 
6 
6 
6 
Thus the graph can be drawn as follows:
(vii)
First prepare a table as follows:
x 
1 
0 
1 
y 
2 
2 
2 
Thus the graph can be drawn as follows:
(viii)
First prepare a table as follows:
x 
1 
0 
1 
y 
y = 5/3 
6 
3y + 5 = 0 
Thus the graph can be drawn as follows:
(ix)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
_{} 
_{} 
Thus the graph can be drawn as follows:
(x)
First prepare a table as follows:
x 
1 
0 
1 
y 
0 
0 
0 
Thus the graph can be drawn as follows:
(xi)
First prepare a table as follows:
x 
0 
0 
0 
y 
1 
0 
1 
Thus the graph can be drawn as follows:
Solution 2
(i)
First prepare a table as follows:
x 
1 
0 
1 
y 
3 
0 
3 
Thus the graph can be drawn as follows:
(ii)
First prepare a table as follows:
x 
1 
0 
1 
y 
1 
0 
1 
Thus the graph can be drawn as follows:
(iii)
First prepare a table as follows:
x 
1 
0 
1 
y 
2 
0 
2 
Thus the graph can be drawn as follows:
(iv)
First prepare a table as follows:
x 
1 
0 
1 
y 
1 
0 
1 
Thus the graph can be drawn as follows:
(v)
First prepare a table as follows:
x 
1 
0 
1 
y 
5 
0 
5 
Thus the graph can be drawn as follows:
(vi)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
0 
_{} 
Thus the graph can be drawn as follows:
(vii)
First prepare a table as follows:
x 
1 
0 
1 
y 
4 
0 
4 
Thus the graph can be drawn as follows:
(viii)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
0 
_{} 
Thus the graph can be drawn as follows:
(ix)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
0 
_{} 
Thus the graph can be drawn as follows:
Solution 3
(i)
First prepare a table as follows:
x 
1 
0 
1 
y 
1 
3 
5 
Thus the graph can be drawn as follows:
(ii)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
1 
_{} 
Thus the graph can be drawn as follows:
(iii)
First prepare a table as follows:
x 
1 
0 
1 
y 
5 
4 
3 
Thus the graph can be drawn as follows:
(iv)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
_{} 
_{} 
Thus the graph can be drawn as follows:
(v)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
_{} 
_{} 
Thus the graph can be drawn as follows:
(vi)
First prepare a table as follows:
x 
1 
0 
1 
y 
2 


Thus the graph can be drawn as follows:
(vii)
The equation will become:
_{}
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
_{} 
_{} 
Thus the graph can be drawn as follows:
(viii)
The equation will become:
_{}
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
_{} 
_{} 
Thus the graph can be drawn as follows:
(ix)
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
_{} 
_{} 
Thus the graph can be drawn as follows:
Solution 4
(i)
To draw the graph of 3x + 2y = 6 follows the steps:
First prepare a table as below:
X 
2 
0 
2 
Y 
6 
3 
0 
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).
(ii)
To draw the graph of 2x  5y = 10 follows the steps:
First prepare a table as below:
X 
1 
0 
1 
Y 
_{} 
2 
_{} 
Now sketch the graph as shown:
From the graph it can verify that the line intersect x axis at (5,0) and y at (0,2).
(iii)
To draw the graph of follows the steps:
First prepare a table as below:
X 
1 
0 
1 
Y 
5.25 
4.5 
3.75 
Now sketch the graph as shown:
From the graph it can verify that the line intersect x axis at (10,0) and y at (0,7.5).
(iv)
To draw the graph of follows the steps:
First prepare a table as below:
X 
1 
0 
1 
Y 
3 


Now sketch the graph as shown:
From the graph it can verify that the line intersect x axis at and y at (0,4.5).
Solution 5
(i)
First draw the graph as follows:
This is an right trinangle.
Thus the area of the triangle will be:
(ii)
First draw the graph as follows:
This is a right triangle.
Thus the area of the triangle will be:
Solution 6
(i)
To draw the graph of y = 3x  1 and y = 3x + 2 follows the steps:
First prepare a table as below:
X 
1 
0 
1 
Y=3x1 
4 
1 
2 
Y=3x+2 
1 
2 
5 
Now sketch the graph as shown:
From the graph it can verify that the lines are parallel.
(ii)
To draw the graph of y = x  3 and y = x + 5 follows the steps:
First prepare a table as below:
X 
1 
0 
1 
Y=x3 
4 
3 
2 
Y=x+5 
6 
5 
4 
Now sketch the graph as shown:
From the graph it can verify that the lines are perpendicular.
(iii)
To draw the graph of 2x  3y = 6 and follows the steps:
First prepare a table as below:
X 
1 
0 
1 
_{} 
_{} 
2 
_{} 
_{} 
_{} 
3 
_{} 
Now sketch the graph as shown:
From the graph it can verify that the lines are perpendicular.
(iv)
To draw the graph of 3x + 4y = 24 and follows the steps:
First prepare a table as below:
X 
1 
0 
1 
_{} 
_{} 
6 
_{} 
_{} 
_{} 
3 
_{} 
Now sketch the graph as shown:
From the graph it can verify that the lines are parallel.
Solution 7
First prepare a table as follows:
X 
1 
0 
1 
Y=x2 
3 
2 
1 
Y=2x+1 
1 
1 
3 
Y=4 
4 
4 
4 
Now the graph can be drawn as follows:
Solution 8
First prepare a table as follows:
X 
1 
0 
1 
Y=2x1 
3 
1 
1 
Y = 2x 
2 
0 
2 
Y=2x+1 
1 
1 
3 
Now the graph can be drawn as follows:
The lines are parallel to each other.
Solution 9
To draw the graph of 3x + 2y = 6 follows the steps:
First prepare a table as below:
X 
2 
0 
2 
Y 
6 
3 
0 
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(xaxis) and Q(yaxis) are (2,0) and (0,3) respectively.
Solution 10
First prepare a table as follows:
X 
1 
0 
1 
Y 
2 

1 
Thus the graph can be drawn as shown:
(i)
For y = 3 we have x = 3
(ii)
For y = 2 we have x = 7
Solution 11
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
3 
_{} 
The graph of the equation can be drawn as follows:
From the graph it can be verify that
If x = 4 the value of y = 0
If x = 0 the value of y = 3.
Solution 12
First prepare a table as follows:
x 
1 
0 
1 
y 
_{} 
5 
_{} 
The graph of the equation can be drawn as follows:
From the graph it can be verified that:
for y = 10, the value of x = 4.
for x = 8 the value of y = 5.
Solution 13
The equations can be written as follows:
y = 2  x
First prepare a table as follows:
x 
y = 2  x 


1 
3 
_{} 
_{} 
0 
2 
_{} 
0 
1 
1 
2 
_{} 
Thus the graph can be drawn as follows:
From the graph it is clear that the equation of lines are passes through the same point.
Coordinate Geometry Exercise Ex. 26(C)
Solution 1
The angle which a straight line makes with the positive direction of xaxis (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 xaxis is θ = 0°
(ii)The inclination of a line perpendicular to xaxis is θ = 90°
(iii) The inclination of a line parallel to yaxis is θ = 90°
(iv) The inclination of a line perpendicular to yaxis is θ = 0°
Solution 3
If θ is the inclination of a line; the slope of the line is tan θ 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 = tan 0° = 0
(ii)Here the inclination of a line is 30°, then θ = 30°
Therefore the slope of the line is m = tan θ = 30° =
(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
Solution 4
If tan θ is the slope of a line; then inclination of the line is θ
(i)Here the slope of line is 0; then tan θ = 0
Now
_{}
Therefore the inclination of the given line is θ = 0°
(ii)Here the slope of line is 1; then tan θ = 1
Now
_{}
Therefore the inclination of the given line is θ = 45°
(iii)Here the slope of line is _{}; then tan θ = √3
Now
_{}
Therefore the inclination of the given line is θ = 60°
(iv)Here the slope of line is ; then
Now
_{}
Therefore the inclination of the given line is θ = 30°
Solution 5
(i)For any line which is parallel to xaxis, the inclination is θ = 0°
Therefore, Slope(m) = tan θ = tan 0° = 0
(ii) For any line which is perpendicular to xaxis, the inclination is θ = 90°
Therefore, Slope(m) = tan θ = tan 90° = ∞(not defined)
(iii) For any line which is parallel to yaxis, the inclination is θ = 90°
Therefore, Slope(m) = tan θ = tan 90° = ∞(not defined)
(iv) For any line which is perpendicular to yaxis, the inclination is θ = 0°
Therefore, Slope(m) = tan θ = tan 0° = 0
Solution 6
Equation of any straight line in the form y = mx + c, where slope = m(coefficient of x) and
yintercept = c(constant term)
(i)_{}
_{}
Therefore,
_{}
_{}
(ii)_{}
_{}
Therefore,
_{}
_{}
(iii) _{}
_{}
Therefore,
_{}
_{}
(iv) _{}
_{}
Therefore,
_{}
_{}
(v) _{}
_{}
Therefore,
_{}
_{}
(vi) _{}
_{}
Therefore,
_{}
_{}
(vii) _{}
_{}
Therefore,
_{}
_{}
Solution 7
(i)Given
Slope is 2, therefore m = 2
Yintercept is 3, therefore c = 3
Therefore,
_{}
Therefore the equation of the required line is y = 2x + 3
(ii)Given
Slope is 5, therefore m = 5
Yintercept is 8, therefore c = 8
Therefore,
_{}
Therefore the equation of the required line is y = 5x + (8)
(iii)Given
Slope is 4, therefore m = 4
Yintercept is 2, therefore c = 2
Therefore,
_{}
Therefore the equation of the required line is y = 4x + 2
(iv)Given
Slope is 3, therefore m = 3
Yintercept is 1, therefore c = 1
Therefore,
_{}
Therefore the equation of the required line is y = 3x  1
(v)Given
Slope is 0, therefore m = 0
Yintercept is 5, therefore c = 5
Therefore,
_{}
Therefore the equation of the required line is y = 5
(vi)Given
Slope is 0, therefore m = 0
Yintercept is 0, therefore c = 0
Therefore,
_{}
Therefore the equation of the required line is y = 0
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 yintercept.
The required yintercept 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 yintercept.
The required yintercept is 6
Solution 10
Given line is
x + y = 5
The graph of the given line is shown below.
From the given line x + y = 5, we get
_{}
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 tan θ then inclination of the line is θ
Now from the equation (A) , we have
_{}
And c = 5
Therefore the required inclination is θ = 135° and yintercept is c = 5