SELINA Solution Class 9 Chapter 26 Co-ordinate Geometry Exercise 26 A

Question 1

For the equation given below; name the dependent and independent variables.
y = ${\displaystyle \frac{4}{3}x}$ - 7

Sol:

y = ${\displaystyle \frac{4}{3}x}$ - 7

Dependent variable is y

Independent variable is x.

Question 1.2

For the equation given below; name the dependent and independent variables.
 x = 9y + 4

Sol:

 x = 9y + 4

Dependent variable is x

Independent variable is y.

Question 1.3

For the equation given below; name the dependent and independent variables.
x = ${\displaystyle \frac{5y+3}{2}}$

Sol:

x = ${\displaystyle \frac{5y+3}{2}}$

Dependent variable is x

Independent variable is y.

Question 1.4

For each equation given below; name the dependent and independent variables.
y = ${\displaystyle \frac{1}{7}}$ (6x + 5)

Sol:

y = ${\displaystyle \frac{1}{7}}$ (6x + 5)

Dependent variable is y

Independent variable is x.

Question 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)

Sol:

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)

Question 3.1

Find the values of x and y if:
(x - 1, y + 3) = (4, 4)

Sol:

Two ordered pairs are equal.
⇒ Therefore their first components are equal and their second components too are separately equal.
(x - 1, y + 3) = (4, 4)
x - 1 = 4 and y + 3 = 4
x = 5 and y = 1

Question 3.2

Find the values of x and y if:
(3x + 1, 2y - 7) = (9, - 9)

Sol:

Two ordered pairs are equal.
⇒ Therefore their first components are equal and their second components too are separately equal.
(3x + 1, 2y - 7) = (9, - 9)
3x + 1 = 9 and 2y - 7 = -9
3x = 8 and 2y = -2
x = ${\displaystyle \frac{8}{3}}$ and y = -1

Question 3.3

Find the values of x and y if:
(5x - 3y, y - 3x) = (4, -4)

Sol:

Two ordered pairs are equal.
⇒ Therefore their first components are equal and their second components too are separately equal.
(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

Question 4

Use the graph given alongside, to find the coordinates of the 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.

Sol:

(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)

Question 5.1 MCQ

State, true or false:
The ordinate of a point is its x-co-ordinate.

Sol:

The ordinate of a point is its x-coordinate. - False.

Question 5.2

State, true or false:
The origin is in the first quadrant.

Sol:

The origin is in the first quadrant. - False.

Question 5.3

State, true or false:
The y-axis is the vertical number line.

Sol:

The y-axis is the vertical number line. - True.

Question 5.4

State, true or false:
Every point is located in one of the four quadrants.

Sol:

Every point is located in one of the four quadrants. - True.

Question 5.5

State, true or false:
If the ordinate of a point is equal to its abscissa; the point lies either in the first quadrant or in the second quadrant.

Sol:

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.

Question 5.6

State, true or false:
The origin (0, 0) lies on the x-axis.

Sol:

The origin (0,0) lies on the x-axis. - True.

Question 5.7

State, true or false:
The point (a, b) lies on the y-axis if b = 0.

Sol:

The point (a,b) lies on the y-axis if b=0. - False

Question 6.1

In 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:
3 - 2x = 7; 2y + 1 = 10 - 2${\displaystyle \frac{1}{2}}$y.

Sol:

3 - 2x = 7; 2y + 1 = 10 - 2${\displaystyle \frac{1}{2}}$y
Now
3 - 2x = 7
3 - 7 = 2x
-4 = 2x
-2 = x
Again
2y + 1 = 10 - 2${\displaystyle \frac{1}{2}}$y
2y + 1 = 10 -${\displaystyle \frac{5}{2}}$y
4y + 2 = 20 - 5y
4y + 5y = 20 - 2
9y = 18
y = 2
∴ The co-ordinates of the point (-2, 2)

Question 6.2

In 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:
${\displaystyle \frac{2\text{a}}{3}-1=\frac{\text{a}}{2};\frac{15-4\text{b}}{7}=\frac{2\text{b}-1}{3}}$.

Sol:

${\displaystyle \frac{2\text{a}}{3}-1=\frac{\text{a}}{2};\frac{15-4\text{b}}{7}=\frac{2\text{b}-1}{3}}$.
Now
${\displaystyle \frac{2\text{a}}{3}-1=\frac{\text{a}}{2}}$
${\displaystyle \frac{2\text{a}}{3}-\frac{\text{a}}{2}=1}$
${\displaystyle \frac{4\text{a}-3\text{a}}{6}=1}$
${\displaystyle \text{a}=6}$
Again
${\displaystyle \frac{15-4\text{b}}{7}=\frac{2\text{b}-1}{3}}$
45 - 12b = 14b - 7
45 + 7 = 14b + 12b
52 = 26b
2 = b
∴ The co-ordinates of the point (6, 2)

Question 6.3

In 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:
${\displaystyle 5x-(5-x)=\frac{1}{2}(3-x);4-3y=\frac{4+y}{3}}$

Sol:

${\displaystyle 5x-(5-x)=\frac{1}{2}(3-x);4-3y=\frac{4+y}{3}}$
Now
${\displaystyle 5x-(5-x)=\frac{1}{2}(3-x)}$
${\displaystyle (5x+x)-5=\frac{1}{2}(3-x)}$
12 x - 10 = 3 - x
12x + x = 3 +10
13x = 13
x = 1
Again
4 - 3y = ${\displaystyle \frac{4+y}{3}}$
12 - 9y = 4 + y
12 - 4 = y + 9y
8 = 10y
${\displaystyle \frac{8}{10}}$ = y
${\displaystyle \frac{4}{5}}$= y
∴ The co-ordinates of the point ${\displaystyle (1,\frac{4}{5})}$

Question 7.1

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:
A(2, 0), B(8, 0) and C(8, 4).

Sol:

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)

Question 7.2

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:
A (4, 2), B(-2, 2) and D(4, -2).

Sol:

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)

Question 7.3

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:
A (- 4, - 6), C(6, 0) and D(- 4, 0).

Sol:

A (- 4, - 6), C(6, 0) and D(- 4, 0)

After plotting the given points A(- 4, - 6), C(6, 0) and D(- 4, 0) on a graph paper; joining D with and A with C. From the graph it is clear that the vertical distance between the points D( - 4, 0) and A(- 4, - 6) is 6 units and the horizontal distance between the points D(- 4, 0) and C(6, 0) is 10 units, therefore the vertical distance between the points C(6, 0) and B must be 6 units and the horizontal distance between the points A(- 4, - 6) and B must be 10 units. Now complete the rectangle ABCD

As is clear from the graph B(6, - 6)

Question 7.4

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:
B (10, 4), C(0, 4) and D(0, -2).

Sol:

 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 D(0, - 2) 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).

Question 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.

Sol:

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 midpoints of the sides AB and CD.
Therefore the coordinates of the mid-point of AB is E(3, 2) and the coordinates of the mid-point of CD are F(- 1, - 4).

Question 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 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.

Sol:

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 midpoints of the sides BC and CD the co-ordinates of the diagonals of the square.
Therefore the coordinates of the mid-point of BC is E(4, 7) and the coordinates of the mid-point of CD is F(1,10) and the coordinates of the diagonals of the square is G(1, 7).

Question 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)

Sol:

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.
(ii) P(-2, -1), Q(-1,-4) and R(-4,1) are non-collinear.

Question 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.

Sol:

Given A(5, -7)

After plotting the given point A(5,-7) on a graph paper. Now let us draw a perpendicular AM from 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)

Question 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.

Sol:

Given that in square ABCD ; A(3,4), B(-2,4) and C(-2,-1)

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

Question 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.

Sol:

Given that in rectangle OABC; point O is the origin and OA = 10 units along x-axis therefore we get O(0,0) and A(0,0). Also it is given that AB = 8 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), B(10,8) and C(0,8)

Question 1.01

Draw the graph for the linear equation given below:
x = 3

Sol:

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:

Question 1.02

Draw the graph for the linear equation given below:
x + 3 = 0

Sol:

First prepare a table as follows:

x	- 3	- 3	- 3
y	- 1	0	1

Thus the graph can be drawn as follows:

Question 1.03

Draw the graph for the linear equation given below:
x - 5 = 0

Sol:

First, prepare a table as follows:

x	5	5	5
y	-1	0	1

Thus the graph can be drawn as follows:

Question 1.04

Draw the graph for the linear equation given below:
2x - 7 = 0

Sol:

The equation can be written as:
x = ${\displaystyle \frac{7}{2}}$
First prepare a table as follows:

x	${\displaystyle \frac{7}{2}}$	${\displaystyle \frac{7}{2}}$	${\displaystyle \frac{7}{2}}$
y	-1	0	1

Thus the graph can be drawn as follows:

Question 1.05

Draw the graph for the linear equation given below:
y = 4

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	4	4	4

Thus the graph can be drawn as follows:

Question 1.06

Draw the graph for the linear equation given below:
y + 6 = 0

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 6	- 6	- 6

Thus the graph can be drawn as follows:

Question 1.07

Draw the graph for the linear equation given below:
y - 2 = 0

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	2	2	2

Thus the graph can be drawn as follows:

Question 1.08

Draw the graph for the linear equation given below:
3y + 5 = 0

Sol:

First prepare a table as follows:

x	- 1	0	1
y	- 6	- 6	- 6

Thus the graph can be drawn as follows:

Question 1.09

Draw the graph for the linear equation given below:
2y - 5 = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{5}{2}}$	${\displaystyle \frac{5}{2}}$	${\displaystyle \frac{5}{2}}$

Thus the graph can be drawn as follows:

Question 1.1

Draw the graph for the linear equation given below:
y = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	0	0	0

Thus the graph can be drawn as follows:

Question 1.11

Draw the graph for the linear equation given below:
x = 0

Sol:

First prepare a table as follows:

x	0	0	0
y	-1	0	1

Thus the graph can be drawn as follows:

Question 2.1

Draw the graph for the linear equation given below:
y = 3x

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 3	0	3

Thus the graph can be drawn as follows:

Question 2.2

Draw the graph for the linear equation given below:
y = - x

Sol:

First prepare a table as follows:

x	- 1	0	1
y	1	0	- 1

Thus the graph can be drawn as follows:

Question 2.3

Draw the graph for the linear equation given below:
y = - 2x

Sol:

First prepare a table as follows:

x	- 1	0	1
y	2	0	- 2

Thus the graph can be drawn as follows:

Question 2.4

Draw the graph for the linear equation given below:
y = x

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 1	0	1

Thus the graph can be drawn as follows:

Question 2.5

Draw the graph for the linear equation given below:
5x+ y = 0.

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	5	0	- 5

Thus the graph can be drawn as follows:

Question 2.6

Draw the graph for the linear equation given below:
x + 2y = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{1}{2}}$	0	-${\displaystyle \frac{1}{2}}$

Thus the graph can be drawn as follows:

Question 2.7

Draw the graph for the linear equation given below:
4x - y = 0

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 4	0	4

Thus the graph can be drawn as follows:

Question 2.8

Draw the graph for the linear equation given below:
3x + 2y = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{3}{2}}$	0	-${\displaystyle \frac{3}{2}}$

Thus the graph can be drawn as follows:

Question 2.9

Draw the graph for the linear equation given below:
x = - 2y

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{1}{2}}$	0	-${\displaystyle \frac{1}{2}}$

Thus the graph can be drawn as follows:

Question 3.1

Draw the graph for the linear equation given below:
y = 2x + 3

Sol:

First, prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{5}{3}}$	3	5

Thus the graph can be drawn as follows:

Question 3.2

Draw the graph for the linear equation given below:
y = ${\displaystyle \frac{2x}{3}-1}$

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{5}{3}}$	-1	${\displaystyle -\frac{1}{3}}$

Thus the graph can be drawn as follows:

Question 3.3

Draw the graph for the linear equation given below:
y = - x + 4

Sol:

First, prepare a table as follows:

x	-1	0	1
y	5	4	3

Thus the graph can be drawn as follows:

Question 3.4

Draw the graph for the linear equation given below:
y = ${\displaystyle 4x-\frac{5}{2}}$

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{13}{2}}$	${\displaystyle -\frac{5}{2}}$	${\displaystyle \frac{3}{2}}$

Thus the graph can be drawn as follows:

Question 3.5

Draw the graph for the each linear equation given below:
y = ${\displaystyle \frac{3x}{2}+\frac{2}{3}}$

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{5}{6}}$	${\displaystyle \frac{2}{3}}$	${\displaystyle \frac{13}{6}}$

Thus the graph can be drawn as follows:

Question 3.6

Draw the graph for the linear equation given below:
2x - 3y = 4

Sol:

First prepare a table as follows:

x	- 1	0	1
y	- 2	${\displaystyle -\frac{4}{3}}$	${\displaystyle -\frac{2}{3}}$

Thus the graph can be drawn as follows

:

Question 3.7

Draw the graph for the linear equation given below:
${\displaystyle \frac{x-1}{3}-\frac{y+2}{2}=0}$

Sol:

The equation will become:
2x - 3y = 8
First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{10}{3}}$	${\displaystyle -\frac{8}{3}}$	-2

Thus the graph can be drawn as follows:

Question 3.8

Draw the graph for the linear equation given below:
x - 3 = ${\displaystyle \frac{2}{5}(y+1)}$

Sol:

The equation will become:
5x - 2y = 17
First prepare a table as follows:

x	- 1	0	1
y	- 11	${\displaystyle -\frac{17}{2}}$	- 6

Thus the graph can be drawn as follows:

Question 3.9

Draw the graph for the linear equation given below:
x + 5y + 2 = 0

Sol:

First prepare a table as follows:

x	- 1	0	1
y	${\displaystyle -\frac{1}{5}}$	${\displaystyle -\frac{2}{5}}$	${\displaystyle -\frac{3}{5}}$

Thus the graph can be drawn as follows:

Question 4.1

Draw the graph for the equation given below:
3x + 2y = 6

Sol:

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 intersects the x-axis at (2,0) and y at (0,3).

Question 4.2

Draw the graph for the equation given below:
2x - 5y = 10

Sol:

To draw the graph of 2x - 5y = 10 follows the steps:
First, prepare a table as below:

X	-1	0	1
Y	${\displaystyle -\frac{12}{5}}$	-2	${\displaystyle -\frac{8}{5}}$

Now sketch the graph as shown

:
From the graph it can verify that the line intersects the x-axis at (5,0) and y at (0,-2).

Question 4.3

Draw the graph for the equation given below:
${\displaystyle \frac{1}{2}x+\frac{2}{3}y=5}$.

Sol:

To draw the graph of ${\displaystyle \frac{x}{2}+\frac{2y}{3}=5}$ 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 the x-axis at (10,0) and y at (0,7.5).

Question 4.4

Draw the graph for the equation given below:
${\displaystyle \frac{2x-1}{3}-\frac{y-2}{5}=0}$

Sol:

To draw the graph of  ${\displaystyle \frac{2x-1}{3}-\frac{y-2}{5}=0}$  follows the steps:
First prepare a table as below:

X	-1	0	1
Y	-3	${\displaystyle \frac{1}{3}}$	${\displaystyle \frac{11}{3}}$

Now sketch the graph as shown:

From the graph it can verify that the line intersects the x-axis at ${\displaystyle (-\frac{1}{10},0)}$ and y at (0,4.5).

Question 5.1

For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
3x - (5 - y) = 7

Sol:

First draw the graph as follows:

This is an right triangle.

Thus the area of the triangle will be:

= ${\displaystyle \frac{1}{2}\times \text{base}\times \text{altitude}}$

=  ${\displaystyle \frac{1}{2}\times 4\times 12}$

=  24 sq.units

Question 5.2

For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
7 - 3 (1 - y) = - 5 + 2x.

Sol:

First draw the graph as follows:

This is a right triangle.
Thus the area of the triangle will be:

A = ${\displaystyle \frac{1}{2}\times \text{base}\times \text{altitude}}$

 = ${\displaystyle \frac{1}{2}\times \frac{9}{2}\times 3}$

= ${\displaystyle \frac{27}{4}}$

= 6.75 sq.units

Question 6.1

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
y = 3x - 1
y = 3x + 2

Sol:

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 = 3x -1	- 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.

Question 6.2

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
y = x - 3
y = - x + 5

Sol:

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 = x - 3	- 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.

Question 6.3

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
2x - 3y = 6
${\displaystyle \frac{x}{2}+\frac{y}{3}=1}$

Sol:

To draw the graph of 2x - 3y = 6 and ${\displaystyle \frac{x}{2}+\frac{y}{3}=1}$ follows the steps:
First prepare a table as below:

X	-1	0	1
Y = ${\displaystyle \frac{2}{3}\times 2}$	${\displaystyle -\frac{8}{3}}$	-2	${\displaystyle -\frac{4}{3}}$
Y = ${\displaystyle -\frac{3}{2}\times +3}$	${\displaystyle \frac{9}{2}}$	3	${\displaystyle \frac{3}{2}}$

Now sketch the graph as shown:

From the graph it can verify that the lines are perpendicular.

Question 6.4

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
3x + 4y = 24
${\displaystyle \frac{x}{4}+\frac{y}{3}=1}$

Sol:

To draw the graph of 3x + 4y = 24 and ${\displaystyle \frac{x}{4}+\frac{y}{3}=1}$ follows the steps:
First prepare a table as below

X	-1	0	1
Y = ${\displaystyle -\frac{3}{4}\times +6}$	${\displaystyle \frac{27}{4}}$	6	${\displaystyle \frac{21}{4}}$
Y = ${\displaystyle -\frac{3}{4}\times +3}$	${\displaystyle \frac{15}{4}}$	3	${\displaystyle \frac{9}{4}}$

Now sketch the graph as shown:

From the graph it can verify that the lines are parallel.

Question 7

On the same graph paper, plot the graph of y = x - 2, y = 2x + 1 and y = 4 from x= - 4 to 3.

Sol:

First, prepare a table as follows:

X	-1	0	1
Y = x - 2	-3	-2	-1
Y = 2x + 1	-1	1	3
Y = 4	4	4	4

Now the graph can be drawn as follows:

Question 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?

Sol:

First, prepare a table as follows:

X	-1	0	1
Y = 2x - 1	-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.

Question 9

The graph of 3x + 2y = 6 meets the x=axis at point P and the y-axis at point Q. Use the graphical method to find the co-ordinates of points P and Q.

SoL:

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 intersects the x-axis at (2,0) and y at (0,3), therefore the coordinates of P(x-axis) and Q(y-axis) are (2,0) and (0,3) respectively.

Question 10

Draw the graph of equation x + 2y - 3 = 0. From the graph, find:
(i) x₁, the value of x, when y = 3
(ii) x₂, the value of x, when y = - 2.

Sol:

First, prepare a table as follows:

X	-1	0	1
Y	2	${\displaystyle \frac{3}{2}}$	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

Question 11

Draw the graph of the equation 3x - 4y = 12.
Use the graph drawn to find:
(i) y₁, the value of y, when x = 4.
(ii) y₂, the value of y, when x = 0.

Sol:

First, prepare a table as follows:

X	- 1	0	1
Y	${\displaystyle -\frac{15}{4}}$	- 3	${\displaystyle -\frac{9}{4}}$

The graph of the equation can be drawn as follows:

From the graph, it can verify that
If x = 4 the value of y = 0
If x = 0 the value of y = - 3.

Question 12

Draw the graph of equation ${\displaystyle \frac{x}{4}+\frac{y}{5}=1}$ Use the graph drawn to find:
(i) x₁, the value of x, when y = 10
(ii) y₁, the value of y, when x = 8.

Sol

First, prepare a table as follows:

X	-1	0	1
Y	${\displaystyle \frac{25}{4}}$	5	${\displaystyle \frac{15}{4}}$

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.

Question 13

Use the graphical method to show that the straight lines given by the equations x + y = 2, x - 2y = 5 and ${\displaystyle \frac{x}{3}+y=0}$ pass through the same point.

Sol:

The equations can be written as follows:
y = 2 - x

y = ${\displaystyle \frac{1}{2}(x-5)}$

y = ${\displaystyle -\frac{x}{3}}$
First prepare a table as follows:

X	Y = 2 - x	Y = ${\displaystyle \frac{1}{2}(x-5)}$	Y = ${\displaystyle -\frac{x}{3}}$
- 1	3	- 3	${\displaystyle \frac{1}{3}}$
0	2	${\displaystyle -\frac{5}{2}}$	0
1	1	- 2	${\displaystyle -\frac{1}{3}}$

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.