Chapter 26 : Coordinate Geometry - Selina Solutions for Class 9 Maths ICSE

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Chapter 26 - Coordinate Geometry Excercise Ex. 26(A)

Question 1

For each equation given below; name the dependent and independent variables.

(i) y = x -7

(ii) x = 9y + 4

(iii) x =

(iv) y = (6x + 5)

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

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)

Solution 2

Let us take the point as

,, ,,,,,,

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

, ,,,,,,

Question 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 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

left parenthesis 5 straight x minus 3 straight y right parenthesis plus left parenthesis 3 straight y minus 9 straight x right parenthesis equals left parenthesis 4 plus left parenthesis minus 12 right parenthesis right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space minus 4 straight x equals minus 8
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight x equals 2 

Putting the value of x in the equation (B), we get

space space space space straight y minus 3 straight x equals minus 4
rightwards double arrow straight y equals space 3 straight x minus 4
rightwards double arrow straight y equals 3 left parenthesis 2 right parenthesis minus 4
rightwards double arrow straight y equals 2

Therefore we get,

 x = 2, y = 2

Question 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 4

(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

State, true or false:

(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.

Solution 5

(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

Question 6

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)

(ii)

(iii) 

Solution 6

(i)

Now

Again

The co-ordinates of the point

(ii)

Now

Again

The co-ordinates of the point

(iii)

Now

Again

The co-ordinates of the point

Question 7

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).

(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 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

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.

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 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)

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.

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 co-ordinates of the diagonals of the square.

Therefore the co-ordinates of the mid-point of is and the co-ordinates of the mid-point of is and the co-ordinates of the diagonals of the square is

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)

Solution 10

After plotting the given points, we have clearly seen from the graph that

(i) , and are collinear.

(ii), and are non-collinear.

Question 11

Plot the point A(5, -7). From point A, draw AM perpendicular to x-axis and AN perpendicular to y-axis. Write the co-ordinates of points M and N.

Solution 11

Given

After plotting the given point on a graph paper. Now let us draw a perpendicular from the point on the x-axis and a perpendicular from the point on the y-axis.

As from the graph clearly we get the co-ordinates of the points and

Co-ordinate of the point is

Co-ordinate of the point is

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.

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

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.

Solution 13

Given that in rectangle ; point is origin and units along x-axis 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 co-ordinates of the vertices are , and

Chapter 26 - Coordinate Geometry Excercise Ex. 26(B)

Question 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 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

-6

-6

-6

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:

Question 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 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:

Question 3

Draw the graph for the each linear equation given below:

(i) y = 2x + 3

(ii)  straight y equals fraction numerator 2 straight x over denominator 3 end fraction minus 1

(iii) y = -x + 4

(iv)  straight y equals 4 straight x minus 5 over 2

(v) straight y equals fraction numerator 3 straight x over denominator 2 end fraction plus 2 over 3

(vi) 2x - 3y = 4

(vii) fraction numerator straight x minus 1 over denominator 3 end fraction minus fraction numerator straight y plus 2 over denominator 2 end fraction equals 0

(viii) straight x minus 3 equals 2 over 5 left parenthesis straight y plus 1 right parenthesis

(ix) x + 5y + 2 =0

Solution 3

(i)

First prepare a table as follows:

x

-1

0

1

y

 

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

minus 4 over 3

minus 2 over 3

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:

Question 4

Draw the graph for each equation given below:

(i) 3x +2y = 6

(ii) 2x - 5y = 10

(iii) 1 half straight x plus 2 over 3 straight y space equals 5

(iv) fraction numerator 2 straight x minus 1 over denominator 3 end fraction minus fraction numerator straight y minus 2 over denominator 5 end fraction equals 0

In each case, find the co-ordinates of the points where the graph (line ) drawn meets the co-ordinates axes.

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 straight x over 2 plus fraction numerator 2 straight y over denominator 3 end fraction equals 3 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 fraction numerator 2 straight x minus 1 over denominator 3 end fraction minus fraction numerator straight y minus 2 over denominator 5 end fraction equals 0 follows the steps:

First prepare a table as below:

X

-1

0

1

Y

-3

1 third

11 over 3

Now sketch the graph as shown:

From the graph it can verify that the line intersect x axis at open parentheses minus 1 over 10 comma 0 close parenthesesand y at (0,4.5).

Question 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 co-ordinates axes:

(i) 3x - (5 - y) = 7

(ii) 7 - 3 (1 - y) = -5 + 2x.

Solution 5

(i)

First draw the graph as follows:

This is an right trinangle.

Thus the area of the triangle will be:

equals 1 half cross times base cross times altitude
equals 1 half cross times 4 cross times 12
equals 24 space sq. units 

(ii)

First draw the graph as follows:

This is a right triangle.

Thus the area of the triangle will be:

straight A equals 1 half cross times base cross times altitude
equals 1 half cross times 9 over 2 cross times 3
equals 27 over 4 equals 6.75 space sq. units

Question 6

For each pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.

(i) y = 3x - 1

y = 3x + 2

(ii) y = x - 3

y = -x + 5

(iii) 2x - 3y = 6

straight x over 2 plus straight y over 3 equals 1

(iv) 3x + 4y = 24

straight x over 4 plus straight y over 3 equals 1

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=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.

(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=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.

 

(iii)

To draw the graph of 2x - 3y = 6 and straight x over 2 plus straight y over 3 equals 1 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 straight x over 4 plus straight y over 3 equals 1 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.

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.

Solution 7

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?

Solution 8

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.

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(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.

Solution 10

First prepare a table as follows:

X

-1

0

1

Y

2

3 over 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.

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.

Question 12

Draw the graph of equation straight x over 4 plus straight y over 5 equals 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.

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.

Question 13

Use the graphical method to show that the straight lines given by the equations x + y = 2, x - 2y = 5 and straight x over 3 plus straight y equals 0pass through the same point.

Solution 13

The equations can be written as follows:

y = 2 - x

straight y equals 1 half left parenthesis straight x minus 5 right parenthesis

straight y equals minus straight x over 3

First prepare a table as follows:

x

y = 2 - x

straight y equals 1 half left parenthesis straight x minus 5 right parenthesis

straight y equals minus straight x over 3

-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.

Chapter 26 - Coordinate Geometry Excercise Ex. 26(C)

Question 1

In each of the following, find the inclination of line AB:

 

(i)

(ii)

(iii)

Solution 1

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°

Question 2

Write the inclination of a line which is:

(i) Parallel to x-axis.

(ii) Perpendicular to x-axis.

(iii) Parallel to y-axis.

(iv) Perpendicular to y-axis.

Solution 2

(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°

Question 3

Write the slope of the line whose inclination is:

(i) 0o(ii) 30o (iii) 45o(iv) 60o

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° = fraction numerator 1 over denominator square root of 3 end fraction

(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

Question 4

Find the inclination of the line whose slope is:

(i) 0(ii) 1(iii) square root of 3(iv) fraction numerator 1 over denominator square root of 3 end fraction

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 square root of 3; then tan θ = √3

Now

Therefore the inclination of the given line is θ = 60°

(iv)Here the slope of line is fraction numerator 1 over denominator square root of 3 end fraction; then tan space straight theta equals fraction numerator 1 over denominator square root of 3 end fraction

Now

Therefore the inclination of the given line is θ = 30°

Question 5

Write the slope of the line which is:

(i) Parallel to x-axis.

(ii) Perpendicular to x-axis.

(iii) Parallel to y-axis.

(iv) Perpendicular to y-axis.

Solution 5

(i)For any line which is parallel to x-axis, the inclination is θ = 0°

Therefore, Slope(m) = tan θ = tan 0° = 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

Question 6

For each of the equation given below, find the slope and the y-intercept:

(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

Solution 6

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)

Therefore,

(ii)

Therefore,

(iii)

Therefore,

(iv)

Therefore,

(v)

Therefore,

(vi)

Therefore,

(vii)

Therefore,

Question 7

Find the equation of the line whose:

(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

Solution 7

(i)Given

Slope is 2, therefore m = 2

Y-intercept 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

Y-intercept 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

Y-intercept 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

Y-intercept 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

Y-intercept is -5, therefore c = -5

Therefore,

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,

Therefore the equation of the required line is y = 0

Question 8

Draw the line 3x + 4y = 12 on a graph paper. From the graph paper. Read the y-intercept of the line.

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 y-intercept.

The required y-intercept is 3

Question 9

Draw the line 2x - 3y - 18 = 0 on a graph paper. From the graph paper read the y-intercept of the line?

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 y-intercept.

The required y-intercept is -6

Question 10

Draw the graph of line x + y = 5. Use the graph paper drawn to find the inclination and the y-intercept of the line.

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 y-intercept is c = 5

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