## ICSE Class 9 Coordinate Geometry Solution New Pattern By Clarify Knowledge

*ICSE Class 9 Coordinate Geometry Solution New Pattern 2022*

## OUR EBOOKS CAN HELP YOU ICSE CLASS 10 BOARD

## CODE IS EASY CAN HELP YOU IN SEMESTER 2

## ICSE Class 9 Statistics Solution Table

- ICSE Class 9 Coordinate Geometry Solution New Pattern By Clarify Knowledge
- OUR EBOOKS CAN HELP YOU ICSE CLASS 10 BOARD
- CODE IS EASY CAN HELP YOU IN SEMESTER 2
- ICSE Class 9 Statistics Solution Table
- Chapter 26 - Coordinate Geometry Exercise Ex. 26(A)
- Chapter 26 - Coordinate Geometry Exercise Ex. 26(B)
- Chapter 26 - Coordinate Geometry Exercise Ex. 26(C)

## Chapter 26 - Coordinate Geometry Exercise 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

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

Therefore we get,

x = 2, y = 2Question 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.

FalseQuestion 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 Exercise 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 = 0Solution 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 = -2ySolution 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)

(iii) y = -x + 4

(iv)

(v)

(vi) 2x - 3y = 4

(vii)

(viii)

(ix) x + 5y + 2 =0Solution 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 |

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)

(iv)

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 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).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:

(ii)

First draw the graph as follows:

This is a right triangle.

Thus the area of the triangle will be:

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

(iv) 3x + 4y = 24

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 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.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) x_{1}, the value of x, when y = 3

(ii) x_{2}, the value of x, when y = -2.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 = 7Question 11

Draw the graph of the equation 3x - 4y = 12.

Use the graph drawn to find:

(i) y_{1}, the value of y, when x = 4.

(ii) y_{2}, 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 . Use the graph drawn to find:

(i) x_{1}, the value of x, when y = 10

(ii) y_{1}, 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 pass through the same point.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.

## Chapter 26 - Coordinate Geometry Exercise 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) 0^{o}(ii) 30^{o }(iii) 45^{o}(iv) 60^{o}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° = √3Question 4

Find the inclination of the line whose slope is:

(i) 0(ii) 1(iii) (iv) 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°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° = 0Question 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 = 0Solution 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 = 0Solution 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 = 0Question 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 3Question 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 -6Question 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