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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 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

, ,,,,,,

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

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)

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

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

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 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)

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

Solution 10

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

(i) , and are collinear.

(ii), and are non-collinear.

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

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 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

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

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:

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).

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

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.

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:

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.

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.

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

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

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.

Coordinate Geometry Exercise Ex. 26(C)

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°

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°

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

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°

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

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,

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

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

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

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