Graphs – Maharashtra Board Class 9 Solutions for Algebra

Graphs – Maharashtra Board Class 9 Solutions for Algebra

AlgebraGeometryScience and TechnologyHindi

Exercise – 5.1

Solution 1:

Solution 2:

1. The x-coordinate of point P is 4.
2. The y-coordinate of point Q is 2
3. The y-coordinate of point R is 1.
4. The x-coordinate of point S is 0.
5. The y-coordinate of point S is -4.

Solution 3:

A(4, 1), B(3, -5), C(4, 0), D(0, 6)

Solution 4:

Solution 5:

Solution 6:

Solution 7:

1. If both the coordinates are positive, then the point lies in the I^st quadrant.
2. If both the coordinates are negative, then the point lies in the III^rd quadrant.
3. If the X coordinate is negative and Y-co-ordinate is positive, then the point lies in the II^nd quadrant.
4. If the X coordinate is positive and Y coordinate is negative, then the point lies in the IV^th quadrant.

Solution 8:

1. The point (x, y) will lie in the IV^th quadrant.
2. The point (x, y) will lie in the II^nd quadrant.
3. The point (x, y) will lie in the I^st quadrant.
4. The point (x, y) will lie in the III^rd quadrant.
5. The point (x, y) will lie on the Y-axis.
6. The point (x, y) will lie on the X-axis.

Solution 9:

Exercise – 5.2

Solution 1:

Solution 2:

The x-coordinates of points are given below.

1. The y-coordinate of point B is zero.
2. Point F has coordinates (-1, -5) and point A has coordinates (3, 2).
3. The line AH is parallel to the Y-axis.
4. The x-coordinate of point P and Q is same.
5. The y-coordinate of point E is 3.
6. The line EQ is parallel to the X-axis.
7. The x-coordinate of point M on line AH is 3.

Solution 3:

Solution 4:

Exercise – 5.3

Solution 1(i):

Solution 1(ii):

Solution 1(iii):

We have x + 5 = 0
i.e. x = -5

Solution 1(iv):

Solution 1(v):

We have 2y + 1 = y + 3
i.e. 2y – y = 3 – 1
i.e. y = 2

Solution 1(vi):

We have 3(x + 1) = 2x – 3
i.e. 3x + 3 = 2x – 3
i.e. 3x – 2x = -3 – 3
i.e. x = – 6

Solution 1(vii):

We have x – 4 = 0
i.e. x = 4

Solution 1(viii):

We have 2y + 3 = 0
i.e. 2y = -3
i.e. y = -1.5

Solution 1(ix):

We have 4x – 6 = 0
i.e. 4x = 6
i.e. x = 1.5

Solution 2(i):

We have 3x + 4 = 2(x + 5)
i.e. 3x + 4 = 2x + 10
i.e. 3x – 2x = 10 – 4
i.e. x = 6

Solution 2(ii):

We have 2x – 7 = 3(x – 2)
i.e. 2x – 7 = 3x – 6
i.e. 3x – 2x = 6 – 7
i.e. x = -1

Solution 3:

Writing the equation of the lines:

1. The equation of line PQ is y = 3.5.
2. The equation of line RS is x = 6.
3. The equation of line CD is x = -3.
4. The equation of line MN is y = -5.
5. The equation of line X’X is y = 0.
6. The equation of line YY’ is x = 0.

Exercise – 5.4

Solution 1:

Solution 2(i):

Given equation is x + 2y = 0.
Rewriting it we get,
2y = -x
i.e. y = -0.5x

Solution 2(ii):

Given equation is 3x – 2y = 0.
Rewriting it we get,
2y = 3x
i.e. y = 1.5x

Solution 2(iii):

Given equation is -3x + 4y = 12.
Rewriting it we get,
4y = 3x + 12
i.e. y = 0.75x + 3

Solution 2(iv):

Solution 2(v):

Solution 3:

Given equation is 3x + 2y = 6.
Rewriting it we get,
2y = -3x + 6
i.e. y = -1.5x + 3

From the graph, it can be clearly seen that the equation 3x + 2y = 6 intersects the y-axis at (0, 3).

Solution 4:

From the graph, it can be clearly seen that the points P, R and Q are collinear.
Also, the line passing through these lines is parallel to the x-axis.

Solution 5(i):

The coordinates of the points P, Q and R are as shown below.

From the above table,
y = -2x + 1

Solution 5(ii):

The co-ordinates of the points M, L and N are as shown below.

From the above table,
y = x – 1