**Graphs – Maharashtra Board Class 9 Solutions for Algebra**

AlgebraGeometryScience and TechnologyHindi

**Exercise – 5.1**

**Solution 1:**

**Solution 2:**

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

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

**Solution 8:**

- The point (x, y) will lie in the IV
^{th}quadrant. - The point (x, y) will lie in the II
^{nd}quadrant. - The point (x, y) will lie in the I
^{st}quadrant. - The point (x, y) will lie in the III
^{rd}quadrant. - The point (x, y) will lie on the Y-axis.
- 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.

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

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