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 Ist quadrant.
- If both the coordinates are negative, then the point lies in the IIIrd quadrant.
- If the X coordinate is negative and Y-co-ordinate is positive, then the point lies in the IInd quadrant.
- If the X coordinate is positive and Y coordinate is negative, then the point lies in the IVth quadrant.
Solution 8:
- The point (x, y) will lie in the IVth quadrant.
- The point (x, y) will lie in the IInd quadrant.
- The point (x, y) will lie in the Ist quadrant.
- The point (x, y) will lie in the IIIrd 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 areĀ 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