ML Aggarwal Coordinate Geometry Exe-19.3 Class 9 ICSE Maths Solutions

Question 1. Solve the following equations graphically: 3x – 2y = 4, 5x – 2y = 0

Answer :

3x-2y = 4 …(i)

⇒ 2y = 3x-4

⇒ y = (3x-4)/2

When x = 0,

y = (3×0 -4)/2 = (0-4)/2 = -4/2 = -2

when x = 2,

y = (3×2 -4)/2 = (6-4)/2 = 2/2 = 1

when x = 4,

y = (3×4 -4)/2 = (12 -4)/2 = 8/2 = 4

Plot the above points on graph. Join them.

5x-2y = 0 …(ii)

⇒ 2y = 5x

⇒ y = 5x/2

When x = 0,

y = 0

When x = 2,

y = 5× 2/2 = 5

When x = -2,

y = 5× -2/2 = -5

Plot the above points on graph. Join them.

It is clear from the graph that the two lines intersect at (-2,-5).

So the solution of the given equations are x = -2 and y = -5.

Question 2. Solve the following pair of equations graphically. Plot at least 3 points for each straight line 2x – 7y = 6, 5x – 8y = – 4.

Answer :

2x-7y = 6 …(i)

⇒ 2x = 7y+6

⇒ x = (7y+6)/2

when y = 0

x = (7×0 +6)/2 = 6/2 = 3

when y = -1

x = (7×-1 +6)/2 = -1/2 = -0.5

when y = -2

x = (7×-2 +6)/2 = -8/2 = -4

Mark the above points on graph. Join them.

5x-8y = -4 …(ii)

⇒ 5x = 8y-4

⇒ x = (8y-4)/5

when y = 0

x = (8×0 -4)/5 = -4/5 = 0.8

when y = 3

x = (8×3 -4)/5 = (24-4)/5 = 20/5 = 4

when y = -2

x = (8×-2 -4)/5 = (-16-4)/5 = -20/5 = -4

Mark the above points on graph. Join them.

It is clear from the graph that the two lines intersect at (-4,-2).

So the solution of the given equations are x = -4 and y = -2.

Question 3. Using the same axes of co-ordinates and the same unit, solve graphically.

x+y = 0, 3x – 2y = 10

Answer :

x+y = 0 …(i)

⇒ y = -x

When x = -3,

y = 3

When x = -2,

y = 2

When x = -1,

y = 1

Mark the above points on graph. Join them.

3x-2y = 10 …(ii)

⇒ 3x = 2y+10

⇒ x = (2y+10)/3

When y = 1

x = (2×1 +10)/3 = 12/3 = 4

When y = -2

x = (2×-2 +10)/3 = 6/3 = 2

When y = 4

x = (2×4 +10)/3 = 18/3 = 6

Mark the above points on graph. Join them.

It is clear from the graph that the two lines intersect at (2,-2).

So the solution of the given equations are x = 2 and y = -2.

Question 4. Take 1 cm to represent 1 unit on each axis to draw the graphs of the equations 4x- 5y = -4 and 3x = 2y – 3 on the same graph sheet (same axes). Use your graph to find the solution of the above simultaneous equations.

Answer :

4x-5y = -4 …(i)

⇒ 4x = 5y-4

⇒ x = (5y-4)/4

When y = 0

x = (5×0 -4)/4 = -4/4 = -1

When y = 2

x = (5×2 -4)/4 = 6/4 = 1.5

When y = -2

x = (5×-2 -4)/4 = -14/4 = -3.5

Mark the above points on graph. Join them.

3x = 2y-3 …(ii)

⇒ x = (2y-3)/3

When y = 0,

x = (2×0 -3)/3 = -3/3 = -1

When y = 3,

x = (2×3 -3)/3 = 3/3 = 1

When y = -3,

x = (2×-3-3)/3 = -9/3 = -3

It is clear from the graph that the two lines intersect at (-1,0).

So the solution of the given equations are x = -1 and y = 0.

Question 5. Solve the following simultaneous equations graphically, x + 3y = 8, 3x = 2 + 2y

Answer :

x+3y = 8 …(i)

⇒ 3y = 8-x

⇒ y = (8-x)/3

When x = 8,

y = (8-8)/3 = 0

when x = 2,

y = (8-2)/3 = 6/3 = 2

when x = 5,

y = (8-5)/3 = 3/3 = 1

Mark the above points on graph. Join them.

3x = 2+2y

⇒ 2y = 3x-2

⇒ y = (3x-2)/2

When x = 2

y = (3×2 -2)/2 = 4/2 = 2

When x = 4

y = (3×4 -2)/2 = 10/2 = 5

When x = -2

y = (3×-2 -2)/2 = -8/2 = -4

It is clear from the graph that the two lines intersect at (2,2).

So the solution of the given equations are x = 2 and y = 2.

Question 6. Solve graphically the simultaneous equations 3y = 5 – x, 2x = y + 3 (Take 2cm = 1 unit on both axes).

Answer :

3y = 5-x

y = (5-x)/3

When x = 5,

y = (5-5)/3 = 0

When x = 2,

y = (5-2)/3 = 3/3 = 1

When x = -1,

y = (5-(-1))/3 =6/3 = 2

Mark the above points on graph. Join them.

2x = y+3 …(ii)

⇒ y = 2x-3

When x = 0,

y = (2×0 -3) = 0-3 = -3

When x = 1,

y = (2×1 -3) = 2-3 = -1

When x = 2,

y = (2×2 -3) = 4-3 = 1

It is clear from the graph that the two lines intersect at (2,1).

So the solution of the given equations are x = 2 and y = 1.

Question 7. Use graph paper for this question.

Take 2 cm = 1 unit on both axes.
(i) Draw the graphs of x +y + 3 = 0 and 3x-2y + 4 = 0. Plot only three points per line.
(ii) Write down the co-ordinates of the point of intersection of the lines.
(iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.

Answer :

(i) x+y+3 = 0 …(i)

⇒ y = -x-3

When x = -3

y = 3-3 = 0

when x = -2

y = 2-3 = -1

when x = -1

y = 1-3 = -2

Mark the above points on graph. Join them.

3x-2y+4 = 0 …(ii)

⇒ 2y = 3x+4

⇒ y = (3x+4)/2

When x = -4

y = (3×-4 +4)/2 = (-12+4)/2 = -8/2 = -4

When x = -2

y = (3×-2 +4)/2 = (-6+4)/2 = -2/2 = -1

When x = 2

y = (3×2 +4)/2 = (6+4)/2 = 10/2 = 5

(ii) The two lines intersect at (-2,-1).

(iii) Measure the distance from origin to the point (-2,-1).

The distance of the point of intersection of the lines from the origin is 4.5 cm.

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Coordinate Geometry Exercise-19.3

ML Aggarwal Class 9 ICSE Maths Solutions

Page 458

Question 8. Solve the following simultaneous equations, graphically :

2x-3y + 2 = 4x+ 1 = 3x – y + 2

Answer :

2x-3y+2 = 4x+1

⇒ 3y = 2x-4x+2-1

⇒ 3y = -2x+1

⇒ y = (-2x+1)/3

When x = -1,

y = (-2×-1 +1)/3 = 3/3 = 1

When x = 2,

y = (-2×2 +1)/3 = -3/3 = -1

When x = 0.5,

y = (-2×0.5 +1)/3 = 0

Mark the above points on graph. Join them.

Consider second equation.

4x+1 = 3x-y+2

⇒ y = 3x-4x+2-1

⇒ y = -x+1

When x = 0

y = 0+1 = 1

When x = 1

y = -1+1 = 0

When x = 2

y = -2+1 = -1

It is clear from the graph that the two lines intersect at (2,-1).

So the solution of the given equations are x = 2 and y = -1.

Question 9. Use graph paper for this question.

(i) Draw the graphs of 3x -y – 2 = 0 and 2x + y – 8 = 0. Take 1 cm = 1 unit on both axes and plot three points per line.
(ii) Write down the co-ordinates of the point of intersection and the area of the traingle formed by the lines and the x-axis.

Answer :

(i) 3x-y-2 = 0 …(i)

y = 3x-2

When x = 0, y = 3×0 -2 = 0-2 = -2

When x = 1, y = 3×1 -2 = 3-2 = 1

When x = 2, y = 3×2 -2 = 6-2 = 4

Mark the above points on graph. Join them.

2x+y-8 = 0 …(ii)

y = -2x+8

When x = 1, y = -2×1 +8 = -2+8 = 6

When x = 2, y = -2×2 +8 = -4+8 = 4

When x = 3, y = -2×3 +8 = -6+8 = 2

(ii) The coordinates of the points of intersection are (2,4).

Area of the triangle formed = ½ ×base×height

= ½ ×3.4×4

= 6.8 sq. units

Hence,

area of the triangle is 6.8 sq. units.

Question 10. Solve the following system of linear equations graphically : 2x -y – 4 = 0, x + y + 1 = 0. Hence, find the area of the triangle formed by these lines and the y-axis.

Answer :

2x-y-4 = 0 …(i)

y = 2x-4

When x = 1, y = 2×1 -4 = 2-4 = -2

When x = 2, y = 2×2 -4 = 4-4 = 0

When x = 3, y = 2×3 -4 = 6-4 = 2

x+y+1 = 0 …(ii)

⇒ y = -x-1

When x = 0, y = 0-1 = -1

When x = -2, y = 2-1 = 1

When x = -1, y = 1-1 = 0

It is clear from the graph that the two lines intersect at (1,-2).

So the solution of the given equations are x = 1 and y = -2.

The area of the triangle formed by these lines and Y axis = ½ × base × height

= ½ ×3×1

= 1.5 sq. units

Hence,

area of the triangle is 1.5 sq. units.

Question 11. Solve graphically the following equations: x + 2y = 4, 3x – 2y = 4

Take 2 cm = 1 unit on each axis. Write down the area of the triangle formed by the lines and the x-axis.

Answer :

x+2y = 4 …(i)

⇒ 2y = 4-x

⇒ y = (4-x)/2

When x = 0, y = (4-0)/2 = 4/2 = 2

When x = 2, y = (4-2)/2 = 2/2 = 1

When x = 4, y = (4-4)/2 = 0/2 = 0

3x-2y = 4 …(ii)

⇒ 2y = 3x-4

⇒ y = (3x-4)/2

When x = 0, y = (3×0 -4)/2 = (0-4)/2 = -4/2 = -2

When x = 2, y = (3×2 -4)/2 = (6-4)/2 = 2/2 = 1

When x = 4, y = (3×4 -4)/2 = (12-4)/2 = 8/2 = 4

It is clear from the graph that the two lines intersect at (2,1).

So the solution of the given equations are x = 2 and y = 1.

The area of the triangle formed by these lines and X axis = ½ ×base×height

= ½ ×2.7×1

= 1.35 sq. units

Hence area of the triangle is 1.35 sq. units.

Question 12. On graph paper, take 2 cm to represent one unit on both the axes, draw the lines : x + 3 = 0, y –  2 = 0, 2x + 3y = 12 .

Write down the co-ordinates of the vertices of the triangle formed by these lines.

Answer :

x+3 = 0 …(i)

⇒ x = -3

The graph of x = -3 will be a line passing through x = -3 parallel to Y axis.

y-2 = 0 …(ii)

⇒ y = 2

The graph of y = 2 will be a line passing through y = 2 parallel to X axis.

2x+3y = 12 …(iii)

⇒ 3y = 12-2x

⇒ y = (12-2x)/3

When x = 0, y = (12- 2×0)/3 = 12/3 = 4

When x = 3, y = (12- 2×3)/3 = (12-6)/3 = 6/3 = 2

When x = 6, y = (12- 2×6)/3 = (12-12)/3 = 0

From the graph, it is clear that the vertices of the triangle formed by the lines are A(-3,2), B(-3,6) and C(3,2).

Question 13. Find graphically the co-ordinates of the vertices of the triangle formed by the lines y = 0, y – x and 2x + 3y= 10. Hence find the area of the triangle formed by these lines.

Answer :

y = 0 …(i)

The graph of y = 0 is the X axis.

y = x ..(ii)

When x = 1, y = 1.

When x = 2, y = 2.

When x = 3, y = 3.

2x+3y = 10 …(iii)

⇒ 3y = 10-2x

⇒ y = (10-2x)/3

When x = 0.5, y = (10- 2×0.5)/3 = (10-1)/3 = 9/3 = 3

When x = 2, y = (10- 2×2)/3 = (10-4)/3 = 6/3 = 2

When x = 5, y = (10- 2×5)/3 = (10-10)/3 = 0

From the graph, it is clear that the vertices of the triangle formed by the lines are A(2,2), B(5,0) and C(0,0).

Area of triangle formed by these lines = ½ × base × height

= ½ ×5×2

= 5 sq. units

Hence,  area of the triangle is 5 sq. units.

—  : End of ML Aggarwal Coordinate Geometry Exe-19.3 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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