ICSE Class 9 Maths Chapter 27 Graphical Solution

27. Graphical Solution

Solution of Simultaneous Linear Equations, Graphically

27.1 Graphs of Linear Equations in Two Variables

An equation of the form \(ax + by + c = 0\) is called a linear equation in two variables with \(x\) and \(y\) as the variables and \(a\), \(b\), \(c\) as the constants.

Steps to draw the graph of a linear equation ax + by + c = 0:

1. Make \(x\) or \(y\), the subject of the equation.

2. Give at least three suitable values to the variable on the right-hand side (the subject of equation being on the left-hand side) and find the corresponding values of the variable on the left-hand side.

3. Construct a table for the different pairs of values of \(x\) and \(y\).

4. Plot at least three ordered pairs (points) from the table on a graph paper.

5. Draw a straight line passing through the points plotted on the graph.

Draw the graph of 3x + y - 4 = 0.

Solution:

\(3x + y - 4 = 0\)

\(\Rightarrow y = -3x + 4\) [Making \(y\), the subject]

Now give at least three different values to the variable \(x\) and find the corresponding values of \(y\).

Let \(x = 1\); then \(y = -3 \times 1 + 4 = 1\)

Let \(x = -1\); then \(y = -3 \times -1 + 4 = 7\)

Let \(x = 0\); then \(y = -3 \times 0 + 4 = 4\)

Table for \(x\) and \(y\) is:

Plot the points (1, 1), (-1, 7) and (0, 4) on a graph paper. Draw a straight line passing through the points plotted on the graph paper.

Use the table given alongside to draw the graph. Use the graph drawn to find the values of 'a' and 'b'. State the linear relation between the variables x and y.

Solution:

Plot the given points (-2, -3), (0, 1) and (1, 3) on a graph paper. Draw a straight line passing through these points.

To find the value of 'a':

Through \(x = 2\), draw a vertical line which meets the graph at a point, say P. Through P, draw a horizontal line which meets the y-axis at \(y = 5\).

The value of unknown 'a' in the table is 5. Ans.

Similarly, through \(y = -7\), draw a horizontal line which meets the graph at a point, say Q. Through Q, draw a vertical line which meets the x-axis at -4.

\(\therefore b = -4\) Ans.

The linear relation between \(x\) and \(y\) is \(y = 2x + 1\). Ans.

Reason:

Let the linear relation between the variables \(x\) and \(y\) be \(y = mx + c\), Since, the graph passes through (-2, -3); substitute \(x = -2\) and \(y = -3\) in \(y = mx + c\). This gives: \(-3 = -2m + c\) .....I

Again, the graph passes through (0, 1); substitute \(x = 0\) and \(y = 1\) in \(y = mx + c\). This gives: \(1 = m \times 0 + c\) i.e., \(c = 1\)

Now, \(-3 = -2m + c\) \(\Rightarrow\) \(-3 = -2m + 1\) i.e., \(m = 2\)

Required relation is: \(y = mx + c\) i.e., \(y = 2x + 1\)

On a graph paper, draw a straight line represented by the equation 2x - 3y + 12 = 0. Use the graph drawn to find the values of m and n so that the points (m, -2) and (3, n) lie on the given straight line.

Solution:

\(2x - 3y + 12 = 0\)

\(\Rightarrow\) \(2x = 3y - 12\)

\(\Rightarrow\) \(x = \frac{3y - 12}{2}\)

Plot the points (-6, 0), (-3, 2) and (0, 4) on a graph paper; then draw a straight line through these points which will represent the given equation 2x - 3y + 12 = 0.

Since, point (m, -2) lies on the straight line drawn, through \(y = -2\) draw a horizontal line which meets the straight line at point A. Through point A, draw a vertical line which meets x-axis at point -9.

\(\therefore m = -9\) Ans.

Also, as (3, n) lies on the straight line drawn, through \(x = 3\) draw a vertical line which meets the straight line at point B. Through point B, draw a horizontal line which meets y-axis at the point 6.

\(\therefore n = 6\) Ans.

Remember:

1. The equation of the x-axis is \(y = 0\); and equation of the y-axis is \(x = 0\).

2. The graph of \(x = a\) (where a is constant) is a straight line parallel to the y-axis and at a distance of 'a' units from the y-axis.

Similarly, the graph of \(y = b\) (a constant) is a straight line parallel to the x-axis and at a distance of 'b' units from it.

Teacher's Note

Graphing linear equations helps students visualize relationships between variables, making abstract algebra concrete and applicable to real-world scenarios like tracking expenses or measuring growth patterns.

Exercise 27(A)

1. Draw the graph for each equation, given below:

(i) \(x = 5\) (ii) \(x + 5 = 0\)

(iii) \(y = 7\) (iv) \(y + 7 = 0\)

(v) \(2x + 3y = 0\) (vi) \(3x + 2y = 6\)

(vii) \(x - 5y + 4 = 0\) (viii) \(5x + y + 5 = 0\)

2. Draw the graph for each equation given below; hence find the co-ordinates of the points where the graph drawn meets the co-ordinate axes:

(i) \(\frac{1}{5}x + \frac{1}{5}y = 1\) (ii) \(\frac{2x + 15}{3} = y - 1\)

3. Draw the graph of the straight line given by the equation 4x - 3y + 36 = 0

Calculate the area of the triangle formed by the line drawn and the co-ordinate axes.

4. Draw the graph of the equation 2x - 3y - 5 = 0

From the graph, find:

(i) \(x_1\), the value of \(x\), when \(y = 7\)

(ii) \(x_2\), the value of \(x\), when \(y = -5\)

5. Draw the graph of the equation 4x + 3y + 6 = 0

From the graph, find:

(i) \(y_1\), the value of \(y\), when \(x = 12\)

(ii) \(y_2\), the value of \(y\), when \(x = -6\)

6. Use the table given below to draw the graph.

From your graph, find the values of 'a' and 'b'. State a linear relation between the variables x and y.

7. Draw the graph obtained from the table below:

Use the graph to find the values of a, b and c. State a linear relation between the variables x and y.

8. A straight line passes through the points (2, 4) and (5, -2). Taking 1 cm = 1 unit; mark these points on a graph paper and draw the straight line through these points. If points (m, -4) and (3, n) lie on the line drawn; find the values of m and n.

9. Draw the graph (straight line) given by equation \(x - 3y = 18\). If the straight line drawn passes through the points (m, -5) and (6, n); find the values of m and n.

10. Use the graphical method to find the value of k, if:

(i) (k, -3) lies on the straight line \(2x + 3y = 1\)

(ii) (5, k - 2) lies on the straight line \(x - 2y + 1 = 0\)

Teacher's Note

Solving graphing exercises builds spatial reasoning and reinforces the connection between algebraic equations and their geometric interpretations on the coordinate plane.

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