**Graph of linear equation ax + by + c = 0 in two variables, where a ≠ and b ≠ 0 **

**(i) Step I :** To obtain the linear equation, let the equation be ax + by + c = 0.**(ii) Step II :** Express y in terms of x to obtain –

y = – **(iii) Step III :** Give any two values to x and calculate the corresponding values of y from the expression in step II to obtain two solutions, say (α_{1}, β_{1}) and

(α_{2}, β_{2}). If possible take values of x as integers in such a manner that the corresponding values of y are also integers.**(iv) Step IV :** Plot points (α_{1}, β_{1}) and (α_{2}, β_{2}) on a graph paper.**(v) Step V :** Join the points marked in step IV to obtain a line. The line obtained is the **graph of the equation** ax + by + c = 0.

Example:Draw the graph of the equation y – x = 2.

Solution: We have, equation of linear equation is y – x = 2

⇒ y = x + 2

When we substitute x = 1 in given equation, we have : y = 1 + 2 = 3

When we substitute x = 3, we have : y = 3 + 2 = 5.

When we substitute x = 5, we have : y = 5 + 2 = 7.

Thus, we have the following table **exhibiting the abscissa and ordinates of points** on the line represented by the given equation.x1;35x357

Plotting the points (1, 3), (3, 5) and (5, 7) on the graph paper and drawing a line joining them, we obtain the graph of the line represented by the given equation as shown in given figure.

Example: Draw a graph of the line x – 2y = 3. From the graph, find the coordinates of the point when (i) x = – 5 (ii) y = 0.

Solution: We have x – 2y = 3 ⇒ y = (x – 3) /2

When x = 1, we have : y = (1 – 3)/2 = –1

When x = –1, we have : y = (–1 – 3)/2 = –2Thus, we have the following table :x1–1x–1–2

Plotting points (1, –1) & (–1, –2) on graph paper & joining them, we get straight line as shown in fig. This line is required graph of equation x – 2y = 3.

To find the **coordinates of the point** when x = –5, we draw a line parallel to y-axis and passing through (–5, 0). This line meets the graph of x – 2y = 3 at a point from which we draw a line parallel to x-axis which crosses y-axis at y = –4. So, the coordinates of the required point are (–5, –4).Since y = 0 on x-axis. So, the required point is the point where the line meets x-axis. From the graph the coordinates of such point are (3, 0).Hence, **required points** are (–5, –4) and (3, 0).

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