Mathematics

Example: Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representing of the pair so formed is : (i) intersecting lines (ii) parallel lines (iii) coincident lines

Solution : We have a equation of line is 2x + 3y – 8 = 0

(i) Another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines is 3x – 2y – 8 = 0

(ii) Another parallel lines to above line is 4x + 6y – 22 = 0

(iii) Another coincident line to above line is 6x + 9y – 24 = 0

Example: Solve the following system of linear equations graphically. Pair of lines are 3x + y – 11 = 0 ; x – y – 1 = 0. Shade the region bounded by these lines and also y-axis. Then, determine the areas of the region bounded by these lines and y-axis.

Solution: We have pair of equation of a lines are 3x + y – 11 = 0 and x – y – 1 = 0

(a) Graph of the equation 3x + y – 11 = 0. We have, 3x + y – 11 = 0

⇒ y = – 3x + 11

When, x = 2, y = –3 × 2 + 11 = 5

When, x = 3, y = – 3 × 3 + 11 = 2

Then, we have the following table :

Plotting the points P (2, 5) and Q(3, 2) on the graph paper and drawing a line joining between them, we get the graph of the equation 3x + y – 11 = 0 as shown in fig.

(b) Graph of the equation x – y – 1 = 0. We have, x – y – 1 = 0

⇒  y = x – 1

When, x = – 1, y = –2

When, x = 3, y = 2

Then, we have the following table :

 Plotting the points R(–1, –2) and S(3, 2) on the same graph paper and drawing a line joining between them, we get the graph of the equation x – y – 1 = 0 as shown in fig.

You can observe that two lines intersect at Q(3, 2). So, x = 3 and y = 2. The area enclosed by the lines represented by the given equations and also the y-axis is shaded. So, the enclosed area = Area of the shaded portion

= Area of DQUT = × base × height

= × (TU × VQ) = × (TO + OU) × VQ =  (11 + 1) 3 = × 12 × 3 = 18 sq.units.

Hence, required area is 18 sq. units.

Example: Draw the graphs of the following equations 2x – 3y = – 6 ; 2x + 3y = 18; y = 2. Find the vertices of the triangles formed and also find the area of the triangle.

Solution: (a) Graph of the equation 2x – 3y = – 6;

We have, 2x – 3y = – 6 ⇒ y = 

When, x = 0, y =  = 2

When, x = 3, y = = 4

Then, we have the following table :

Plotting the points P(0, 2) and Q(3, 4) on the graph paper and drawing a line joining between them we get the graph of the equation 2x – 3y = – 6 as shown in fig.

(b) Graph of the equation 2x + 3y = 18;

We have 2x + 3y = 18 ⇒ y= 

When, x = 0, y = = 6

When, x = – 3, y = = 8

Then, we have the following table :

Plotting the points R(0, 6) and S(–3, 8) on the same graph paper and drawing a line joining between them, we get the graph of the equation 2x + 3y = 18 as shown in fig.

(c) Graph of the equation y = 2. It is a clear fact that y = 2 is for every value of x. We may take the points T (3, 2), U(6, 2) or any other values. Then, we get the following table :

Plotting the points T(3, 2) and U(6, 2) on the same graph paper and drawing a line joining between them, we get the graph of the equation y = 2 as shown in fig.

From the fig., we can observe that the lines taken in pairs intersect each other at points Q(3, 4), U (6, 2) and P(0, 2). These form the three vertices of the triangle PQU. 

To find area of the triangle so formed The triangle is so formed is PQU (see fig.)

In the DPQU 

QT (altitude) = 2 units

and PU (base) = 6 units

so, area of DPQU = (base × height) 

=  (PU × QT) = × 6 × 2 sq. untis = 6 sq. units.