Graphs of Linear Equations and Area of the Triangle

Video Explanation

Question

Draw the graphs of the following equations and find the vertices of the triangle so obtained. Also find the area of the triangle:

\[ 2x – 3y + 6 = 0 \]

\[ 2x + 3y – 18 = 0 \]

\[ y – 2 = 0 \]

Solution

Step 1: Write the Equations in Convenient Form

Equation (1):

\[ 2x – 3y + 6 = 0 \Rightarrow y = \frac{2}{3}x + 2 \]

Equation (2):

\[ 2x + 3y – 18 = 0 \Rightarrow y = 6 – \frac{2}{3}x \]

Equation (3):

\[ y – 2 = 0 \Rightarrow y = 2 \]

Step 2: Prepare Tables of Values

For Equation (1): \(y = \frac{2}{3}x + 2\)

x	y
0	2
3	4

For Equation (2): \(y = 6 – \frac{2}{3}x\)

x	y
0	6
3	4

For Equation (3): \(y = 2\)

x	y
0	2
6	2

Step 3: Graphical Representation

Plot the above points on the same Cartesian plane and join them to obtain the three straight lines.

The three lines intersect to form a triangle.

Step 4: Vertices of the Triangle

The vertices of the triangle are:

• Intersection of \(y = 2\) and \(y = \frac{2}{3}x + 2\): (0, 2)
• Intersection of \(y = 2\) and \(y = 6 – \frac{2}{3}x\): (6, 2)
• Intersection of \(y = \frac{2}{3}x + 2\) and \(y = 6 – \frac{2}{3}x\): (3, 4)

Step 5: Area of the Triangle

Base of the triangle = distance between (0, 2) and (6, 2) = 6 units

Height of the triangle = vertical distance between y = 2 and y = 4 = 2 units

\[ \text{Area} = \frac{1}{2} \times 6 \times 2 = 6 \]

Answer

Vertices of the triangle are:

• (0, 2)
• (6, 2)
• (3, 4)

Area of the triangle = 6 square units.

Conclusion

By drawing the graphs of the given equations, the required triangle is obtained and its area is 6 square units.