Area of Triangle Formed by Three Lines

Video Explanation

Question

Find the area of the triangle formed by the lines \(2x + 3y = 12\), \(x – y – 1 = 0\), and \(x = 0\).

Solution

Step 1: Find Points of Intersection

Intersection of \(x = 0\) and \(2x + 3y = 12\):

\[ 2(0) + 3y = 12 \Rightarrow y = 4 \Rightarrow (0,4) \]

Intersection of \(x = 0\) and \(x – y – 1 = 0\):

\[ 0 – y – 1 = 0 \Rightarrow y = -1 \Rightarrow (0,-1) \]

Intersection of \(2x + 3y = 12\) and \(x – y – 1 = 0\):

From \(x – y = 1 \Rightarrow x = y + 1\)

Substitute into first equation:

\[ 2(y+1) + 3y = 12 \]

\[ 2y + 2 + 3y = 12 \Rightarrow 5y = 10 \Rightarrow y = 2 \]

\[ x = y + 1 = 3 \Rightarrow (3,2) \]

Step 2: Vertices of Triangle

\[ (0,4),\ (0,-1),\ (3,2) \]

Step 3: Apply Area Formula

Using determinant formula:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| \]

\[ = \frac{1}{2} | 0(-1 – 2) + 0(2 – 4) + 3(4 – (-1)) | \]

\[ = \frac{1}{2} | 3 \times 5 | = \frac{15}{2} \]

\[ = 7.5 \]

Final Answer

\[ \text{Area} = 7.5 \text{ sq. units} \]