Q.Compute the area bounded by the lines $x + 2y = 2$, $y - x = 1$ and $2x + y = 7$.

Concept: Area Under Curve — the region is a triangle formed by three intersecting lines.

Step 1 – Find intersection points

Solve pairwise:

• $x + 2y = 2$ and $y - x = 1$: adding gives $3y = 3 \Rightarrow y = 1$, then $x = 0$. Point $A(0,1)$.
• $y - x = 1$ and $2x + y = 7$: subtract first from second: $(2x+y) - (y-x) = 7-1 \Rightarrow 3x = 6 \Rightarrow x = 2$, then $y = 3$. Point $B(2,3)$.
• $x + 2y = 2$ and $2x + y = 7$: multiply first by 2: $2x + 4y = 4$, subtract second: $(2x+4y) - (2x+y) = 4-7 \Rightarrow 3y = -3 \Rightarrow y = -1$, then $x = 4$. Point $C(4,-1)$.

Step 2 – Compute area using determinant formula

Area of triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$: …

Concept and Intuition

When three non-parallel lines are drawn in a plane, they typically enclose a triangle — unless two are parallel or all three meet at a single point. The problem asks for the area bounded by these three lines, which means the region common to all three constraints. That region is a triangle whose vertices are the pairwise intersections of the lines.

The standard approach: find the three intersection points, then use the determinant (shoelace) formula for the area of a triangle given its vertices. This is cleaner than trying to integrate, since the boundaries are straight lines.

Step-by-step solution

1. Write the equations in a convenient form

We have:

2. Find the intersection points (vertices)

• Intersection of $L_1$ and $L_2$: Substitute $y = x + 1$ into $x + 2y = 2$:

Then $y = 0 + 1 = 1$.

Vertex $A = (0, 1)$.

• Intersection of $L_2$ and $L_3$: Substitute $y = x + 1$ into $2x + y = 7$:

Then $y = 2 + 1 = 3$.

Vertex $B = (2, 3)$.

• Intersection of $L_1$ and $L_3$: Solve the system:

Multiply the first equation by 2: $2x + 4y = 4$. Subtract the second equation:

Then from $x + 2(-1) = 2$, we get $x = 4$.

Vertex $C = (4, -1)$.

So the three vertices are $A(0,1)$, $B(2,3)$, $C(4,-1)$.

3. Compute the area using the shoelace formula …

Method: Area enclosed by three straight lines

When three non-parallel lines bound a region, that region is a triangle; the cleanest route is to locate its three corners and apply the coordinate-geometry area formula.

Steps

Step 1: Confirm a triangle is actually formed.

Check that no two lines share the same slope — parallel lines would leave the region open and unbounded. Three different slopes guarantee three distinct corners.

Step 2: Find the vertices by solving the lines in pairs.

Each corner is the intersection of two of the lines, so solve all three pairs simultaneously. This gives vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$.

Step 3: Apply the shoelace (determinant) formula.

$A = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|.$ …

Common Mistakes

Mistake 1: Setting up integrals instead of using the triangle.

Why it's wrong: three non-parallel lines enclose a triangle, so once the vertices are known the area is a single shoelace computation — integrating is longer and error-prone here. Correct approach: find the three pairwise intersections, then apply $\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$.

Mistake 2: Pairing the wrong lines when finding vertices. …