Math Problem Statement

As shown in the figure, line l1: y = 2x + 3 intersects with l2: y = —x + 6 at point B. Line l1 intersects with the x-axis at point D and with the y-axis at point E; line l2 intersects with the x-axis at point C and with the y-axis at point A. The area of the quadrilateral OCBE is

Solution

To find the area of the quadrilateral OCBEOCBE formed by the intersection points of the lines l1:y=2x+3l_1: y = 2x + 3 and l2:y=x+6l_2: y = -x + 6 with the axes, let's follow these steps:

Step 1: Find the intersection point BB of l1l_1 and l2l_2.

Since BB lies on both lines y=2x+3y = 2x + 3 and y=x+6y = -x + 6, equate the two equations:

2x+3=x+62x + 3 = -x + 6 Solve for xx: 2x+x=633x=3x=12x + x = 6 - 3 \quad \Rightarrow \quad 3x = 3 \quad \Rightarrow \quad x = 1 Now, substitute x=1x = 1 into either equation (say y=2x+3y = 2x + 3): y=2(1)+3=5y = 2(1) + 3 = 5 Thus, the coordinates of BB are: B(1,5)B(1, 5)


Step 2: Find the intercepts of the lines with the axes.

  1. For line l1:y=2x+3l_1: y = 2x + 3:

    • x-intercept DD: Set y=0y = 0. 0=2x+32x=3x=320 = 2x + 3 \quad \Rightarrow \quad 2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2} So, D(32,0)D\left(-\frac{3}{2}, 0\right).
    • y-intercept EE: Set x=0x = 0. y=2(0)+3=3y = 2(0) + 3 = 3 So, E(0,3)E(0, 3).
  2. For line l2:y=x+6l_2: y = -x + 6:

    • x-intercept CC: Set y=0y = 0. 0=x+6x=60 = -x + 6 \quad \Rightarrow \quad x = 6 So, C(6,0)C(6, 0).
    • y-intercept AA: Set x=0x = 0. y=0+6=6y = -0 + 6 = 6 So, A(0,6)A(0, 6).

Step 3: Find the vertices of the quadrilateral OCBEOCBE.

  • O: Origin (0,0)(0, 0)
  • C: (6,0)(6, 0)
  • B: (1,5)(1, 5)
  • E: (0,3)(0, 3)

Step 4: Use the Shoelace Formula to find the area.

The Shoelace Formula for the area of a polygon with vertices (x1,y1),(x2,y2),,(xn,yn)(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n) is:

Area=12i=1n1(xiyi+1yixi+1)+(xny1ynx1)\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|

For the quadrilateral OCBEOCBE with vertices O(0,0)O(0, 0), C(6,0)C(6, 0), B(1,5)B(1, 5), and E(0,3)E(0, 3), the formula becomes:

Area=12(00+65+13+00)(06+01+50+30)\text{Area} = \frac{1}{2} \left| (0 \cdot 0 + 6 \cdot 5 + 1 \cdot 3 + 0 \cdot 0) - (0 \cdot 6 + 0 \cdot 1 + 5 \cdot 0 + 3 \cdot 0) \right|

Simplify:

Area=120+30+3+0(0+0+0+0)=1233=332\text{Area} = \frac{1}{2} \left| 0 + 30 + 3 + 0 - (0 + 0 + 0 + 0) \right| = \frac{1}{2} \left| 33 \right| = \frac{33}{2}


Step 5: Final Answer

The area of the quadrilateral OCBEOCBE is:

332square units.\boxed{\frac{33}{2}} \, \text{square units}.


Do you want further details on any step?

Here are 5 related questions:

  1. How do we determine the area of polygons with more than four sides?
  2. Can the Shoelace Theorem be applied to 3D objects?
  3. How does the area change if one of the lines has a different slope?
  4. What are the conditions for two lines to be perpendicular?
  5. How do we find the point of intersection between two nonlinear curves?

Tip: The Shoelace Formula works only when vertices are arranged in a cyclic order (clockwise or counterclockwise).

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Coordinate Geometry
Area of Polygons

Formulas

Shoelace Formula
Equation of a Line (y = mx + b)

Theorems

Shoelace Theorem

Suitable Grade Level

Grades 9-12