The area of the figure formed by the lines ax + by + c = 0, ax – by + c = 0, ax + by – c = 0 and ax – by – c = 0 is
Finding the Area of the Figure Formed by Four Lines
The problem asks for the area of the geometric figure formed by the intersection of four given lines: \({\rm{ax}} + {\rm{by}} + {\rm{c}} = 0\), \({\rm{ax}} - {\rm{by}} + {\rm{c}} = 0\), \({\rm{ax}} + {\rm{by}} - {\rm{c}} = 0\), and \({\rm{ax}} - {\rm{by}} - {\rm{c}} = 0\).
Analyzing the Given Lines
Let's rewrite the equations to observe their relationships:
- Line 1 (L1): \({\rm{ax}} + {\rm{by}} = -{\rm{c}}\)
- Line 2 (L2): \({\rm{ax}} - {\rm{by}} = -{\rm{c}}\)
- Line 3 (L3): \({\rm{ax}} + {\rm{by}} = {\rm{c}}\)
- Line 4 (L4): \({\rm{ax}} - {\rm{by}} = {\rm{c}}\)
We can see that:
- L1 (\({\rm{ax}} + {\rm{by}} = -{\rm{c}}\)) and L3 (\({\rm{ax}} + {\rm{by}} = {\rm{c}}\)) are parallel lines because they have the same coefficients for \({\rm{x}}\) and \({\rm{y}}\) (\({\rm{a}}\) and \({\rm{b}}\)) but different constant terms.
- L2 (\({\rm{ax}} - {\rm{by}} = -{\rm{c}}\)) and L4 (\({\rm{ax}} - {\rm{by}} = {\rm{c}}\)) are also parallel lines for the same reason.
Since the figure is formed by two pairs of parallel lines, it is a parallelogram. Let's find the vertices of this parallelogram by finding the intersection points of these lines.
Method 1: Finding the Vertices and Calculating Area
The vertices of the figure are the points where pairs of lines intersect. We need to solve the equations simultaneously.
- Intersection of L1 and L2:
\begin{align*} {\rm{ax}} + {\rm{by}} &= -{\rm{c}} \quad &(1) \\ {\rm{ax}} - {\rm{by}} &= -{\rm{c}} \quad &(2)\end{align*} Adding (1) and (2): \(2{\rm{ax}} = -2{\rm{c}} \implies {\rm{x}} = -{\rm{c}}/{\rm{a}}\).
Substituting \({\rm{x}} = -{\rm{c}}/{\rm{a}}\) into (1): \({\rm{a}}(-{\rm{c}}/{\rm{a}}) + {\rm{by}} = -{\rm{c}} \implies -{\rm{c}} + {\rm{by}} = -{\rm{c}} \implies {\rm{by}} = 0\). Assuming \({\rm{b}} \ne 0\), we get \({\rm{y}} = 0\).
Vertex 1: \(( -{\rm{c}}/{\rm{a}}, 0)\) - Intersection of L1 and L4:
\begin{align*} {\rm{ax}} + {\rm{by}} &= -{\rm{c}} \quad &(1) \\ {\rm{ax}} - {\rm{by}} &= {\rm{c}} \quad &(4)\end{align*} Adding (1) and (4): \(2{\rm{ax}} = 0 \implies {\rm{x}} = 0\).
Substituting \({\rm{x}} = 0\) into (1): \({\rm{a}}(0) + {\rm{by}} = -{\rm{c}} \implies {\rm{by}} = -{\rm{c}}\). Assuming \({\rm{b}} \ne 0\), we get \({\rm{y}} = -{\rm{c}}/{\rm{b}}\).
Vertex 2: \((0, -{\rm{c}}/{\rm{b}})\) - Intersection of L3 and L4:
\begin{align*} {\rm{ax}} + {\rm{by}} &= {\rm{c}} \quad &(3) \\ {\rm{ax}} - {\rm{by}} &= {\rm{c}} \quad &(4)\end{align*} Adding (3) and (4): \(2{\rm{ax}} = 2{\rm{c}} \implies {\rm{x}} = {\rm{c}}/{\rm{a}}\).
Substituting \({\rm{x}} = {\rm{c}}/{\rm{a}}\) into (3): \({\rm{a}}({\rm{c}}/{\rm{a}}) + {\rm{by}} = {\rm{c}} \implies {\rm{c}} + {\rm{by}} = {\rm{c}} \implies {\rm{by}} = 0\). Assuming \({\rm{b}} \ne 0\), we get \({\rm{y}} = 0\).
Vertex 3: \(({\rm{c}}/{\rm{a}}, 0)\) - Intersection of L3 and L2:
\begin{align*} {\rm{ax}} + {\rm{by}} &= {\rm{c}} \quad &(3) \\ {\rm{ax}} - {\rm{by}} &= -{\rm{c}} \quad &(2)\end{align*} Adding (3) and (2): \(2{\rm{ax}} = 0 \implies {\rm{x}} = 0\).
Substituting \({\rm{x}} = 0\) into (3): \({\rm{a}}(0) + {\rm{by}} = {\rm{c}} \implies {\rm{by}} = {\rm{c}}\). Assuming \({\rm{b}} \ne 0\), we get \({\rm{y}} = {\rm{c}}/{\rm{b}}\).
Vertex 4: \((0, {\rm{c}}/{\rm{b}})\)
The vertices of the figure are \((-{\rm{c}}/{\rm{a}}, 0)\), \((0, -{\rm{c}}/{\rm{b}})\), \(({\rm{c}}/{\rm{a}}, 0)\), and \((0, {\rm{c}}/{\rm{b}})\). These points lie on the axes and are symmetric about the origin. This confirms the figure is a rhombus (a special type of parallelogram).
The lengths of the diagonals of the rhombus are:
- Diagonal 1 (along the x-axis, connecting \((-{\rm{c}}/{\rm{a}}, 0)\) and \(({\rm{c}}/{\rm{a}}, 0)\)): length \( = |{\rm{c}}/{\rm{a}} - (-{\rm{c}}/{\rm{a}})| = |2{\rm{c}}/{\rm{a}}|\)
- Diagonal 2 (along the y-axis, connecting \((0, -{\rm{c}}/{\rm{b}})\) and \((0, {\rm{c}}/{\rm{b}})\)): length \( = |{\rm{c}}/{\rm{b}} - (-{\rm{c}}/{\rm{b}})| = |2{\rm{c}}/{\rm{b}}|\)
The area of a rhombus is given by the formula \( \frac{1}{2} \times (\text{length of diagonal 1}) \times (\text{length of diagonal 2}) \).
Area \( = \frac{1}{2} \times |2{\rm{c}}/{\rm{a}}| \times |2{\rm{c}}/{\rm{b}}| = \frac{1}{2} \times \frac{2|{\rm{c}}|}{|{\rm{a}}|} \times \frac{2|{\rm{c}}|}{|{\rm{b}}|} = \frac{4{\rm{c}}^2}{2|{\rm{ab}}|} = \frac{2{\rm{c}}^2}{|{\rm{ab}}|} \).
Assuming \({\rm{a}}\) and \({\rm{b}}\) have the same sign or the context implies \({\rm{ab}} > 0\) in the denominator, the area is \( \frac{2{\rm{c}}^2}{{\rm{ab}}} \).
Method 2: Using the Parallelogram Area Formula
The area of the parallelogram formed by the lines \({\rm{a}}_1{\rm{x}} + {\rm{b}}_1{\rm{y}} + {\rm{c}}_1 = 0\), \({\rm{a}}_1{\rm{x}} + {\rm{b}}_1{\rm{y}} + {\rm{c}}_2 = 0\), \({\rm{a}}_2{\rm{x}} + {\rm{b}}_2{\rm{y}} + {\rm{d}}_1 = 0\), and \({\rm{a}}_2{\rm{x}} + {\rm{b}}_2{\rm{y}} + {\rm{d}}_2 = 0\) is given by the formula:
\( \text{Area} = \frac{|({\rm{c}}_1 - {\rm{c}}_2)({\rm{d}}_1 - {\rm{d}}_2)|}{|{\rm{a}}_1{\rm{b}}_2 - {\rm{a}}_2{\rm{b}}_1|} \)
Let's identify the coefficients and constants from the given lines:
- Pair 1 (parallel lines): \({\rm{ax}} + {\rm{by}} + {\rm{c}} = 0\) and \({\rm{ax}} + {\rm{by}} - {\rm{c}} = 0\). Here, \({\rm{a}}_1 = {\rm{a}}\), \({\rm{b}}_1 = {\rm{b}}\), \({\rm{c}}_1 = {\rm{c}}\), \({\rm{c}}_2 = -{\rm{c}}\). The difference in constants is \({\rm{c}}_1 - {\rm{c}}_2 = {\rm{c}} - (-{\rm{c}}) = 2{\rm{c}}\).
- Pair 2 (parallel lines): \({\rm{ax}} - {\rm{by}} + {\rm{c}} = 0\) and \({\rm{ax}} - {\rm{by}} - {\rm{c}} = 0\). Here, \({\rm{a}}_2 = {\rm{a}}\), \({\rm{b}}_2 = -{\rm{b}}\), \({\rm{d}}_1 = {\rm{c}}\), \({\rm{d}}_2 = -{\rm{c}}\). The difference in constants is \({\rm{d}}_1 - {\rm{d}}_2 = {\rm{c}} - (-{\rm{c}}) = 2{\rm{c}}\).
- The determinant term in the denominator: \({\rm{a}}_1{\rm{b}}_2 - {\rm{a}}_2{\rm{b}}_1 = ({\rm{a}})(-{\rm{b}}) - ({\rm{a}})({\rm{b}}) = -{\rm{ab}} - {\rm{ab}} = -2{\rm{ab}}\).
Now, plug these values into the formula:
\( \text{Area} = \frac{|(2{\rm{c}})(2{\rm{c}})|}{|-2{\rm{ab}}|} = \frac{|4{\rm{c}}^2|}{|-2{\rm{ab}}|} = \frac{4{\rm{c}}^2}{2|{\rm{ab}}|} = \frac{2{\rm{c}}^2}{|{\rm{ab}}|} \).
Again, assuming the denominator is \({\rm{ab}}\) representing \(|{\rm{ab}}|\) or \({\rm{ab}} > 0\), the area is \( \frac{2{\rm{c}}^2}{{\rm{ab}}} \).
Conclusion
Both methods yield the same result for the area of the figure formed by the given lines. The figure is a rhombus centered at the origin, and its area is \( \frac{2{\rm{c}}^2}{{\rm{ab}}} \).
The final answer is \( \frac{{2{{\rm{c}}^2}}}{{{\rm{ab}}}}\) .
Revision Table - Area of Figure from Lines
| Concept | Description | Formula/Method |
|---|---|---|
| Given Lines | Four lines: \({\rm{ax}} \pm {\rm{by}} \pm {\rm{c}} = 0\) | \({\rm{ax}} + {\rm{by}} = \pm {\rm{c}}\), \({\rm{ax}} - {\rm{by}} = \pm {\rm{c}}\) |
| Figure Shape | Parallelogram (specifically a rhombus) | Two pairs of parallel lines |
| Vertices | Intersection points | \((-{\rm{c}}/{\rm{a}}, 0), (0, -{\rm{c}}/{\rm{b}}), ({\rm{c}}/{\rm{a}}, 0), (0, {\rm{c}}/{\rm{b}})\) |
| Diagonal Lengths | Distance between opposite vertices | \(|2{\rm{c}}/{\rm{a}}|\) and \(|2{\rm{c}}/{\rm{b}}|\) |
| Area (Rhombus) | Half product of diagonals | \( \frac{1}{2} |d_1 d_2| = \frac{2{\rm{c}}^2}{|{\rm{ab}}|} \) |
| Area (Parallelogram) | Formula based on coefficients | \( \frac{|({\rm{c}}_1 - {\rm{c}}_2)({\rm{d}}_1 - {\rm{d}}_2)|}{|{\rm{a}}_1{\rm{b}}_2 - {\rm{a}}_2{\rm{b}}_1|} = \frac{2{\rm{c}}^2}{|{\rm{ab}}|} \) |
Additional Information - Geometry Concepts
When dealing with areas of figures formed by lines in coordinate geometry, several concepts are useful:
- Parallel Lines: Two lines \({\rm{A}}_1{\rm{x}} + {\rm{B}}_1{\rm{y}} + {\rm{C}}_1 = 0\) and \({\rm{A}}_2{\rm{x}} + {\rm{B}}_2{\rm{y}} + {\rm{C}}_2 = 0\) are parallel if \({\rm{A}}_1{\rm{B}}_2 - {\rm{A}}_2{\rm{B}}_1 = 0\). In simpler terms, their slopes (\(-{\rm{A}}_1/{\rm{B}}_1\) and \(-{\rm{A}}_2/{\rm{B}}_2\)) are equal.
- Perpendicular Lines: Two lines are perpendicular if the product of their slopes is -1 (provided neither line is vertical). The slope of \({\rm{ax}} + {\rm{by}} + {\rm{c}} = 0\) is \(-{\rm{a}}/{\rm{b}}\) and the slope of \({\rm{ax}} - {\rm{by}} + {\rm{c}} = 0\) is \({\rm{a}}/{\rm{b}}\). Their product is \(( -{\rm{a}}/{\rm{b}}) \times ({\rm{a}}/{\rm{b}}) = -{\rm{a}}^2/{\rm{b}}^2\). If this equals -1, then \({\rm{a}}^2 = {\rm{b}}^2\), meaning \(|{\rm{a}}| = |{\rm{b}}|\). In this specific case, if \(|{\rm{a}}| = |{\rm{b}}|\), the rhombus becomes a square.
- Distance between Parallel Lines: The distance between two parallel lines \({\rm{Ax}} + {\rm{By}} + {\rm{C}}_1 = 0\) and \({\rm{Ax}} + {\rm{By}} + {\rm{C}}_2 = 0\) is \( \frac{|{\rm{C}}_1 - {\rm{C}}_2|}{\sqrt{{\rm{A}}^2 + {\rm{B}}^2}} \).
- Area of Parallelogram (Base x Height): The area can also be calculated as base times height. We could find the distance between one pair of parallel lines (the height) and the length of a side formed by the intersection of two non-parallel lines (the base). However, the diagonal method for a rhombus or the determinant formula for a parallelogram formed by specific parallel lines is often more direct.
- Area using Vertices (Shoelace Formula): For a polygon with vertices \(({\rm{x}}_1, {\rm{y}}_1), ({\rm{x}}_2, {\rm{y}}_2), \dots, ({\rm{x}}_{\rm{n}}, {\rm{y}}_{\rm{n}})\) in order, the area is \( \frac{1}{2} |({\rm{x}}_1{\rm{y}}_2 + {\rm{x}}_2{\rm{y}}_3 + \dots + {\rm{x}}_{\rm{n}}{\rm{y}}_1) - ({\rm{y}}_1{\rm{x}}_2 + {\rm{y}}_2{\rm{x}}_3 + \dots + {\rm{y}}_{\rm{n}}{\rm{x}}_1)| \). This formula could also be applied to the four vertices we found as a third method to verify the area.
Understanding the geometric shape formed by the equations is key to solving these types of coordinate geometry problems efficiently.
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