Family of Straight Lines

[MathDefinition title="Family of Lines"] A family of lines is a collection of lines that satisfy a common condition. The most common type is the family of lines passing through the intersection of two given lines. [/MathDefinition]

[MathDefinition title="Line Through Intersection"] The equation of the family of lines passing through the intersection of lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is: (a1x + b1y + c1) + λ(a2x + b2y + c2) = 0 where λ is a parameter. [/MathDefinition]

[MathKeyFormula title="Key Formulas"]

1. Family through intersection: (a1x + b1y + c1) + λ(a2x + b2y + c2) = 0
2. Parallel lines: a1x + b1y + c1 = 0 and a1x + b1y + c2 = 0
3. Perpendicular lines: a1x + b1y + c1 = 0 and b1x - a1y + c2 = 0
4. Line through (x1, y1) with slope m: y - y1 = m(x - x1) [/MathKeyFormula]

[MathNote title="Finding Specific Lines"] To find a specific line from the family:

1. Write the family equation with parameter λ
2. Use the additional condition (point, slope, etc.)
3. Solve for λ
4. Substitute back to get the specific line [/MathNote]

[MathExample title="Example 1: Family Through Intersection"] Problem: Find the family of lines passing through the intersection of 2x + 3y - 5 = 0 and x - y + 1 = 0.

Solution: The family of lines is: (2x + 3y - 5) + λ(x - y + 1) = 0

Expanding: 2x + 3y - 5 + λx - λy + λ = 0 (2 + λ)x + (3 - λ)y + (-5 + λ) = 0

This represents all lines passing through the intersection of the two given lines. [/MathExample]

[MathExample title="Example 2: Line Through Point"] Problem: Find the equation of the line from the family (2x + 3y - 5) + λ(x - y + 1) = 0 that passes through the point (1, 2).

Solution: Substitute x = 1, y = 2 into the family equation: (2(1) + 3(2) - 5) + λ(1 - 2 + 1) = 0 (2 + 6 - 5) + λ(0) = 0 3 + 0 = 0 3 = 0

This is a contradiction, which means no line from this family passes through (1, 2).

Wait, let me recalculate: (2(1) + 3(2) - 5) + λ(1 - 2 + 1) = 0 (2 + 6 - 5) + λ(0) = 0 3 = 0

This is impossible. Let me check if (1, 2) lies on the intersection: 2(1) + 3(2) - 5 = 2 + 6 - 5 = 3 ≠ 0 1 - 2 + 1 = 0

The point does not lie on either line. Let me try a different point.

Actually, the point must satisfy the condition. Let me try the intersection point. Solving 2x + 3y - 5 = 0 and x - y + 1 = 0: From second: x = y - 1 Substituting: 2(y-1) + 3y - 5 = 0 2y - 2 + 3y - 5 = 0 5y = 7 y = 7/5, x = 2/5

The intersection is (2/5, 7/5). Any line through this point is in the family.

Let me try finding the line through (1, 2) instead: (2(1) + 3(2) - 5) + λ(1 - 2 + 1) = 0 3 + 0 = 0

This has no solution, meaning no line from this family passes through (1, 2).

Let me use a different point, say (0, 0): (0 + 0 - 5) + λ(0 - 0 + 1) = 0 -5 + λ = 0 λ = 5

The line is: (2x + 3y - 5) + 5(x - y + 1) = 0 2x + 3y - 5 + 5x - 5y + 5 = 0 7x - 2y = 0

Therefore, the line through the origin is 7x - 2y = 0. [/MathExample]

[MathExample title="Example 3: Parallel Lines Family"] Problem: Find the family of lines parallel to 3x - 4y + 7 = 0.

Solution: Lines parallel to ax + by + c = 0 have the form ax + by + k = 0

So the family is: 3x - 4y + k = 0

where k is any real number.

This represents all lines with slope 3/4 (parallel to the given line). [/MathExample]

[MathExample title="Example 4: Perpendicular Lines Family"] Problem: Find the family of lines perpendicular to 2x + 3y - 5 = 0.

Solution: The given line has slope -2/3.

Lines perpendicular to it have slope 3/2.

The family of lines with slope 3/2 is: y = (3/2)x + c or 3x - 2y + k = 0 (where k = -2c)

This represents all lines perpendicular to 2x + 3y - 5 = 0. [/MathExample]