Math Problem Statement

Grade 6 question using segment addition and subtration properties .On a straight line xy place in order the points A C B D with AB=CD and AC=BD Let point I be midpoint of BC and midpoint of AD. . Trace the straight line d perpendicular to xy at C let m be an arbitrary point of d . Is ma=mb. Draw graph and justify

Solution

To solve this problem, let's break it down step by step using the properties of segment addition and subtraction.

Given Information:

  1. Points A, C, B, D are on a straight line in the order A, C, B, D.
  2. AB=CDAB = CD
  3. AC=BDAC = BD
  4. Point I is the midpoint of BC and the midpoint of AD.
  5. Line d is perpendicular to xy at point C, and point M is an arbitrary point on line d.

Steps to Solve:

  1. Set Up the Points:

    • Place the points on a coordinate system.
    • Let A(0,0)A(0, 0), C(a,0)C(a, 0), B(a+b,0)B(a + b, 0), and D(a+b+c,0)D(a + b + c, 0), where bb is the length of segment AB and cc is the length of segment CD.
    • Given AB=CDAB = CD, b=cb = c.
  2. Find Coordinates for Points:

    • Since II is the midpoint of both segments:
      • II for BC: I=(a+(a+b)2,0)=(a+b2,0)I = \left(\frac{a + (a + b)}{2}, 0\right) = \left(a + \frac{b}{2}, 0\right)
      • II for AD: I=(0+(a+b+c)2,0)=(a+b+c2,0)I = \left(\frac{0 + (a + b + c)}{2}, 0\right) = \left(\frac{a + b + c}{2}, 0\right)
      • Set b=cb = c implies that a+b2=a+2b2a + \frac{b}{2} = \frac{a + 2b}{2}.
  3. Draw the Perpendicular Line:

    • At point C(a,0)C(a, 0), draw line dd which is vertical, meaning MM can be any point of the form (a,y)(a, y) where yy is not equal to 0.
  4. Show that MA=MBMA = MB:

    • The distances from MM to AA and BB can be calculated:
      • Distance MAMA: MA=(a0)2+(y0)2=a2+y2MA = \sqrt{(a - 0)^2 + (y - 0)^2} = \sqrt{a^2 + y^2}
      • Distance MBMB: MB=(a(a+b))2+(y0)2=(b)2+y2=b2+y2MB = \sqrt{(a - (a + b))^2 + (y - 0)^2} = \sqrt{(-b)^2 + y^2} = \sqrt{b^2 + y^2}
    • Since AB=CDAB = CD (or b=cb = c), and with b=cb = c, MA=MBMA = MB.

Conclusion:

Since both distances are equal for any arbitrary point MM on line dd, we conclude that MA=MBMA = MB.

Graph:

To visualize this, you can draw:

  • A horizontal line with points A, C, B, and D marked at appropriate distances.
  • A vertical line at point C.
  • Points A, B, and M marked clearly.

Justification:

The geometric properties of midpoints and the equal distances from a perpendicular point to two points on a straight line show that the distances MAMA and MBMB remain equal.

Questions for Further Understanding:

  1. How would the distances change if bcb \neq c?
  2. What happens if point M lies on the x-axis instead of line d?
  3. Can you identify similar properties in other geometric figures?
  4. How would you prove the midpoint theorem using different shapes?
  5. What role does symmetry play in this problem?

Tip:

When solving geometric problems, always visualize them with a sketch to better understand relationships and properties.

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Math Problem Analysis

Mathematical Concepts

Segment Addition
Subtraction Properties
Midpoints
Perpendicular Lines
Distance Formula

Formulas

Midpoint formula (M = ((x1 + x2)/2, (y1 + y2)/2))
Distance formula (d = √((x2 - x1)^2 + (y2 - y1)^2))

Theorems

Midpoint Theorem
Properties of Perpendicular Lines

Suitable Grade Level

Grade 6