Question

The figure below shows parallelograms ABCE, PQRS, and TNLM on a coordinate plane.
• Points S, P, Q, and Rare midpoints of ABCE
• T, N, L, and Mare midpoints of PQRS
Write the equation of line ML. Show your work.

Answer 1

Answer:

ML = √[¼ (EC)²]

Step-by-step explanation:

ML² = ½(QR)²

QR² = ½(EC)²

so, ML² = ¼ (EC)²

=> ML = √[¼ (EC)²]

Answer 2

The equation of the line ML is given by $ML = \frac{AC}{2\sqrt{2} }$   .

Concept:

1. We will find the length of SE and ER.
2. Then we will find the length of SR by using the Pythagoras theorem in Δ ESR.
3. Then, we will find the length of MR and RL and hence, again using the Pythagoras theorem in Δ MRL and find the length of ML.

How to solve the given question?

• Since, S is the midpoint of AE.
             $SE = \frac{AE}{2}$
  Similarly, $RE = \frac{CE}{2}$
• In Δ SER, ∠E = 90°,
  ∴ By Pythagoras theorem, SR² = SE² + ER²
  ∴ $SR^2 = (\frac{AE}{2})^2 + ( \frac{RE}{2})^2 = \frac{AE^2 + CE^2}{4}$   ...... (1)
  ∴ Similarly, $RQ^2 = \frac{BC^2 + CE^2}{4}$          ............ (2)
  

• Since, M & L are the midpoints of SR and RQ.
             $MR = \frac{SR}{2}$ &  $RL = \frac{RQ}{2}$
  
• In Δ LMR, ∠R = 90°,
  ∴ By Pythagoras theorem, ML² = RM² + RL²
  ∴ $ML^2 = (\frac{SR}{2})^2 + ( \frac{RQ}{2})^2 = \frac{SR^2 + RQ^2}{4}$
  
  Substituting value of SR and RQ from 1 and 2.
  $ML^2 =\frac{1}{4} [ \frac{AE^2 + CE^2}{4} + \frac{BC^2 + CE^2}{4} ]$
  
  As ABCE is parallelogram, BC = AE
  ∴ $ML^2 =\frac{1}{4} [ \frac{2AE^2 + 2CE^2}{4} ] = [ \frac{AE^2 + CE^2}{8} ]$
  $ML = \frac{AC}{2\sqrt{2} }$

Thus, the equation of the line ML is given by $ML = \frac{AC}{2\sqrt{2} }$   .

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