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Sophie [7]
3 years ago
9

A food truck operator is parked in a lot at the corner of two streets. She wants to be equidistant from both streets. Should she

park her truck on a perpendicular bisector, an angle bisector, a median, or an altitude?a) perpendicular bisectorb) angle bisectorc) mediand) altitude
Mathematics
1 answer:
alekssr [168]3 years ago
7 0

Answer:

Angle bisector

Step-by-step explanation:

median isn't applicable in this case as the roads from the streets are inclined at an angle.

altitude refers to height which is also not applicable

The perpendicular bisector is the locust of points equidistant from two points,

in this question the street are not seen as points but as lines which forms an angle and the bisection of this angle forms a locus where she can park her car. If she parks her car anywhere on the angular bisector of the two streets, she would be at equal distance from both streets.

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Answer:

3. \overline{PM}\cong \overline{ON} Distances between two parallel lines \overline{MN} \ and\ \overline{PO}

4. \overline{AC} = \overline{CE}: Reason; Corresponding part of ΔACB and ΔDCE

C is the midpoint of \overline{AE}: Reason;  \overline{AC} = \overline{CE}: Definition of midpoint

Step-by-step explanation:

3. A parallelogram is defined as a quadrilateral with two opposite sides equal and parallel and having equal opposite interior angles

MNOP is a parallelogram: Reason; Given

\overline{PM}\left |  \right |\overline{ON} : Reason; Opposite sides of a parallelogram

∠NOM ≅ ∠OMP: Reason Alternate interior angles

\overline{MN}\left |  \right |\overline{PO}: Reason; Opposite sides of a parallelogram

∠NMO ≅ ∠MOP: Reason Alternate interior angles

\overline{PM}\cong \overline{ON} Distances between two parallel lines \overline{MN} \ and\ \overline{PO}

4. \overline{AB}\left |  \right |\overline{DE} : Reason; Given

∠EAB ≅ ∠AED: Reason; Alternate int. ∠s Thm

∠ABC ≅ ∠EDB : Reason; Alternate int. ∠s Thm

C is the midpoint of  \overline{BD}: Reason; Given

\overline{BC} = \overline{CD}: Reason; Definition of midpoint

Therefore, ΔACB ≅ ΔDCE: Reason Angle Angle Side (AAS) Theorem

\overline{AC} = \overline{CE}: Reason; Corresponding part of ΔACB and ΔDCE

C is the midpoint of \overline{AE}: Reason; Definition of midpoint

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