Assume that photo coordinates (in mm) of points a and b are Xa=65.35, Ya=74.88 and Xb=31.45, Yb= -55.50. What is the photo distance ab?

Question

Assume that photo coordinates (in mm) of points a and b are Xa=65.35, Ya=74.88 and Xb=31.45, Yb= -55.50. What is the photo distance ab?

(a) 134.7

(b) 39.05

(c) 162.4

(d) 92.68

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Adalynn 2 weeks 2021-09-12T08:30:33+00:00 1 Answer 0

Answers ( )

    0
    2021-09-12T08:32:26+00:00

    Answer:

    Option A (134.7mm)

    Step-by-step explanation:

    Let’s find the distance, but first we need to remember that the distance between two points with coordinates (Xa,Ya) and (Xb,Yb) is defined by:

    distance = \sqrt{(Xb-Xa)^{2} + (Yb-Ya)^{2}  }

    From the situation we notice that:

    Xb=31.45 and Xa=65.35, as well as:

    Yb=-55.50 and Ya=74.88

    Using the previous equation we have:

    distance = \sqrt{(31.45-65.35)^{2} + (-55.50-74.88)^{2}  }

    distance = \sqrt{(-33.9)^{2} + (-130.38)^{2}  }

    distance = \sqrt{1149.21 + 16998.9444}

    distance = \sqrt{18148.1544}

    distance = 134.7151mm

    In conclusion, the distance between points (65.35,74.88) and (31.45,-55.50) is 134.7151mm, which is option A (134.7mm).

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