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

Respuesta :

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:

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

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:

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

[tex]distance = \sqrt{(-33.9)^{2} + (-130.38)^{2}  }[/tex]

[tex]distance = \sqrt{1149.21 + 16998.9444}[/tex]

[tex]distance = \sqrt{18148.1544}[/tex]

[tex]distance = 134.7151mm[/tex]

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