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

Answer 1

Option A (134.7mm)

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 = √(1149.21 + 16998.9444)`

`distance = √(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).