How do you find the magnitude of YZ given Y(5,0) and Z(7,6)?

Question

Sadie Ackerman · Accepted Answer

# bb(vec(YZ)) = ( (2), (6) ) \ \ # and # \ \ abs(bb(vec(YZ))) = 2sqrt(10) #

We have #Y# and #Z# with coordinates #(5,0)# and #(7,6)# respectively. So in vector notation we can write: # bb(vec(OY)) = ( (5), (0) ) \ \ # and # \ \ bb(vec(OZ)) = ( (7), (6) ) # We can calculate #abs(bb(vec(YZ)))# in several ways: Method 1: Using the coordinates along, we can apply pythagoras theorem: # YZ^2 = (7-5)^2 + (6-0)^2 # # \ \ \ \ \ \ \ = 2^2 + 6^2 # # \ \ \ \ \ \ \ = 4+36 # # \ \ \ \ \ \ \ = 40 # And so #YZ = sqrt(40) = 2sqrt(10) # Method 2: Using vector notation we can calculate the vector #bb(vec(YZ))# and then calculate its magnitude. We have: # bb(vec(YZ)) = bb(vec(OZ)) - bb(vec(OY)) # # \ \ \ \ \ \ = ( (7), (6) ) - ( (5), (0) ) # # \ \ \ \ \ \ = ( (7-5), (6-0) ) # # \ \ \ \ \ \ = ( (2), (6) ) # And so: # abs(bb(vec(YZ))) = sqrt(2^2+6^2) # # \ \ \ \ \ \ \ \ = 2sqrt(10) #, as before.

Greyson Brown · Answer

undefined You can find the magnitude of YZ using the distance formula, which states that the distance between two points (x1, y1) and (x2, y2) in a coordinate plane is given by the formula:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Given Y(5,0) and Z(7,6), plug the coordinates into the formula:

$$ d = \sqrt{(7 - 5)^2 + (6 - 0)^2} $$

$$ d = \sqrt{(2)^2 + (6)^2} $$

$$ d = \sqrt{4 + 36} $$

$$ d = \sqrt{40} $$

$$ d = 2\sqrt{10} $$

So, the magnitude of YZ is $ 2\sqrt{10} $.