Given point (-8,8) how do you find the distance of the point from the origin, then find the measure of the angle in standard position whose terminal side contains the point?

1 Answer
Mar 19, 2018

Measure of the angle #theta = 135^@ # or #((3pi)/4)^c#

Explanation:

Distance formula : # d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

Given point #A (-8, 8)# and Origin #O (0,0)#

#vec(OA) = r = sqrt ((-8)^2 + 8^2) = sqrt 128 = 8sqrt2#

#x = r cos theta#, #y = r sin theta#

#- 8 = 8 sqrt2 cos theta#

#cos theta = -cancel8 / (cancel8 sqrt2) = -1/sqrt2# or #theta =color(purple)( - pi/4 = (pi - pi/4) = (3pi)/4#

Similarly, #8 sqrt2 sin theta = 8#

#sin theta = 8 / (8 sqrt2) = 1/sqrt2# or #theta = pi/4# or = #pi - pi/4) = (3pi)/4#

Slope of the line #m = tan theta = -8 / 8 = -1 #

#x# #(-)# and #y# #(+)# in II quadrant.

Hence #theta = tan ^-1 (-8/8) = tan^-1 -1 = -1 = (3pi)/4 #

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