The terminal ray of a 300° angle lies in the This angle measures i radians. quadrant.
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Well, we know the line is perpendicular to that one above.... what is the slope of that one anyway? well, notice, the equation is already in slope-intercept form 
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so, we're looking for the equation of a line perpendicular to that one, now, since that one has a slope of 4/3, a perpendicular line will have a negative reciprocal slope to that one,
so, what is the equation of a line whose slope is -3/4 and runs through -4,9?
![\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ &&(~ -4 &,& 9~) \end{array} \\\\\\ % slope = m slope = m\implies -\cfrac{3}{4} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-9=-\cfrac{3}{4}[x-(-4)] \\\\\\ y-9=-\cfrac{3}{4}(x+4)](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%5C%5C%0A%26%26%28~%20-4%20%26%2C%26%209~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%20m%5Cimplies%20-%5Ccfrac%7B3%7D%7B4%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-9%3D-%5Ccfrac%7B3%7D%7B4%7D%5Bx-%28-4%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay-9%3D-%5Ccfrac%7B3%7D%7B4%7D%28x%2B4%29)
now, the x-intercept for any function is found by zeroing out the "y" and solving for "x", thus
x = 8, y = 0 ( 8 , 0 )
so, we're looking for the equation of a line perpendicular to that one, now, since that one has a slope of 4/3, a perpendicular line will have a negative reciprocal slope to that one,
so, what is the equation of a line whose slope is -3/4 and runs through -4,9?
now, the x-intercept for any function is found by zeroing out the "y" and solving for "x", thus
x = 8, y = 0 ( 8 , 0 )
