while the rectangular coordinates show the position of a point along the #x# or #y# axis, polar coordinates show the distance of the point from #(0,0)# and the angle that the point makes with the #x#-axis.
the rectangular coordinates here are #(-sqrt2, sqrt2)#. if a horizontal line was drawn, then a vertical line, the horizontal line would extend to a point #sqrt2# to the left of #(0,0)# and the vertical line would extend from this new point to the point #sqrt2# from the left of #(0,0)# and #sqrt2# above #(0,0)#.
however, this can be drawn as one line. the ends of this line are at the ends of the lines where they do not meet. this forms a triangle with three sides that meet at three vertices.
since the Cartesian graph is right-angled, the triangle is also right-angled.
this means that we can use Pythagoras' theorem to solve for the length of the new line, which is the hypotenuse.
#a^2 + b^2 = c^2#
#a# is the distance along the #x#-axis, which is #-sqrt2#.
#b# is the distance along the #y#-axis, which is #sqrt2#
#c# is the length of the new line.
#(-sqrt2)^2 + (sqrt2)^2 = 2 + 2 = 4#
#c^2 = 4#
#c = +-2#
since the point is to the left of #(0,0)#, the line is also travelling left of #(0,0)#. this means that the length of line #c# is negative.
#c = -2#, so the distance of the point from #(0,0)# is #-2#.
meanwhile, we can again use two sides of the triangle to find the size of the angle between the line #c# and the #x#-axis. the angle can be labelled #theta#.
the #x#-length is #-sqrt2#. this is the adjacent to the angle.
the #y#-length is #sqrt2#. this is the opposite to the angle.
#tan theta = O/A#
here, #tan theta = (sqrt2)/(-sqrt2)#, which is #-1#.
then the inverse tan function could be applied to find #theta:#
#theta = tan^-1(-1) = -45^@#.
the coordinates are #(-2, 45^@)#
