Video Transcript
The point π§ satisfies the argument
of π§ minus six over π§ minus six π equals π by four. Sketch the locus of π§ on an Argand
diagram.
We notice that the equation in this
question looks a lot like the argument of π§ minus π§ one over π§ minus π§ two
equals π. Now, the locus of π§ in this case
is an arc of a circle which subtends an angle of π between the points represented
by π§ one and π§ two. And of course this is sketched in a
counterclockwise direction from π§ one to π§ two. Comparing the general form to the
equation in our question, and we find that π§ one must be equal to six and π§ two
must be equal to six π. On an Argand diagram, these are
points represented by the Cartesian coordinates six, zero and zero, six,
respectively. We also see that π is equal to π
by four. And we know if π is less than π
by two, then our locus represents a major arc.
We also know that the endpoints are
not included in our locus, but here we do have a bit of a problem. We know our locus is the arc of a
circle. But without knowing the center of
the circle, we canβt use this information to find the correct arc. In fact, it could be this major arc
or this major arc. But we do know that the locus is
drawn in a counterclockwise direction from π§ one, well thatβs six, zero, to π§ two,
which is zero, six. For that to be the case, we have to
use this arc on the right, and we see that the locus is shown. We can also choose to add the angle
of π by four radians here.
