Answer: Choice (3). Draw a circle with center A and radius equal to CB.
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Explanation:
Your teacher wasn't clear with this, but I'm assuming the distance from B to C is greater than half of AB. Otherwise, the bisection wouldn't be possible.
Assuming this is the case, drawing a circle centered at A with the same radius, will allow us to intersect the two circles at points D and E as shown below. Segment DE is the perpendicular bisector of AB. We can prove this through the use of congruent triangles. Segment DE cuts AB into two smaller equal pieces FA and FB, so FA = FB.
I assume the first equation is supposed to be
and not
As an augmented matrix, this system is given by
Multiply through row 3 by 1/2:
Add -1(row 2) to row 3:
Multiply through row 3 by 1/5:
Add -2(row 3) to row 1, and add 3(row 3) to row 2:
Add -3(row 2) to row 1:
Multiply through row 1 by -1:
Add -2(row 1) to row 2:
Multipy through row 2 by -1:
The solution to the system is then