b) The main diagonal is the perpendicular bisector of the segment between A and C. AC has slope 3/5, so your line will have slope -5/3. The midpoint of AC is
.. ((2, 4) +(-3, 1))/2 = (-1/2, 5/2)
In point-slope form the line is
.. y = (-5/3)(x +1/2) +5/2
or in standard form,
.. 5x +3y = 5
c) You want to find a point on the line parallel to the line through AC that goes through the point P and is the same distance that P is from the line of part (b).
The image point will be on line
.. y = 3/5(x +4) -1
The distance from P to the line of part (b) is given by
.. d = |5*(-4) +3*(-1) -5|/√(5^2 +3^2) = |-28|/√34 = (28/34)√34 ≈ 4.802
We can find the x-coordinate of the image of P from
.. d = 28/√34 = |5x +3(3/5(x +4) -1) -5|/√34
.. 28 = |(34/5)x -4/5|
.. 28.8/6.8 = x = 4 4/17 ≈ 4.235
The y-coordinate is then
.. y = (3/5)(4 1/17 +4) -1 = 3 16/17 ≈ 3.941
Q = (4 4/17, 3 16/17) ≈ (4.235, 3.941)