Exercise 1: Line-Line Intersection of Two Diagonals — Possible Solution
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SETTING UP THE PROBLEM
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Line 1 passes through P1=(1,1) and P2=(5,5) - this is the line y=x.
Line 2 passes through P3=(1,5) and P4=(5,1) - this is the line
y = -x + 6 (it descends from (1,5) to (5,1)).

SOLVING GEOMETRICALLY FIRST, AS A SANITY CHECK
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Setting the two line equations equal: x = -x + 6, so 2x = 6, x = 3,
and since y=x, y=3 as well. The expected intersection point is (3,3).

APPLYING THIS CHAPTER'S OWN FORMULA
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Using x1=1,y1=1, x2=5,y2=5, x3=1,y3=5, x4=5,y4=1:

    denom = (x1-x2)(y3-y4) - (y1-y2)(x3-x4)
          = (1-5)(5-1) - (1-5)(1-5)
          = (-4)(4) - (-4)(-4)
          = -16 - 16
          = -32

    t = [(x1-x3)(y3-y4) - (y1-y3)(x3-x4)] / denom
      = [(1-1)(5-1) - (1-5)(1-5)] / -32
      = [0 - 16] / -32
      = -16 / -32
      = 0.5

    intersection = (x1 + t*(x2-x1), y1 + t*(y2-y1))
                 = (1 + 0.5*(5-1), 1 + 0.5*(5-1))
                 = (1 + 2, 1 + 2)
                 = (3, 3)

RESULT
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The intersection point is (3, 3), matching the independent geometric
sanity check exactly.

WHY THIS WORKS AS AN ANSWER
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The answer solves the problem two independent ways - once via simple
geometric reasoning about the two lines' own equations, and once by
mechanically applying this chapter's own formula step by step - and
confirms both arrive at the identical result, rather than presenting
only the formula's output without any independent check.