Geometry & Figures

Depth · one concept from first principles · step 2 of 4

Rotate a line about its fixed point

Every line y=ax-1 passes through (0, -1), whatever a is. Find all slopes a for which this line touches the segment from A(-2, 4) to B(-5, 1).
FunctionsExpressions
Your answer
How to solve
Strategy Draw a Diagram — Setting x=0 shows every such line goes through P(0,-1), so changing a just rotates the line about P. Draw it: the line sweeps across the segment between the position through B and the position through A. Find those two boundary slopes by substituting each endpoint, then take everything in between.
1STEP 1

Find the fixed point

Put x=0: the a term vanishes, so every line in the family passes through the same point.

x=0 → y=a · 0-1=-1
2STEP 2

Slope when the line hits A

Substitute A(-2, 4) into y=ax-1.

4=-2a-1 → a=-5/2
3STEP 3

Slope when the line hits B

Substitute B(-5, 1) into y=ax-1.

1=-5a-1 → a=-2/5
4STEP 4

Sweep between the two positions

The segment is left of the y-axis, so turning the line from A to B sweeps every point, with slope running from -5/2 to -2/5.

-5/2 ≤ a ≤ -2/5
Answer
-5/2 ≤ a ≤ -2/5
Try a=-1 (inside the range): y=-x-1 meets segment AB, which lies on y=x+6, where -x-1=x+6, so x=-7/2, between -5 and -2. Try a=-3 (outside): -3x-1=x+6 gives x=-7/4, which is not between -5 and -2, so it misses.
Takeaway

When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

  • Find the fixed point
  • Slope when the line hits A
  • Slope when the line hits B
  • Sweep between the two positions
Where next?
Another one like thissuggested

A parent dashboard for the family lives at sensimlab.com.

▶ Practice — 12 problems