External Division: A Point Beyond the Segment
Same formula with a minus: a point dividing PQ externally in ratio m:n lies OUTSIDE the segment.
Equation
y = x
Graph
Table
| x | y |
|---|---|
| -10 | -10 |
| -9 | -9 |
| -8 | -8 |
| -7 | -7 |
| -6 | -6 |
| -5 | -5 |
| -4 | -4 |
| -3 | -3 |
| -2 | -2 |
| -1 | -1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
| 4 | 4 |
| 5 | 5 |
| 6 | 6 |
| 7 | 7 |
| 8 | 8 |
| 9 | 9 |
| 10 | 10 |
| 11 | 11 |
| 12 | 12 |
| 13 | 13 |
| 14 | 14 |
| 15 | 15 |
What this lesson covers
What you do
You shape the function y = m*x + c and watch the graph answer.
Challenges to clear
- The external 2:1 division of (2, 1) and (6, 2) is ((2·6−1·2)/(2−1), (2·2−1·1)/(2−1)) = (10, 3). Make the line pass through it.
- Anchor the other end: pass through (2, −1) as well — slope ½, beyond the segment.
- The 3:1 external division of (3, −2) and (5, 1) lands at (9, 4) — beyond the segment, not inside it. Pass through it.
- Now anchor the other end at (3, −2) as well. External points still sit on the same straight line.
Check yourself
How does external division differ from internal?
- The denominator becomes m - n instead of m + n; the dividing point sits OUTSIDE the segment PQ. — correct
- The endpoints are swapped.
- The ratio is always negative for external division.
- External division uses only the midpoint formula twice.
Think about it
- External division pushes the point PAST the second endpoint — the formula swaps plus to minus: (m·x2 − n·x1)/(m − n).