Coordinate Geometry · Mathematics

Minimum PA + PB: Reflection Trick 3D Simulation

One moving point, three puzzles, on a sheet of 3D graph paper — the simulation of our “One moving point” video. A(1, 3) and B(5, 1); P slides along the x-axis by itself. Brute force first: the PA + PB graph draws as P sweeps and its dip is not the middle. Then the mirror: reflect A, join A′B, and P lands on the dip at (4, 0) with PA + PB = 4√2 and equal 45° angles — the path light takes. Then the triangle inequality gives the largest PA − PB = √20 at x = 7, and the homework: PA − PB has no least value, it only falls towards −4. Drag A, B and P.

Interactive virtual lab · no download · works on phones and laptops

Chapter: Locus & Coordinate Geometry
For: Class 10–12 · JEE Main · JEE Advanced
Time: ~20 min

Mirror Trick: Moving Point 3D Lab is a Pro simulation

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How to use it

1Pick a module (4); P slides along the axis by itself — drag P to hold it, drag A or B to move them.
2In the mirror module press the step buttons in order: reflect A, join A′B, put P there.
3The graph under the scene is the brute-force check; use Predict, Derive, Questions, Practise and Compare.

What to notice

Reflect $A$ in the axis: $P A = P A^{'}$ , so $P A + P B = P A^{'} + P B \geq A^{'} B$ — least where line $A^{'} B$ meets the axis.
At that point the two lines make equal angles with the axis: angle of incidence = angle of reflection.
If $A$ and $B$ are on opposite sides, no mirror is needed: segment $A B$ already crosses the axis.
$P A - P B \leq A B$ , with equality where line $A B$ meets the axis (on the far side of $B$ ).
A function can approach a value forever without reaching it: $P A - P B \to x_{A} - x_{B}$ has no minimum here.

For teachers

A ready-made teaching kit comes with this lab

Lesson plan
24 min, 4 steps
Ask the class
4 questions with answers
Quick quiz
5 questions, printable
Key formulas
4 formulas

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About the Mirror Trick: Moving Point 3D Lab simulation

The Mirror Trick 3D Lab is the simulation of our “One moving point” video. On a sheet of graph paper in 3D, A(1, 3) is a blue ball and B(5, 1) an orange one, and a purple ball P slides along the x-axis by itself, joined to both. The first module is brute force: as P sweeps, the graph of PA + PB draws underneath, and its single dip is at x = 4 — not halfway between A and B, where most people guess.

The second module finds that dip with no calculus, in three steps you press yourself: reflect A in the axis to A′(1, −3), so PA = PA′ for every P; join A′ to B with a straight line, the shortest path; and put P where it crosses the axis. The least PA + PB is A′B = √32 = 4√2 ≈ 5.66, and a ray of light running A → P → B shows the two 45° angles are equal. The third module uses the triangle inequality: PA − PB ≤ AB, reached where line AB meets the axis, at x = 7, giving √20. The last is the video's homework: PA − PB never has a least value here, because equality would need A between P and B; it only falls towards x_A − x_B = −4 far to the left, which you can watch by stretching the graph to x = −1000. Drag A, B and P anywhere.

Also covers: minimum of pa + pb, pa + pb minimum, minimise pa + pb, point on x axis minimum distance, reflection method minimum distance, shortest path reflection, image of a point in x axis, maximum of pa - pb, |pa - pb| maximum, triangle inequality, heron's shortest path, law of reflection geometry, locus coordinate geometry jee, straight lines jee, minimum and maximum distance moving point.

Frequently asked questions

How do you find the point P on the x-axis that minimises PA + PB?

If A and B are on the same side of the axis, reflect one of them, say A, in the axis to get A′. Every P on the axis is as far from A as from A′, so PA + PB = PA′ + PB, which is least when P lies on the straight segment A′B. P is where A′B crosses the axis, and the least value is the length A′B. If A and B are on opposite sides, the segment AB already crosses the axis and the least value is AB.

Why is the answer not the midpoint?

The point nearer the axis pulls P towards itself. For A(1, 3) and B(5, 1), PA + PB is 5.84 at the midpoint x = 3 but 5.66 at x = 4, which is the true minimum. The lab draws the whole curve so you can see the dip is off-centre.

What is the maximum of |PA − PB| for P on a line?

By the triangle inequality |PA − PB| ≤ AB, with equality when P, A and B are collinear with P outside the segment. So the maximum is AB, reached where line AB meets the given line — provided it does meet it (if AB is parallel to the line, the value is only approached). For points on opposite sides, reflect one point first.

Does PA − PB always have a minimum?

No. For A(1, 3) and B(5, 1) the bound PA − PB ≥ −AB would need A between P and B on a line, which never happens for P on the x-axis. PA − PB keeps decreasing as P moves left and approaches x_A − x_B = −4 without ever reaching it: it has an infimum but no minimum.

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