两点间距离公式liǎng diǎn jiān jùlí gōngshì

Core Concept

The distance formula (两点间距离公式) gives the straight-line distance between any two points in the coordinate plane. It is the single most frequently used formula in CSCA analytic geometry — it underpins the equation of a circle, the definitions of the ellipse and hyperbola, and every "find the length of segment $AB$" question.

Mathematical Definition

For points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$:

$|P_1P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Why It Works

The formula is the Pythagorean theorem in disguise. Draw a right triangle with the segment $P_1P_2$ as its hypotenuse and legs parallel to the axes:

• Horizontal leg: $|x_2 - x_1|$
• Vertical leg: $|y_2 - y_1|$

Then $|P_1P_2|^2 = |x_2-x_1|^2 + |y_2-y_1|^2$. Because both legs are squared, the absolute-value signs are unnecessary — which is why the formula is usually written without them.

The Three-Dimensional Version

In space, with $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$, add one more squared term:

$|P_1P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

This appears in CSCA space-coordinate questions.

Related Formulas

Midpoint Formula

The midpoint $M$ of segment $P_1P_2$ is the coordinate-wise average:

$M\left(\frac{x_1 + x_2}{2}, \ \frac{y_1 + y_2}{2}\right)$

Distance from a Point to a Line

The distance from $P(x_0, y_0)$ to the line $Ax + By + C = 0$ is:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

This is a separate formula and a separate exam favourite — note the absolute value in the numerator, which many students drop.

Distance from the Origin

A useful special case when $P_1$ is the origin:

$|OP| = \sqrt{x^2 + y^2}$

Worked Applications

Application 1: Classifying a Triangle

Is the triangle with vertices $A(0,0)$, $B(4,0)$, $C(2, 2\sqrt{3})$ equilateral?

• $|AB| = \sqrt{4^2 + 0^2} = 4$
• $|AC| = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = 4$
• $|BC| = \sqrt{(2-4)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = 4$

All three sides equal $4$, so the triangle is equilateral.

Application 2: Deriving the Circle Equation

A circle is the set of points at fixed distance $r$ from centre $C(a, b)$. Writing that condition with the distance formula:

$\sqrt{(x-a)^2 + (y-b)^2} = r$

Squaring both sides gives the standard equation $(x-a)^2 + (y-b)^2 = r^2$. The circle equation is the distance formula.

CSCA Practice Problems

💡 Note: These problems follow the CSCA exam syllabus and Chinese standardized test formats.

Example 1: Basic (Difficulty ★☆☆☆☆)

The distance between $A(-2, 1)$ and $B(1, 5)$ is:

Options:

• A. $3$
• B. $4$
• C. $5$
• D. $7$

Detailed Solution:

$|AB| = \sqrt{(1 - (-2))^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Recognising the $3$-$4$-$5$ triple saves time. CSCA questions overwhelmingly use Pythagorean triples ($3,4,5$ / $5,12,13$ / $6,8,10$) so the answer comes out as a whole number — if your answer is an ugly surd, re-check the subtraction.

Answer: C

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Example 2: Intermediate (Difficulty ★★★☆☆)

Point $P$ lies on the $x$-axis and is equidistant from $A(1, 2)$ and $B(5, 6)$. Find the coordinates of $P$.

Options:

• A. $(5, 0)$
• B. $(6, 0)$
• C. $(7, 0)$
• D. $(8, 0)$

Detailed Solution:

1. Since $P$ lies on the $x$-axis, write $P(x, 0)$.

2. Set $|PA| = |PB|$. Square both sides immediately to clear the radicals: $(x-1)^2 + (0-2)^2 = (x-5)^2 + (0-6)^2$

3. Expand both sides: $x^2 - 2x + 1 + 4 = x^2 - 10x + 25 + 36$

4. The $x^2$ terms cancel — this is what makes the equation linear and easy: $-2x + 5 = -10x + 61$

5. Solve: $8x = 56 \implies x = 7$

6. Verify by substituting back: $|PA|^2 = (7-1)^2 + 2^2 = 36 + 4 = 40$ and $|PB|^2 = (7-5)^2 + 6^2 = 4 + 36 = 40$. ✓

Answer: C

Method note: Any point equidistant from $A$ and $B$ lies on the perpendicular bisector of $AB$. You could also find that line and intersect it with $y = 0$ — but squaring the distance equation is faster when only one intersection is needed.

Common Mistakes

❌ Mistake 1: Forgetting to take the square root

$(x_2-x_1)^2 + (y_2-y_1)^2$ is the square of the distance. Half-finished answers like "$25$" instead of "$5$" are the single most common error on this topic.

❌ Mistake 2: Subtracting coordinates across the pair

Writing $\sqrt{(x_2 - y_1)^2 + (y_2 - x_1)^2}$ mixes the axes. Always subtract $x$ from $x$, and $y$ from $y$.

❌ Mistake 3: Dropping the absolute value in point-to-line distance

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2+B^2}}$ can produce a negative numerator without the bars — and a negative distance is meaningless.

Study Tips

1. ✅ Understand it as the Pythagorean theorem, not as a formula to memorise
2. ✅ Square both sides early when setting two distances equal — it kills the radicals
3. ✅ Learn the common triples ($3,4,5$ / $5,12,13$) to spot clean answers
4. ✅ Keep the midpoint formula next to it — they appear together in perpendicular-bisector questions
5. ✅ Substitute your answer back into the original condition to verify

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💡 Exam Tip: 求平面两点间的距离 is the highest-frequency knowledge point in CSCA analytic geometry. It is rarely the whole question — it is usually step 1 of a longer circle or conic problem, so speed here buys time later.