Distance Formula

The Distance Formula

How far apart are two points $A(x_1, y_1)$ and $B(x_2, y_2)$? Drop perpendiculars to form a right triangle: the horizontal leg is $(x_2 - x_1)$ and the vertical leg is $(y_2 - y_1)$. By the Pythagoras theorem, the distance is the hypotenuse:

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

Key Point: It does not matter which point you call first — squaring removes any sign issue. So $(x_2 - x_1)^2 = (x_1 - x_2)^2$.

[Board Important] This formula is the heart of the chapter. Write it down first, then substitute carefully — most marks here are for correct substitution.

Distance from the Origin

The distance of a point $P(x, y)$ from the origin $O(0, 0)$ is a special case with $x_1 = y_1 = 0$:

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

Think of it this way: it's just the distance formula with the origin as the second point.

Key Point: $OP = \sqrt{x^2 + y^2}$ — memorise this shortcut; it appears in many 1-mark questions.

[Board Important] For example, the distance of $(3, 4)$ from the origin is $\sqrt{9 + 16} = \sqrt{25} = 5$.

Using Distance to Identify Shapes

Distances let us classify triangles and quadrilaterals:

• Equilateral triangle: all three sides equal.
• Isosceles triangle: exactly two sides equal.
• Right triangle: the three sides satisfy the Pythagoras relation $a^2 + b^2 = c^2$.
• Square: all four sides equal AND the two diagonals equal.
• Rhombus: all four sides equal, but diagonals unequal.
• Rectangle: opposite sides equal AND diagonals equal.
• Parallelogram: opposite sides equal (diagonals generally unequal).

Key Point: To name a quadrilateral, compute all four sides and both diagonals, then compare.

[Board Important] A frequent 3-mark question: 'Show that the points … form a square/rhombus/right triangle.' Always justify with the side and diagonal lengths.

Collinearity Using Distances

Three points $A$, $B$, $C$ are collinear (lie on one straight line) if the longest distance equals the sum of the other two — for example $AC = AB + BC$.

This is because the only way to travel from $A$ to $C$ through $B$ with no detour is along a straight line.

Key Point: Collinear ⇒ one distance = sum of the other two. (We'll see a faster area-based test in Section 4.)

[JEE Tip] The distance method for collinearity is reliable but slow; the zero-area test is quicker for three given points.

Solved Examples

Example 1: Basic distance

Find the distance between $A(1, 2)$ and $B(4, 6)$.

Solution:

1. $AB = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2}$.
2. $= \sqrt{9 + 16} = \sqrt{25} = 5$.

Final Answer: $AB = 5$ units.

Takeaway: $(3, 4, 5)$ shows up constantly — recognise Pythagorean triples.

Example 2: Distance from origin

Find the distance of $P(-6, 8)$ from the origin.

Solution:

1. $OP = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64}$.
2. $= \sqrt{100} = 10$.

Final Answer: 10 units.

Takeaway: Use $OP = \sqrt{x^2 + y^2}$ directly.

Example 3: Find an unknown coordinate

The distance between $(2, -3)$ and $(10, y)$ is 10. Find $y$.

Solution:

1. $\sqrt{(10-2)^2 + (y+3)^2} = 10 \Rightarrow 64 + (y+3)^2 = 100$.
2. $(y+3)^2 = 36 \Rightarrow y + 3 = \pm 6$.
3. $y = 3$ or $y = -9$.

Final Answer: $y = 3$ or $y = -9$.

Takeaway: Square both sides; expect two answers from $\pm$.

Example 4: Point equidistant from two points

Find a point on the x-axis equidistant from $A(2, -5)$ and $B(-2, 9)$.

Solution:

1. Let the point be $P(x, 0)$.
2. $PA^2 = PB^2$: $(x-2)^2 + 25 = (x+2)^2 + 81$.
3. $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$.
4. $-8x = 56 \Rightarrow x = -7$.

Final Answer: $(-7, 0)$.

Takeaway: Equidistant ⇒ set squared distances equal; the $x^2$ terms cancel.

Example 5: Isosceles triangle

Show that $A(5, -2)$, $B(6, 4)$, $C(7, -2)$ form an isosceles triangle.

Solution:

1. $AB = \sqrt{1^2 + 6^2} = \sqrt{37}$.
2. $BC = \sqrt{1^2 + (-6)^2} = \sqrt{37}$.
3. $AC = \sqrt{2^2 + 0^2} = 2$.
4. $AB = BC \neq AC$.

Final Answer: Isosceles (two equal sides).

Takeaway: Two equal sides ⇒ isosceles.

Example 6: Right triangle test

Do $A(0, 0)$, $B(6, 0)$, $C(0, 8)$ form a right triangle?

Solution:

1. $AB = 6$, $AC = 8$, $BC = \sqrt{36 + 64} = 10$.
2. Check: $6^2 + 8^2 = 36 + 64 = 100 = 10^2 = BC^2$.

Final Answer: Yes, right-angled at $A$.

Takeaway: Verify the Pythagoras relation with the longest side as hypotenuse.

Example 7: Collinearity by distances

Are $A(1, 1)$, $B(3, 3)$, $C(5, 5)$ collinear?

Solution:

1. $AB = \sqrt{4 + 4} = 2\sqrt2$; $BC = \sqrt{4 + 4} = 2\sqrt2$; $AC = \sqrt{16 + 16} = 4\sqrt2$.
2. $AB + BC = 2\sqrt2 + 2\sqrt2 = 4\sqrt2 = AC$.

Final Answer: Yes, collinear.

Takeaway: If one distance equals the sum of the other two, the points are collinear.

Example 8: Equilateral triangle

Show that $A(0, 0)$, $B(4, 0)$, $C(2, 2\sqrt3)$ form an equilateral triangle.

Solution:

1. $AB = 4$.
2. $AC = \sqrt{4 + 12} = \sqrt{16} = 4$.
3. $BC = \sqrt{(2-4)^2 + (2\sqrt3)^2} = \sqrt{4 + 12} = 4$.
4. All sides equal 4.

Final Answer: Equilateral (all sides 4).

Takeaway: All three sides equal ⇒ equilateral.

Example 9: Equidistant point on y-axis

Find a point on the y-axis equidistant from $A(6, 5)$ and $B(-4, 3)$.

Solution:

1. Let $P(0, y)$. Then $PA^2 = PB^2$.
2. $36 + (y-5)^2 = 16 + (y-3)^2$.
3. $36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9$.
4. $61 - 10y = 25 - 6y \Rightarrow 36 = 4y \Rightarrow y = 9$.

Final Answer: $(0, 9)$.

Takeaway: A point on the y-axis has $x = 0$; set squared distances equal.

Example 10: Radius of a circle

The centre of a circle is $(2, 3)$ and a point on it is $(5, 7)$. Find the radius.

Solution:

1. Radius = distance from centre to the point.
2. $r = \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Final Answer: Radius $= 5$ units.

Takeaway: The radius is just the distance from the centre to any point on the circle.