NDA Maths · Lines

Distance, Section & Locus

Distances between points, from a point to a line, and between parallel lines; the section formula for dividing a segment; and locus equations from a geometric condition.

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Why this matters

These three formulas underpin half the chapter. The point-to-line distance and the section formula appear constantly, and 'locus' questions are just a geometric condition translated into an equation.

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Concept 1 of 3: Distance: point-point, point-line, parallel lines

Three distances recur: between two points (Pythagoras), from a point to a line (perpendicular distance using the normalised equation), and between two parallel lines (difference of constants over the same normaliser).

Definition

Two points: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .
Point to line: distance from $(x_0,y_0)$ to $ax+by+c=0$ is $\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ .
Parallel lines $ax+by+c_1=0$ , $ax+by+c_2=0$ : $\dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}$ (make the $a,b$ match first).
Distance measured PARALLEL to a given direction ("the distance from $P$ to the line $L$ , measured along a line making angle $\theta$ with the x-axis") is NOT the perpendicular distance: it is the length of the segment from $P$ to $L$ along that direction. Parametrise $P+t(\cos\theta,\sin\theta)$ , substitute into $L$ , solve for $t$ ; $|t|$ is the answer, and it equals $\dfrac{|ax_0+by_0+c|}{|a\cos\theta+b\sin\theta|}$ . The perpendicular distance is its minimum over $\theta$ .

Distance formulas

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\qquad \dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\qquad \dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}

Worked example

Find the distance from

(1,2)

to the line

3x+4y-10=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2018 · Q64Easy

Example 1 · Lines · Distance, Section, and Locus

What is the distance between

3x+4y=9

and

6x+8y=15

Put the line in $ax+by+c=0$ form first, and divide by $\sqrt{a^2+b^2}$ — not by $a+b$

Two things go wrong with point-to-line distance. (1) The line must be written as

ax+by+c=0

(everything on one side) before substituting — applying the formula to

y=mx+c

ax+by=c

as-is gives the wrong numerator. (2) The denominator is

\sqrt{a^2+b^2}

, not

a^2+b^2

a+b

. For parallel lines, the

a,b

coefficients of both lines must be made identical first, or

|c_1-c_2|

is meaningless.

Drill 10 more on distance: point-point, point-line, parallel lines

Concept 2 of 3: Section formula and ratios

The point dividing a segment in a given ratio is a weighted average of the endpoints. Run it backwards to find the ratio in which a point (or a line) divides a segment, or forwards to find midpoints and divisions.

Definition

Forward: the point dividing $P(x_1,y_1)$ and $Q(x_2,y_2)$ internally in ratio $m:n$ is $\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$ . The midpoint is the $1:1$ case. External division uses $m:-n$ .
Backwards — the $k:1$ trick. When the ratio is what you are looking for, write it as $k:1$ rather than $m:n$ . You may always do this, because $m:n$ is the same ratio as $\tfrac{m}{n}:1$ — a ratio has only one unknown in it, not two. The dividing point becomes $\left(\dfrac{kx_2+x_1}{k+1},\dfrac{ky_2+y_1}{k+1}\right)$ ; impose whatever the question says about that point and solve the single equation for $k$ . This is the move behind almost every "in what ratio does … divide …" question.

Section formula and midpoint

\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\qquad \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Worked example

Find the point dividing

(1,2)

and

(4,8)

internally in ratio

2:1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q56Moderate

Example 2 · Lines · Distance, Section, and Locus

What is the ratio in which the point

C\!\left(-\dfrac{2}{7}, -\dfrac{20}{7}\right)

divides the line joining the points

A(-2, -2)

and

B(2, -4)

In $m:n$ the weight $m$ multiplies the far endpoint $x_2$ — mind the cross-pairing

For a point dividing

P_1P_2

in ratio

m:n

(with

m

the part nearer

P_2

), the x-coordinate is

\dfrac{mx_2+nx_1}{m+n}

: the ratio's first number

m

pairs with the second point's coordinate

x_2

. Writing

\dfrac{mx_1+nx_2}{m+n}

swaps the weights and divides in ratio

n:m

instead — the point lands on the wrong side. Also remember the

m+n

in the denominator.

Drill 9 more on section formula and ratios

Concept 3 of 3: Locus from a condition

A locus is the set of points satisfying a rule. Let the moving point be

(x,y)

, write the geometric condition algebraically, and simplify — the equation that survives is the locus. 'Equidistant from two points' always gives the perpendicular bisector.

Definition

The standard move: set $P=(x,y)$ , translate the stated condition into algebra, and reduce. The equation left standing is the locus.
Equidistant from two points $A,B$ : $PA^2=PB^2$ ⇒ the perpendicular bisector of $AB$ , a straight line. Equidistant from two lines ⇒ the pair of angle bisectors between them.
From a varying parameter — the other common species, and the one students miss. Sometimes the moving point is not defined by a distance at all: something in the figure varies, and the point rides along with it. Write both $x$ and $y$ in terms of that parameter, then eliminate the parameter between the two equations. The relation between $x$ and $y$ that survives is the locus — and the parameter must not appear in your answer.

Worked example

Find the locus of points equidistant from

A(1,2)

and

B(3,4)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2022 · Q89Easy

Example 3 · Lines · Distance, Section, and Locus

What is the equation of the locus of the mid-point of the line segment obtained by cutting the line

x+y=p

(where

p

is a real number) by the coordinate axes?

Equidistant from two points gives a line (perp. bisector); equidistant from two lines gives the angle bisectors

Don't conflate the two 'equidistant' loci. Equidistant from two points

A,B

is the perpendicular bisector of

AB

— a single straight line. Equidistant from two lines is the pair of angle bisectors between them. Setting up

PA^2=PB^2

(distances to points) when the question means distance-to-lines, or vice versa, produces the wrong locus entirely.

Drill 5 more on locus from a condition

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Distance: point-point, point-line, parallel lines
Distance formulas
$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\qquad \dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\qquad \dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}$
Section formula and ratios
Section formula and midpoint
$\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\qquad \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Watch out for (3)

Put the line in $ax+by+c=0$ form first, and divide by $\sqrt{a^2+b^2}$ — not by $a+b$
→ Distance: point-point, point-line, parallel lines
In $m:n$ the weight $m$ multiplies the far endpoint $x_2$ — mind the cross-pairing
→ Section formula and ratios
Equidistant from two points gives a line (perp. bisector); equidistant from two lines gives the angle bisectors
→ Locus from a condition

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