## Distance Formula, Section Formula

Subject: Optional Mathematics

#### Overview

The distance formula is used to determine the distance, d, between two points. If the coordinates of the two points are (x1, y1) and (x2, y2), the distance equals the square root of x2 − x1 squared + y2 − y1 squared. The distance formula is derived by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. The hypotenuse of the triangle is the distance between the two points.

#### Distance Formula

The calculation of a distance between any two points on a plane surface can be done.

##### Derivation of distance formula

Let P(x1,y1) and Q(x2,x2) be two given points in the coordinate plane. Draw perpendicular PL from P to the x-axis. Then , OL = x1 and PL = y1. Draw perpendicular QM from Q to the x-axis.

Then, OM = x2and QM =y2

Again, draw perpendicular PR from P to the line segment QM.

Then, PR = LM = OM-OL =x2-x1

QR = QM - NM = QM - PL = y2-y1

From right-angled ΔPRQ, by Pythagoras theorem,

PQ2 = PR2+ QR2 = (x2-x1)2 + (y2-y1)2

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

(Only square root is to be taken because PQ being the distance beyween two points is positive)

The distance of the point (x,y) from the origin (0,0) =$\sqrt{(x-0)^2 + (y-0)^2}$ = $\sqrt{x^2+y^2}$

Remarks

1. The formula remains the same if the point P (x1,y1) and Q(x2,y2) are taken in different quadrants.
2. If a point lies on the x-axis, its ordinate is zero. Therefore, any point on x-axis can be taken as (x,0).
3. If a point lies on Y-axis, its abscissa is zero. therefore, any point on y-axis can be taken as (0,y).
4. To prove that a quadrilateral is a:
a) rhombus, show that all sides are equal.
b) square, show that all are equal and diagonals are also equal.
c) parallelogram, show that opposite sides are equal.
d) rectangle, show that opposite sides are equal and diagonals are also equal.
5. To prove that a triangle is a :
a) scalene, show that none of the sides are equal.
b) isosceles, show that two sides are equal.
c) equilateral, show that all sides are equal.
d) right-angled triangle, show that square on one side equals to the sum of squares of the other two sides.

#### Section Formula

##### Formula for internal division

To find the coordinates of the point which divides internally the line joining two points (x1,y1) and (x2,y2) in the given ratio m1:m2.

Let A(x, y) and B(x1, y1) be two given points. Let the point P (x2, y2) divide the line joining AB internally in the ratiom1: m2

Then AP : PB = m1: m2

Draw perpendiculars AL, PN and BM from A, P and B respectively to the x-axis. Then, OL = x1, ON = x, OM = x2, AL = y1, PN = y and BM = y2. Again draw respectively AQ and PR from A and P to the line segments PN and BM respectively.

The AQ =LN = ON -OL = x - x1 PR = NM = OM - ON = x2-x
PQ = PN - QN = PN -AL = y - y1 BR = BM - RM = BM - PN = y2 - y

In Δs PQA and BRA, ∠PQA = ∠ BRP, ∠QAP = ∠RPB and ∠APQ = ∠PBR

So, Δs PQA and BRP are similar.

Then,
$\frac{AP}{PB}$
= $\frac{AQ}{PR}$
= $\frac{QP}{RB}$
= $\frac{m_1}{m_2}$
= $\frac{x-x_1}{x_2-x}$
= $\frac{y-y_1}{y_2y}$

Taking first two relation, we get,
or, $\frac{m_1}{m_2}$ = $\frac{x-x_1}{x_2-x}$
or, m2x - m2x1 = m1x2-m1x
or, m2x + m1x = m1x2 + m2x1
or, x(m1+m2) = m1x2+m2x1
or, x = $\frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Similarly, from the relations

$\frac{m_1}{m_2}$ = $\frac{y-y_1}{y_2-y}$
we get
y = $\frac{m_1y_2 + m_2y_1}{m_1 + m_2}$
∴ The coordinates of P are $\frac{m_1x_2 + m_2x_1}{m_1 +m_2}$,$\frac{m_1y_2 + m_2y_1}{m_1+m_2}$

##### Formula for external division

If the point P(x, y) divides AB externally in the ratio of m1:m2 then the divided segment BP is measured in opposite direction and hence m2 is taken as negative.
$\therefore$ The section formulae for external division is,
(x, y) = ($\frac{m_1x_2 - m_2x_1}{m_1 - m_2}$),($\frac{m_1y_2 - m_2y_1}{m_1 - m_2}$)

##### Mid-point Formula

In special case, the midpoint formulae is also used'.
m1:m2 = 1:1 i.e. m1 = m2
$\therefore$ x = $\frac{x_1 + x_2}{2}$ and y = $\frac{y_1 + y_2}{2}$
Thus, co-ordinates P(x, y) are P($\frac{x_1 + x_2}{2}$,$\frac{y_1 + y_2}{2}$) which is called mid-point formulae.

##### K-formula

If the point P(x, y) divides the joining two points A(x1, y1) and B(x2, y2) in the ratio k : 1, then
x = $\frac{k . x_2 + 1 . x_1}{k + 1}$ = $\frac{kx_2 + x_1}{k + 1}$
and, y = $\frac{k . y_2 + 1 . y_1}{k + 1}$ = $\frac{ky_2 + y_1}{k + 1}$

∴ Coordinate of P are ($\frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1}$)
For the problems in which it is required to find the ratio when a given point divides the join of two given points, it is convenient to take the ratio k : 1.

##### Centroid formula

Let P(x1, y1), Q(x2, y2) and R(x3, y3) are the vertices of a triangle PQR. Let S, T and U are the mid - points of sides QR, RP and PQ respectively. Then, PS, QT and RU are called medians of the triangle PQR. If these medians intersect each other at a point N, then N is called the centroid of the triangle PQR.
Since S is the middle point of the side QR, then its coordinates are ($\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}$)
From plane geometry, we know that the centroid of a triangle divides the median in the ratio 2 : 1. Now, if coordinates of N are (x, y), then by section formula.
x = $\frac{2 . \frac{x_2 + x_3}{2} + 1 . x_1}{2 + 1}$ = $\frac{x_1 + x_2 + x_3}{3}$ and y = $\frac{2 . \frac{y_2 + y_3}{2} + 1 . y_1}{2 + 1}$ = $\frac{y_1 + y_2 + y_3}{3}$

Hence, coordinates of N are ($\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3}$)

##### Things to remember

The formula remains the same if the points P (x1,y1) and Q (x2,y2) are taken in different quadrants.

If a point lies on x-axis, its ordinate is zero. Therefore, any point on x-axis can be taken as (x,0).

If a point lies on y-axis, its abscissa is zero. Therefore, any point on y-axis can be taken as (0,y).

To prove that a quadrilateral is a

1) rhombus, show that all sides are equal.

2) square, show that all sides are equal and diagonals are equal.

3) parallelogram, show that opposite sides are equal.

4) rectangle, show that opposite sides are equal and diagonals are equal.

To prove that a triangle is a

1) scalene, show that none of the sides are equal.

2) isosceles, show that two sides are equal.

3) equilateral, show that all sides are equal.

4) right-angled triangle, show that square on one side equals

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Distance Formula, Section Formula ##### Distance formula | Analytic geometry | Geometry | Khan Academy ##### Using The Distance Formula Or Pythagorean Theorem To Find The Distance Between Two Points

Let , the given points be
(x1 , y1) = (4 , 3) amd (x2 , y2) = (5 ,6)
ratio m1:m2 = 2 : 3 or m1 = 2 and m2 = 3
x = $\frac{m1x2+m2x1}{m1+m2}$ and y = $\frac{m1y2 + m2y1}{m1 + m2}$

$\therefore$ x = $\frac{2.5+3.4}{2+3}$ and y = $\frac{2.6 + 3.3}{2 + 3}$

= $\frac{10 + 12}{5}$ = $\frac{22}{5}$ and y = $\frac{12+9}{5}$ = $\frac{21}{5}$
$\therefore$ (x , y) = $\frac{22}{5}$ , $\frac{21}{5}$ Ans.

Let the fiven points be

(x1 y1)=(4,3) and (x2 y2)=(5,6)

ratio m1:m2=2:3 or m1=2 and m2=3

Using section formula

x=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$ and y=$\frac{m_1y_2+m_2y_1}{m_1+m_2}$

x=$\frac{2.5+3.4}{2+3}$ and y=$\frac{2.6+3.3}{2+3}$

=$\frac{10+12}{5}$ =$\frac{22}{5}$ and y=$\frac{12+9}{5}$ =$\frac{21}{5}$

∴(x,y)=($\frac{22}{5}$,$\frac{21}{5}$).Ans.

Let the fiven points be

(x1 y1)=(-3,-4) and (x2 y2)=(-8,7)

ratio m1:m2=3:4 or m1=3 and m2=4

Using section formula

x=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$ and y=$\frac{m_1y_2+m_2y_1}{m_1+m_2}$

x=$\frac{3×-8+4×-3}{3+4}$ ,y=$\frac{3×7+4×-4}{3+4}$

=$\frac{-24-12}{7}$ , y=$\frac{21-16}{7}$

∴(x,y)=($\frac{-36}{7}$,$\frac{5}{7}$).Ans.

Let the fiven points be

(x1 y1)=(2,-3) and (x2 y2)=(-6,5)

ratio m1:m2=4:5

Using section formula

x=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$ and y=$\frac{m_1y_2+m_2y_1}{m_1+m_2}$

x=$\frac{4 ×-6+5×2}{4+5}$ ,y=$\frac{4 × 5 +5 ×-3}{4+5}$

=$\frac{-24+10}{9}$ , y=$\frac{20-15}{9}$

=$\frac{-14}{9}$, $\frac{5}{9}$

∴(x,y)=($\frac{-14}{9}$,$\frac{5}{9}$).Ans.

Let the given points be

(x1 y1)=(-3,-4) and (x2 y2)=(-8,7)

ratio m1:m2==3:(-4) or m1=3 and m2=-4

Using section formula

x=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$ and y=$\frac{m_1y_2+m_2y_1}{m_1+m_2}$

x=$\frac{3×-8+(-4)×-3}{3+(-4)}$ ,y=$\frac{3×7+(-4)×-4}{3+(-4)}$

=$\frac{-24+12}{3-4}$ , y=$\frac{21+16}{3-4}$

=$\frac{-12}{-1}$, =$\frac{37}{-1}$

= 12 , =-37

∴(x,y)=(12,-37).Ans.

• Let the given points be

(x1 y1)=(1,-2) and (x2 y2)=(4,7)

Internal ratio m1:m2=1:2

Using formula,

x=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$ and y=$\frac{m_1y_2+m_2y_1}{m_1+m_2}$

or, x=$\frac{1 × 4 + 2×1 }{1+2}$ ,y=$\frac{1 ×7+2×-2}{1+2}$

=$\frac{4+2}{3}$ , =$\frac{7-4}{3}$

=$\frac{6}{3}$, =$\frac{3}{3}$

= 2 , =1

∴(x,y)=The required point=(2,1).Ans.

• Let the given points be

(x1 y1)=(1,-2) and (x2 y2)=(4,7)

Let (x,y) divides externally in the ratio m1:m2=2:3 or m1=2 and m2=3

Using formula,

x=$\frac{m_1x_2-m_2x_1}{m_1-m_2}$ and y=$\frac{m_1y_2-m_2y_1}{m_1-m_2}$

or, x=$\frac{2 × 4 - 3 ×1 }{2-3}$ ,y=$\frac{2 ×7-3×-2}{2-3}$

=$\frac{8-3}{-1}$ , =$\frac{14+6}{-1}$

=$\frac{5}{-1}$, =$\frac{20}{-1}$

= -5 , =-20

∴The required point=(-5,-20).Ans.

Here, Let the given points be (x1,y1)= (4,-5) and (x2,y2) = (6,3) and m1:m2=2:5

Let (x,y) divides the given points internally then

Using formula,

x=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$ and y=$\frac{m_1y_2+m_2y_1}{m_1+m_2}$

x=$\frac{2×6+5×4}{2+5}$ ,y=$\frac{2×3+5×-5}{2+5}$

=$\frac{12+20}{7}$ , y=$\frac{6-25}{7}$

=$\frac{32}{7}$, =$\frac{-19}{7}$

∴The required point =($\frac{32}{7}$,$\frac{-19}{7}$).Ans.

If (x1,y1) divides externally then using formula

x'=$\frac{m_1x_2-m_2x_1}{m_1-m_2}$ and y'=$\frac{m_1y_2-m_2y_1}{m_1-m_2}$

x'=$\frac{2×6-5×4}{2-5}$ ,y'=$\frac{2×3-5 ×-5)}{2-5}$

=$\frac{12-20}{-3}$ , y=$\frac{6+25}{-3}$

=$\frac{-8}{-3}$, =$\frac{31}{-3}$

∴The required point=$\frac{8}{3}$, =$\frac{-31}{3}$ Ans.

Here,

Let the given points be (x1,y1) =A(-3,-6) and (x2,y2) = B(1,-2) If (x,y) be the mid point, then using

x=$\frac{x_1+x_2}{2}$ and y=$\frac{y_1+y_2}{2}$

∴ x =$\frac{-3+1}{2}$ and y=$\frac{-6-2}{2}$

=$\frac{-2}{2}$, =$\frac{-8}{2}$

∴x=-1, =-4

∴(x,y)=(-1,-4).Ans.

Here , given P(x1 , y1) = (2 , 3) and Q(x2 , y2) = (4 , 3)
Using formula , PQ = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}$
Distance , PQ = $\sqrt{(4-2)^{2} + (3-3)^{2}}$ = $\sqrt{2^{2} + 0^{2}}$ = $\sqrt{2^{2}}$ = 2.
PQ = 2 units. Ans.

Here,

Let the mid point of M(1,4) and N(x,y) be (-2,2)

-2=$\frac{1+x}{2}$ and 2=$\frac{4+y}{2}$

or, -4=1+x, 4=4+y

or, -4=-1=x, 4-4=y

or, -5=x, 0=y

∴(x,y)=(-5,0).Ans.

Here given , P(x1, ,y1) = (-1 , 3) and Q(x2 , y2) = (5 , 1)
Using formula , PQ = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}$
Distance PQ = $\sqrt{(5 -(-1)^{2} + (1 -3)^{2}}$ = $\sqrt{6^{2} + 2^{2}}$ = $\sqrt{36 + 4}$
$\therefore$ PQ = $\sqrt{40}$ = $\sqrt{4 \times 10}$ = 2$\sqrt{10}$ Units. Ans.

Here,

Let one end (x1y1)=(4,4) they let other end =(x2y2)

Midpoint (x,y)=(-2,2)

Using formula,

x=$\frac{x_1+x_2}{2}$ and y=$\frac{y_1+y_2}{2}$

-2=$\frac{4+x_2}{2}$, 2=$\frac{y_1+y_2}{2}$

or, -4=4+x2 4=4+y2

or, -4-4=x2 0=y2

or, -8=x2, 0=y2

∴(x2y2)=(-8,0).Ans.

Here given , P(x1 , y1) = (1 , -2) and Q(x2, , y2) = (-2 , 2)
Using formula , PQ = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}$
Distance PQ = $\sqrt{(-1 -3)^{2} + [-1 - (-1)]^{2}}$
= $\sqrt{(-4)^{2} + (-1 = 1)^{2}}$
= $\sqrt{4^{2} + 0^{2}}$ = $\therefore$ PQ = $\sqrt{4^{2}}$ = 4 units. Ans.

Here given , P(x1, y1) = (-6 , 7) and Q(x2 , y2) = (-1 , -5)
Using formula , PQ = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}$
Distance PQ = $\sqrt{[-1 -(-6)]^{2} + (-5 -7)^{2}}$
= $\sqrt{-1 ^{2} + 6^{2} + (-12)^{2}}$ = $\sqrt{5 ^{2} + 144}$
= $\sqrt{25 + 144}$ = $\sqrt{169}$
$\therefore$ PQ = 13 Units Ans.

Here given points,A(2,-4) and B(-3,6)

Let the points on x-axis be P(x,0) which divides AB in the ratio k:1 then,

Using formula y=$\frac{m_1y_2+m_2y_1}{m_1+m_2}$

0=$\frac{k×6+1×-4}{k+1}$

or, 0×(k+1)=6k-4

or, 6k-4=0

or, 6k=4, or k=$\frac{4}{6}$=$\frac{2}{3}$

∴This shows that x-axis divides the line AB internally in the ratio 2:3.Ans.

Here given , P(x1 , y1) = (4 , 3) and Q(x2 , y2) = (3 , -6)
Using formula , PQ = $\sqrt{(x2 - x1)^{2} + (y2 - y1) ^{2}}$
Distance PQ = $\sqrt{(-2 , -4)^{2} + (2 - 3 )^{2}}$ = $\sqrt{(-6)^{2} + (-1)^{2}}$
= $\sqrt{36 + 1}$ = $\sqrt{37}$ Units. Ans.

Here , given P(x1 , y1) = (-2 , 6) and Q(x2 , y2) = (3 , -6)
Using formula , PQ = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}$
Distance PQ = $\sqrt{[ 3 - (-2)^{2} + (-6 -6)^{2}}$
= $\sqrt{(3 + 2)^{2} + (-12)^{2}}$ = $\sqrt{5 ^{2} + 144}$
= $\sqrt{25 + 144}$ = $\sqrt{169}$ = 13 Units. Ans.

Here, Let (3,y) be a point in the line PQ with co-ordinates P(7,-3) and Q(-2,-5) and as it divided PQ in the ratio k:1 then,

Using formula, x=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$

3=$\frac{k×-2+1×7}{k+1}$ or, 3k +3=-2k+7

or, 3k +2k=7-3,or, 5k = 4 ∴k=$\frac{4}{5}$

∴Ratio m1:m2=4:5

Again using y =$\frac{m_1x_2+m_2x_1}{m_1+m_2}$

or, y=$\frac{4 ×-5+5 ×-3}{4+5}$ =$\frac{-20-15}{9}$

=$\frac{-35}{9}$

∴ Requierd point (x,y)=(3,$\frac{-35}{9}$).Ans.

Here, given A(0 , 0) and B(3 , -4)
$\therefore$ AB = $\sqrt{(3 - 0)^{2} + (4 - 0)^{2}}$ = $\sqrt{9 + 16}$ = $\sqrt{25}$ = 5 units.

Here given , A(0 , 0) amd C(-3 , 4)
$\therefore$ AC = $\sqrt{(-3 -0)^{2} + (4 - 0)^{2}}$
= $\sqrt{(-3)^{2} + (4)^{2}}$
= $\sqrt{9 + 16}$
= $\sqrt{25}$ = 5 units. Ans.

Here given , A(0 , 0) and C (-3 , 4)
$\therefore$ AC = $\sqrt{(-3 - 0)^{2}+(4 - 0)^{2} }$
= $\sqrt{(-3)^{2} + (4)^{2}}$ = $\sqrt{9 + 16}$ = $\sqrt{25}$ = 5 units. Ans.

Let the point Q(0,y) on y-axis divides AB in the ratio k:1,then,

Using formula y=$\frac{m_1x_2+m_2x_1}{m_1+m_2}$

0=$\frac{k×-3+1×2}{1+2}$

or, 0=3k+2

or, 3k=2∴k=$\frac{2}{3}$

∴y-axis divides internally in the ratio 2:3.Ans.

Let , ABC be a triangle whose vertices are A(2 , 4) B(6 , 4) , C(6 , 7)
Using distance formula ,
AB = $\sqrt{(6 - 2)^{2} + (4 - 4)^{2}}$ = $\sqrt{4^{2}}$ = 4

BC = $\sqrt{(6 - 6)^{2} + (7 - 4)^{2}}$ = $\sqrt{3^{2}}$ = 3

CA= $\sqrt{(2 -6)^{2} + (4 - 7)^{2}}$ = $\sqrt{16 + 9}$ = $\sqrt{25}$ = 5

Again here ,
AC $^2$ = AB$^2$ + BC$^2$
or , 5$^2$ = 4$^2$ + 3$^2$
$\therefore$ 25 = 25
Since , h$^2$ = p + b$^2$ is satisfied so by pythagoras theorem $\triangle$ ABC is a right angled triangle.

Here,A(-2,1) and B(4,3) be any two points. Using distance formula,we get (AB)2 =(x2-x1)2+(y2-y1)2 =(-2-4)2 + (1-3)2 =(-6)2 +(-2)2 =36+4 =40