Note on Distance Formula, Section Formula

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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.

source:www.sciencehq.com
source:www.sciencehq.com

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
formula for internal division
Internal division formula Source:www.onlinemath4all.com

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
External division formula
External division formula Source:www.onlinemath4all.com

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
Mid-point Formula
Mid-point Formula Source:www.studyblue.com

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
Centroid
Centroid
Source:math.tutorvista.com

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 (\(\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3}\))

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 

  

   

.

Very Short Questions

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 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 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 , 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, 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.

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

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trisection


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tell me about tirsection


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Ajk

Could you provide some examples for locus's problems pleasant.


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