Notes on Co-ordinate geometry: Introduction | Grade 12 > Business Math > Co-ordinate Geometry | KULLABS.COM

Co-ordinate geometry: Introduction

Notes, Exercises, Videos, Tests and Things to Remember on Co-ordinate geometry: Introduction

Please scroll down to get to the study materials.

Registration open for Special Scholarships

  • Note
  • Things to remember

Co-ordinate Geometry:

Introduction:

A co-ordinate is referred to an ordered pair of numbers in Geometry. Firstly introduced by a French philosopher and mathematician Rene Descartes (1596-1650) in the 17th century, co-ordinate geometry represents the position of points in the plane and algebraic equations for lines and curves, using an ordered pair of numbers.
These co-ordinates and plane are also referred as Cartesian Co-ordinates and Cartesian Co-ordinate Plane respectively.

Rectangular Co-ordinate System:

With point O as the origin, let XOX’ and YOY’ be two perpendicular lines that are intersected at the origin in the plane. Then, let A be a point and AB be the perpendicular line drawn from A on XOX’.

Here, AB and OB are called the Rectangular Cartesian Co-ordinates and the lines XOX’ and YOY’ are called x-axis and y-axis respectively.

v

Likewise, XOY, YOX’, X’OY’ and Y’OX are the four quadrants which are the divided co-ordinate plane by the axes and are known as the 1st, 2nd, 3rd and 4th quadrants respectively.

The lengths OB and AB are called abscissa (x-co-ordinate) and ordinate (y-co-ordinate) respectively of the point A and the co-ordinates of the point A is denoted by an ordered pair (x, y).

v

Distance between Two points:

v

Let A(x1, y1) and B(x2, y2) be two points. Draw AC and BD as perpendiculars from A and B on the x-axis. Then, draw AM perpendicular from A on BD.

Now,

AM = CD = OD – OC = x2 – x1

BM = BD – MD = BD – AC = y2 – y1

Then, from the right angled triangle AMB, we get,

AB² = AM² + BM²

or, AB² = (x2 – x1)2+ (y2 – y1)2

So, AB = \(\sqrt {(x2-x1)2+(y2-y1)2}\)

Hence, the formula to find the distance between any two points is:

Distance (d) = \(\sqrt {(x2-x1)2+(y2-y1)2}\)

Example:

Find the distance between the points (1, 3) and (2, 4).

Solution:

Given, (x1, y1) = (1, 3) and (x2, y2) = (2, 4)

We know,
(d) = \(\sqrt {(x2-x1)2+(y2-y1)2}\)

So, (d) = \(\sqrt {(2-1)2+(4-3)2}\)

= \(\sqrt {12 + 12}\)

= \(\sqrt {2}\)units

Slope of a Straight line joining Two points:

The slope of a straight line is referred to as the tangent of the angle which is made by the line with the positive x-axis. It is denoted by ‘m’. Therefore, if θ be the angle made, then the slope of the line is tan θ.

m = tanθ

v

Let A(x1, y1) and B(x2, y2) be two points. When joining AB, the line meets x-axis at C such that ∠QCB = θ. Then draw the lines AP and BQ perpendicular to each other on the x-axis. Likewise, draw the perpendicular AR from A on BQ making ∠RAB = θ.

Now,

AR = PQ = OP – OC = x2 – x1

BR = BQ – RQ = BQ – AP = y2 – y1

From the right angled triangle ARB,

tan = \(\frac {QL}{PL}\)= \(\frac {y2–y1}{x2–x1}\)= m

Example:

Find the slope of the line joining (3, -2) and (-1, 1).

Solution:

Given, (x1, y1) = (3, -2) and (x2, y2) = (-1, 1)

We know,
m = \(\frac {y2 - y1}{x2 - x1}\)

= \(\frac {1 + 2}{-1 - 3}\)

= - \(\frac {3}{4}\)

Internal Division:
Let PQ be a line and O be the point. If O lies within PQ then O divides PQ into 2 parts i.e. PO and OQ. In such case, O is said to have divided PQ internally in the ratio PO:OQ.

v

Co-ordinates of a point Dividing a Line Joining Two points in a Given Ratio:

First, let C(x, y) divide a line joining A(x1, y1) and B(x2, y2) in the ratio m1:m2.

Case I: Internal Division

Draw the perpendiculars AF, CG and BH on the x-axis, meeting BH in E and FA produced in D.

v

Then, from the triangles CAD and EBC, we get;

\(\frac {DC}{CE}\) = \(\frac {AC}{CB}\)

or, \(\frac {FG}{GH}\) = \(\frac {m1}{m2}\)

or, \(\frac {OG – OF}{OH – OG}\) = \(\frac {m1}{m2}\)

or, \(\frac {x – x1}{x2 – x}\) = \(\frac {m1}{m2}\)

X = \(\frac {m1x2 + m2x1}{m1 + m2}\)

Similarly,

Taking \(\frac {DA}{BE}\) = \(\frac {AC}{CB}\) = \(\frac {m1}{m2}\) , we get,

Y = \(\frac {m1y2 + m2y1}{m1 + m2}\)

Thus, the co-ordinates of C are ( \(\frac {m1x2 + m2x1}{m1 + m2}\) , \(\frac {m1y2 + m2y1}{m1 + m2}\) )

Note: if C is the middle point of AB, we get m1 = m2,

So, x = \(\frac {x1 + x2}{2}\) and y = \(\frac {y1 + y2}{2}\)

External Division:
Let PQ be a line given and R be a point. Here, if R lies outside the line and if on producing PQ, it meets R, then PQ is said to be divided by R externally in the ratio of PR:RQ i.e. O divides PQ into two parts; PQ and RQ.

v

Case II: External Division

Draw the perpendiculars AF, BG and CH from A, B and C respectively on the x-axis. Then, draw line CD parallel to x-axis while meeting BG produced in E and FA produced in D.

v

Then, from the triangles DAC and EBC, we get;

\(\frac {DC}{EC}\) = \(\frac {AB}{BC}\) = \(\frac {m1}{m2}\)

or, \(\frac {FH}{GH}\) = \(\frac {m1}{m2}\)

or, \(\frac {OH – OF}{OH – OG}\) = \(\frac {m1}{m2}\)

or, \(\frac {x – x1}{x – x2}\) = \(\frac {m1}{m2}\)

x = \(\frac {m1x2 – m2x1}{m1 – m2}\)

Similarly,

Taking \(\frac {DA}{EB}\) = \(\frac {AC}{BC}\) = \(\frac {m1}{m2}\) , we get,

Y = \(\frac {m1y2 – m2y1}{m1 – m2}\)

Thus, the co-ordinates of B are ( \(\frac {m1x2 – m2x1}{m1 – m2}\) , \(\frac {m1y2 – m2y1}{m1 – m2}\) )

Note: The external division co-ordinates are obtained from those of internal division changing m2 to – m2.

Example:

A point internally dividing the line joining the points (2, 5) and (6, 15) has the ratio of 2:3. Find its co-ordinates.

Solution:

Let (x, y) be the point dividing the line joining the points (2, 5) and (6, 15).

Given,

(x1, y1) = (2, 5)

(x2, y2) = (6, 15)

m1:m2 = 2:3 internally,

We have,

x = \(\frac {m1 x2 + m2 x1}{m1 + m2}\)

= \(\frac {2 x 6 + 3 x 2}{2 + 3}\)

= \(\frac {18}{5}\)

y = \(\frac {m1y2 + m2y1}{m1 + m2}\)

= \(\frac {2 x 15 + 3 x 5}{2 + 3}\)

=\(\frac {45}{5}\) = 9

Therefore, the required co-ordinates of the point are (18/5, 9).

Note: if the points were externally divided, we’d get,

x = \(\frac {m1x2 – m2x1}{m1 – m2}\)

= \(\frac {2 x 6 – 3 x 2}{2 – 3}\)

= \(\frac {6}{– 1}\) = – 6

y = \(\frac {m1y2 – m2y1}{m1 – m2}\)

= \(\frac {2 x 15 – 3 x 5}{2 – 3}\)

= \(\frac {15}{– 1}\) = – 15

Hence, the required co-ordinates of the point are (– 6, – 15)

Notes:

  • Parallelogram: Opposite sides are equal or parallel.
  • Rectangle: Opposite sides, as well as length of the diagonals, are equal.
  • Square: With one right angle, all sides are equal.
  • Isosceles triangle: Base angles are equal with any equal two sides.
  • Equilateral triangle: All sides and all angles are equal.
  • Right angled triangle: (Square of hypotenuse) Sum of the squares of the remaining two sides.
  • Rhombus: All sides are equal.
  • Median: It’s a line that joins the vertex of a triangle to the mid-points of its opposite side.

(Tamang, Pant, & G.C, 2016)


Bibliography

Tamang, G., Pant, N., & G.C, P. B. (2016). Business Mathematics. Putalisadak: Asmita Publication.



Co-ordinates geometry:

  • Introduction
  • Rectangular Co-ordinate System
  • Distance between Two points
  • Slope of a Straight line joining two points
  • Co-ordinates of a point dividing a line joining two points in a given ratio; Case I and Case II
.

Questions and Answers

Click on the questions below to reveal the answers

0%

ASK ANY QUESTION ON Co-ordinate geometry: Introduction

No discussion on this note yet. Be first to comment on this note