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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 17^{th} 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.
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 1^{st}, 2^{nd}, 3^{rd} and 4^{th} 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).
Distance between Two points:
Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) 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 = x_{2} – x_{1}
BM = BD – MD = BD – AC = y_{2} – y_{1}
Then, from the right angled triangle AMB, we get,
AB² = AM² + BM²
or, AB² = (x_{2} – x_{1})^{2}+ (y_{2} – y_{1})^{2}
So, AB = \(\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)
Hence, the formula to find the distance between any two points is:
Distance (d) = \(\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)
Example:
Find the distance between the points (1, 3) and (2, 4).
Solution:
Given, (x_{1}, y_{1}) = (1, 3) and (x_{2}, y_{2}) = (2, 4)
We know,
(d) = \(\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)
So, (d) = \(\sqrt {(2-1)^{2}+(4-3)^{2}}\)
= \(\sqrt {1^{2} + 1^{2}}\)
= \(\sqrt {2}\)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
Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) 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 = x_{2} – x_{1}
BR = BQ – RQ = BQ – AP = y_{2} – y_{1}
From the right angled triangle ARB,
tan = \(\frac {QL}{PL}\)= \(\frac {y_{2}–y_{1}}{x_{2}–x_{1}}\)= m
Example:
Find the slope of the line joining (3, -2) and (-1, 1).
Solution:
Given, (x_{1}, y_{1}) = (3, -2) and (x_{2}, y_{2}) = (-1, 1)
We know,
m = \(\frac {y_{2} - y_{1}}{x_{2} - x_{1}}\)
= \(\frac {1 + 2}{-1 - 3}\)
= - \(\frac {3}{4}\)
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.
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(x_{1}, y_{1}) and B(x_{2}, y_{2}) in the ratio m_{1}:m_{2}.
Case I: Internal Division
Draw the perpendiculars AF, CG and BH on the x-axis, meeting BH in E and FA produced in D.
Then, from the triangles CAD and EBC, we get;
\(\frac {DC}{CE}\) = \(\frac {AC}{CB}\)
or, \(\frac {FG}{GH}\) = \(\frac {m_{1}}{m_{2}}\)
or, \(\frac {OG – OF}{OH – OG}\) = \(\frac {m_{1}}{m_{2}}\)
or, \(\frac {x – x_{1}}{x_{2} – x}\) = \(\frac {m_{1}}{m_{2}}\)
X = \(\frac {m_{1}x_{2} + m_{2}x_{1}}{m_{1 }+ m_{2}}\)
Similarly,
Taking \(\frac {DA}{BE}\) = \(\frac {AC}{CB}\) = \(\frac {m_{1}}{m_{2}}\) , we get,
Y = \(\frac {m_{1}y_{2} + m_{2}y_{1}}{m_{1 }+ m_{2}}\)
Thus, the co-ordinates of C are ( \(\frac {m_{1}x_{2} + m_{2}x_{1}}{m_{1 }+ m_{2}}\) , \(\frac {m_{1}y_{2} + m_{2}y_{1}}{m_{1 }+ m_{2}}\) )
Note: if C is the middle point of AB, we get m_{1} = m_{2},
So, x = \(\frac {x_{1} + x_{2}}{2}\) and y = \(\frac {y_{1} + y_{2}}{2}\)
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.
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.
Then, from the triangles DAC and EBC, we get;
\(\frac {DC}{EC}\) = \(\frac {AB}{BC}\) = \(\frac {m_{1}}{m_{2}}\)
or, \(\frac {FH}{GH}\) = \(\frac {m_{1}}{m_{2}}\)
or, \(\frac {OH – OF}{OH – OG}\) = \(\frac {m_{1}}{m_{2}}\)
or, \(\frac {x – x_{1}}{x – x_{2}}\) = \(\frac {m_{1}}{m_{2}}\)
x = \(\frac {m_{1}x_{2} – m_{2}x_{1}}{m_{1 }– m_{2}}\)
Similarly,
Taking \(\frac {DA}{EB}\) = \(\frac {AC}{BC}\) = \(\frac {m_{1}}{m_{2}}\) , we get,
Y = \(\frac {m_{1}y_{2} – m_{2}y_{1}}{m_{1 }– m_{2}}\)
Thus, the co-ordinates of B are ( \(\frac {m_{1}x_{2} – m_{2}x_{1}}{m_{1 }– m_{2}}\) , \(\frac {m_{1}y_{2} – m_{2}y_{1}}{m_{1 }– m_{2}}\) )
Note: The external division co-ordinates are obtained from those of internal division changing m_{2} to – m_{2}.
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,
(x_{1}, y_{1}) = (2, 5)
(x_{2}, y_{2}) = (6, 15)
m_{1}:m_{2} = 2:3 internally,
We have,
x = \(\frac {m_{1} x_{2} + m_{2 }x_{1}}{m_{1 }+ m_{2}}\)
= \(\frac {2 x 6 + 3 x 2}{2 + 3}\)
= \(\frac {18}{5}\)
y = \(\frac {m_{1}y_{2} + m_{2}y_{1}}{m_{1 }+ m_{2}}\)
= \(\frac {2 x 15 + 3 x 5}{2 + 3}\)
=\(\frac {45}{5}\) = 9
Therefore, the required co-ordinates of the point are (18/5, 9).
Note: if the points were externally divided, we’d get,
x = \(\frac {m_{1}x_{2} – m_{2}x_{1}}{m_{1 }– m_{2}}\)
= \(\frac {2 x 6 – 3 x 2}{2 – 3}\)
= \(\frac {6}{– 1}\) = – 6
y = \(\frac {m_{1}y_{2} – m_{2}y_{1}}{m_{1 }– m_{2}}\)
= \(\frac {2 x 15 – 3 x 5}{2 – 3}\)
= \(\frac {15}{– 1}\) = – 15
Hence, the required co-ordinates of the point are (– 6, – 15)
Notes:
(Tamang, Pant, & G.C, 2016)
Bibliography
Tamang, G., Pant, N., & G.C, P. B. (2016). Business Mathematics. Putalisadak: Asmita Publication.
Co-ordinates geometry:
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