Īccording to Cajori (1906, page 185), “The term abscissa occurs for the first time in a Latin work of 1659, written by Stefano degli Angeli (1623-1697), a professor of mathematics in Rome.” A footnote attributes this information to Moritz Cantor.Ībscissa was used in Latin by Gottfried Wilhelm Leibniz (1646-1716) in “De linea ex lineis numero infinitis ordinatim ductis inter se concurrentibus formata.,” Acta Eruditorum 11 1692, 168-171 (Leibniz, Mathematische Schriften, Abth. Cavalieri in his Indivisibles, by Stefano degli Angeli (1623-1697), a professor of mathematics in Rome, and by others.”Ībscissa is found in Latin in 1656 in Exercitationum Mathematicarum by Frans van Schooten. In the more general sense of a ‘distance’ it was used earlier by B. The technical use of ‘abscissa’ is observed in the eighteenth century by C. Īccording to Cajori (1919, page 175), “The words ‘abscissa’ and ‘ordinate’ were not used by Descartes. Join A, B, C and D ABCD is the required trapezium.From the "Earliest Known Uses of Some of the Words of Mathematics" pages. Similarly plot the point B, C and D taking abscissa as – 2, –2 and 4 and ordinates as 3, – 5, and –7 respectively. Solution: Plot the point A taking its abscissa as 4 and ordinate as 6. PQRS is the required rectangle.Įxample 6: Draw a trapezium ABCD in which vertices A, B, C and D are (4, 6), (–2, 3), (–2, –5) and Similarly, plot the points Q, R and S taking abscissa as –5, –5 and 1 and ordinates as 4, – 3 and –3 respectively. Solution: Plot the point P by taking its abscissa 1 and ordinate – 4. This is the required triangle.Įxample 5: Draw a rectangle PQRS in which vertices P, Q, R and S are (1, 4), (–5, 4), (–5, –3) and Similarly, plot points B and C taking abscissa 2 and –2 and ordinates – 2 and 2 respectively. Solution: Plot the point A by taking its abscissa O and ordinate = 2. (iii) Co-ordinates of point P = (abscissa, ordinate)Įxample 3: Write down the (i) abscissa (ii) ordinate (iii) Co-ordinates of P, Q, R and S as given in the figure.Įxample 4: Draw a triangle ABC where vertices A, B and C are (0, 2), (2, – 2), and (–2, 2) respectively. (ii) Ordinate of the point P = MP = ON = b Solution: (i) Abscissa of the point P = – NP = –OM = – a (i) Abscissa (ii) ordinate (iii) Co-ordinates of point P given in the following figure. (iii) Co-ordinates of the point P = (Abscissa, ordinate) = (3, 4) Solution: (i) Abscissa = PN = OM = 3 units In the fourth quadrant, for a point, the abscissa is positive and the ordinate is negative.Ĭartesian Coordinate System Example Problems With SolutionsĮxample 1: From the adjoining figure find.In the third quadrant, for a point, both abscissa and ordinate are negative.In the second quadrant, for a point, abscissa is negative and ordinate is positive.In the first quadrant, both co-ordiantes i.e., abscissa and ordinate of a point are positive.These axes are called the co-ordinate axes.Ī quadrant is 1/4 part of a plane divided by co-ordinate axes. The plane is called the cartesian plane or the coordinate plane or the xy-plane. So, the plane consists of axes and quadrants. The axes divide the plane into four parts. The distance of the point P from x-axis is called its ordinate. The distance of the point P from y-axis is called its abscissa. Therefore, the coordinates of origin are (0, 0). It has zero distance from both the axes so that its abscissa and ordinate are both zero. It is point O of intersection of the axes of co-ordinates. In the figure OX and OY are called as x-axis and y-axis respectively and both together are known as axes of co-ordinates. OM (or NP) and ON (or MP) are called the x-coordinate (or abscissa) and y-coordinate (or ordinate) of the point P respectively. Let O be the fixed point called the origin and XOX’ and YOY’, the two perpendicular lines through O, called Cartesian or Rectangular co-ordinates axes.ĭraw PM and PN perpendiculars on OX and OY respectively. In Cartesian co-ordinates the position of a point P is determined by knowing the distances from two perpendicular lines passing through the fixed point.
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