Introduction for descartes Cartesian coordinate system:
In Cartesian coordinates system,name Cartesian is derived from the name of french mathematician and philosopher René Descartes (Latin: Cartesius) from the Euclidean geometry.Cartesian coordinate system which consist of analytic geometry, calculus, and cartography. Descartes introduces the new point or plane on a surface by two intersecting axes.In analytical geometry solving we have point coordinates with the formulas for finding the required parameters and in this section we have the points and the problems.
Descartes Cartesian Coordinates System:
In descartes Cartesian coordinates system we have the point geometry and the formulas for finding the parameters.In a Cartesian coordinates we have many points such as collinear point coordinates, equidistant point coordinates and mid point coordinates.
Collinear points:
Collinear points are a point when three or more points lies on same line.
Midpoint:
Mid point is a halfway point where the line segment divided into two equal parts.
Equidistant point:
In a line segment a point is equal length from other points which are in congruent then the point are equidistant point.I like to share this second order differential equation with you all through my article.
Problems in Descartes Cartesian Coordinates System:
Example 1:
Find the Cartesian coordinates distance between the points A(2,6) and B (2,3).
Solution:
Let assume "d" be the distance between A and B. (x1,y1)= (2,6), (x2,y2)= (2,3)
Then d (A, B) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
= `sqrt((2-2)^2 +(3-6)^2)`
= `sqrt(0^2+(-3)^2)`
= `sqrt(0+9)`
=`sqrt9`
=3
Example 2:
Determine the mid point coordinates of a line segment joining given points A(1,8) and B(-1,-5)
Solution:
The required mid point is
Formula = ` ((x_1+x_2)/(2))`,`((y_1+ y_2)/(2))` here, (x1, y1) = (1,8),(x2, y2) = (-1,-4)
= `((1-1)/(2))``((8-4)/(2)) `
= ` (0/2)`` (4/2)`
= (0,2)
Example 3:
Find the centroid coordinates of a triangle whose vertices's points are given (0, -3), (0,6) and (3,3).
Solution:
(x1 y1) = (0,-3), (x2 y2) = (0,6), (x3 y3) =(3,3)
Formula for centroid = ` ((x_1+ x_2+ x_3)/(3))` ,`(( y_1+ y_2+ y_3)/(3))`
The centroid of the triangle = `((0+0+3)/3)`,`((-3+6+3)/3)`
=`(3/3),(6/3)`
=(1,2)
In Cartesian coordinates system,name Cartesian is derived from the name of french mathematician and philosopher René Descartes (Latin: Cartesius) from the Euclidean geometry.Cartesian coordinate system which consist of analytic geometry, calculus, and cartography. Descartes introduces the new point or plane on a surface by two intersecting axes.In analytical geometry solving we have point coordinates with the formulas for finding the required parameters and in this section we have the points and the problems.
Descartes Cartesian Coordinates System:
In descartes Cartesian coordinates system we have the point geometry and the formulas for finding the parameters.In a Cartesian coordinates we have many points such as collinear point coordinates, equidistant point coordinates and mid point coordinates.
Collinear points:
Collinear points are a point when three or more points lies on same line.
Midpoint:
Mid point is a halfway point where the line segment divided into two equal parts.
Equidistant point:
In a line segment a point is equal length from other points which are in congruent then the point are equidistant point.I like to share this second order differential equation with you all through my article.
Problems in Descartes Cartesian Coordinates System:
Example 1:
Find the Cartesian coordinates distance between the points A(2,6) and B (2,3).
Solution:
Let assume "d" be the distance between A and B. (x1,y1)= (2,6), (x2,y2)= (2,3)
Then d (A, B) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
= `sqrt((2-2)^2 +(3-6)^2)`
= `sqrt(0^2+(-3)^2)`
= `sqrt(0+9)`
=`sqrt9`
=3
Example 2:
Determine the mid point coordinates of a line segment joining given points A(1,8) and B(-1,-5)
Solution:
The required mid point is
Formula = ` ((x_1+x_2)/(2))`,`((y_1+ y_2)/(2))` here, (x1, y1) = (1,8),(x2, y2) = (-1,-4)
= `((1-1)/(2))``((8-4)/(2)) `
= ` (0/2)`` (4/2)`
= (0,2)
Example 3:
Find the centroid coordinates of a triangle whose vertices's points are given (0, -3), (0,6) and (3,3).
Solution:
(x1 y1) = (0,-3), (x2 y2) = (0,6), (x3 y3) =(3,3)
Formula for centroid = ` ((x_1+ x_2+ x_3)/(3))` ,`(( y_1+ y_2+ y_3)/(3))`
The centroid of the triangle = `((0+0+3)/3)`,`((-3+6+3)/3)`
=`(3/3),(6/3)`
=(1,2)
