Analytical and Problem Solving

Introduction for analytical geometry and problem solving

In analytic geometry, the plane is given a coordinate system, by which every point has a pair of real number coordinates. The most common coordinate system to use is the Cartesian coordinate system, where each point has an x-coordinate representing its horizontal position, and a y-coordinate representing its vertical position. (Source from Wikipedia). Here we are going to learn, solving problems in analytical geometry.

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Basic principles used in analytical geometry and solving problems


Here we are going to see some basic principles for solving problems in analytical geometry.

Coordinates in analytical geometry

Coordinate system is a system uses set of numbers and point in a XY plane to represent a geometrical shape. For example, to indicate a line in a XY plane we need two points.

Equations of lines and curves in analytical geometry

We use equations to represent lines and curves to denote them in a coordinate plane. For example, the equation of a line is given by, y = mx + b and this type of line equations are first order equations. The equations of curves given by second order equations. For example, the equation of a circle is given by, `x^2``y^2` = `r^2` , and the equation of a parabola is given by `y^2` = 2ax.

Distance between two points in analytical geometry

Let (x1, y1) and (x2, y2) be two points in a coordinate plane. The distance between these two points are given by,

d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

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Example problems in analytical geometry


Here we are going to learn solving problems in analytical geometry.

Problem 1

Find the slope of the line containing the points (2, 3) and (-1, 5).

Solution

Slope of the line = `(y2 - y1)/(x2 - x1)`

= `(5 - 3)/(-1 - 2)`

= `2/-3`

So the slope of the line containing (2, 3) and (-1, 5) is -`2/3`

Problem 2

Find the distance between the points (2, 3) and (-1, 5).

Solution

The distance between the given points = `sqrt((-1-2)^2 + (5-3)^2)`

= `sqrt((-3)^2 + 2^2)`

= `sqrt(9 + 4)`

= `sqrt(13)`

So the distance between the points (2, 3) and (-1, 5) is `sqrt(13)`