LEAVE A COMMENT FOR US

Home > Algebra > Coordinate Geometry

Coordinate Geometry

Coordinate geometry is a mathematical technique that allows any linear function to be represented by a pair of numbers or coordinates. It uses algebraic methods to solve geometrical problems.

Every linear function or equation can be represented by y = mx + c where m and c are numerical values for a straight line.

 

Distance Formula

In a number plane, if a point A is represented by (x1, y1) then x1 and y1 are called the coordinates the point A. Now, let us consider point A (x1, y1) and point B (x2, y2). We can find the distance between the points A and B using Pythagoras rule which states that

AB= AC+ BC2   distance_formula

AB2(x_2~-~x_1)^2~+~(y_2~-~y_1)^2

AB~=~sqrt{(x_2~-~x_1)^2~+~(y_2~-~y_1)^2}

This is called distance formula.

The distance AB between A (x1, y1) and B (x2, y2) is give by

AB~=~d~=~sqrt{(x_2~-~x_1)^2~+~(y_2~-~y_1)^2}

Now let us look at a few examples.

Example 1: Find the distance between (3, 5) and (4, 8).

Here x= 3, y= 5, x= 4, y= 8. Applying the distance formula –

d~=~sqrt{(x_2~-~x_1)^2~+~(y_2~-~y_1)^2}

~=~sqrt{(4~-~3)^2~+~(8~-~5)^2}

~=~sqrt{1^2~+~3^2}

= sqrt{10} units

 

Example 2: Which is closer to (4, 0) – the point A (1, 3) or B (-2, 2)?

With A (1, 3), x= 1, y= 3 and with B (-2, 2), x= 4, y= 0

Distance of given point (4, 0) from A is

d_1~=~sqrt{(4~-~1)^2~+~(0~-~3)^2}

= sqrt{3^2~+~(-3)^2}

= 3sqrt{2} units

=  4.2426 units

 

Distance of given point (4,0) from B is x= -2, y= 2

d_2~=~sqrt{(4-(-2))^2~+~(0~-~2)^2}

= sqrt{6^2~+~2^2}

= sqrt{36~+~4}

= sqrt{40}

= 2sqrt{10}

= 6.3245 units

Since d1 is lesser than d2, the point A is closer to (4, 0) than point B.

Example 3: Prove that the triangle ABC is an isosceles triangle

Given A (2, 0), B (-2, 0), and C (-0, +2).

Distance AB~=~d~=~sqrt{(x_2~-~x_1)^2~+~(y_2~-~y_1)^2}

A(2, 0) : x1– = 2,  y= 0

B(-2, 0) : x2– = -2,  y= 0

AB = sqrt{(-2-2)^2~+~(0)^2}

  =  sqrt{(-4)^2}  =  4

Similarly we can find distances AC and BC.

To get AC –

A(2, 0) : x1– = 2,  y= 0

C(-0, +2) : x2– = -0,  y= +2

AC  =  ~sqrt{(-0~-~2)^2~+~(2~-~0)^2}

~=~sqrt{(2)^2~+~(2)^2}

 =  sqrt{8}~~=~2sqrt{2} units

To get BC –

B(-2, 0) : x1– = -2,  y= 0

C(-0, +2) : x2– = -0,  y= +2

BC  =  sqrt{(-0+2)^2~+~(+2-0)^2}~~~=~sqrt{(2)^2~+~(2)^2}

=  2sqrt{2} units

In an isosceles triangle, two sides are of equal length. In triangle ABC, sides AC and BC have equal length (2sqrt{2} units). Hence triangle ABC is an isosceles triangle.

Example 4: A is the point (-2, 7) and B (5, -2). M is halfway between A and B. How far is M from A?

distance_formula_for_line_segment

x coordinate of M is halfway between x co-ordinates of A and B

x~= ~{-2~+~5}/2~=~~3/2

and y coordinate of M is halfway between y co-ordinates of A and B

y~= ~{7~-~2}/2~=~~5/2

Therefore the coordinates of the midpoint is M(3/2,~5/2)

Distance of M(3/2,~5/2) from A (-2, 7) is

d  =  sqrt{(-2~-~3/2)^2~+~(7~-~5/2)^2}

  =  sqrt{(-7/2)^2~+~(9/2)^2}

  =  sqrt{49/4~+~81/4}

  =  sqrt{130/4}

  =  sqrt{65/2}  units