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Straight line can be defined as a simple geometrical figure. even though it is simple it is one of the important concept in geometryconcepts . We come across the concept of straight lines in our day to day experience. Sides of all polygons are straight lines. The path of the shortest distance between two points is a straight line.

The graph of a linear function is a straight line. In fact the name ‘linear’ is used in the name because of this fact. The function of a straight line is expressed in several forms depending upon the context. The orientation of a straight line in a grid depicts the nature of a linear function.

*A horizontal line represents a function but not a vertical line.*

As a special case of ratio, the coordinates of the mid point of a line segment with end points as (x_{1}, y_{1}) and (x_{2}, y_{2}) are given by,

x = ½( x_{1} + x_{2}) and y = ½( y_{1} + y_{2})

The area of the triangle ABC

A = Area of trapezoid DABE + Area of trapezoid EBCF - Area of trapezoid DACF

= ½ (y_{2 }+ y_{1})(x_{2} – x_{1}) + ½ (y_{3 }+ y_{2})(x_{3} – x_{2}) - ½ (y_{3}+ y_{1})(x_{3} – x_{1})

On simplification, A = ½ [ x_{1} (y_{3 }– y_{2) }+_{ }x_{2} (y_{1 }– y_{3) }+ x_{3}(y_{2 }– y_{1})]

What happens when the points A, B and C are in the same line (collinear) ?

In such a case A = 0 or, x_{1} (y_{3 }– y_{2}) + x_{2} (y_{1 }– y_{3}) + x_{3}(y_{2 }– y_{1}) = 0

*This brings an important concept that three points with coordinates (x _{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) will be *

* x _{1} (y_{3 }– y_{2}) + x_{2} (y_{1 }– y_{3}) +_{ } x_{3}(y_{2 }– y_{1}) = 0*

Example 1

The coordinates of points A and B are (-1, -3) and (2, 1) respectively. What is the distance between A and B?

Solution

As per the distance formula,

AB^{2 }= [(2) –(-1)]^{2} + [(1) – (-3)]^{2} = (3^{2} + 4^{2}) = (9 + 16) = 25

Hence AB = 5

The distance between the points A and B is 5 units.

Example 2

The coordinates of points P and Q are (4, 3) and (6, 7) respectively. If R is the midpoint of the straight line joining P and Q, what are the coordinates of R?

Solution

The coordinates of the mid point of a line segment with end points as (x_{1}, y_{1}) and (x_{2}, y_{2})

are given by, x = ½( x_{1} + x_{2}) and y = ½( y_{1} + y_{2})

x_{1} = 4, x_{2 }= 6, y_{1} = 3 and y_{2 }= 7

Therefore the coordinate of the mid point R is,

x = ½( x_{1} + x_{2}) = ½( 4 + 6) = 5 and y = ½( y_{1} + y_{2}) = = ½( 3 + 7) = 5

Problem 1

The coordinates of points P, Q and R are (2, -1), (4, 3) and (6, 7) respectively. Show that the points P, Q and R are collinear.

Solution

Three points with coordinates (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) will be collinear if,

x_{1} (y_{3 }– y_{2}) + x_{2} (y_{1 }– y_{3}) +_{ } x_{3}(y_{2 }– y_{1}) = 0

x_{1} = 2, x_{2 }= 4, x_{3} = 6, y_{1} = -1, y_{2 }= 3 and y_{3 }= 7

x_{1} (y_{3 }– y_{2}) + x_{2} (y_{1 }– y_{3}) +_{ } x_{3}(y_{2 }– y_{1}) = 2(7_{ }– 3)_{ }+ 4 (-1_{ }– 7) +_{ } 6[(3_{ }– (-1)]

= 2(4) + 4(-8) + 6(4)

= 8 – 32 + 24 = 0

Hence the points P, Q and R collinear.

Problem 2

Prove that the line joining any point on y-axis to any point on x-axis always forms a right triangle.

Solution

Assume that O is the origin, A is any point on y-axis and B be any point on x-axis.

The coordinates of A will be (0, y) and of B will be (x, 0).

Using distance formula,

OA^{2} = (0 – 0)^{2} + (y – 0)^{2} = y^{2}

OB^{2} = (x – 0)^{2} + (0 – 0)^{2} = x^{2}

AB^{2} = (0 – x)^{2} + (y – 0)^{2} = x^{2 }+ y^{2}

Therefore, AB^{2} = OA^{2} + OB^{2}

Hence the triangle AOB is a right triangle.