In this post we will discuss about equations of a line in different form. Every equation is related to other equation and can be derived from one other. In the post I have derived the different forms of equations of a straight line. I have started from an equation which is called two-point form and end with the normal form. normal form is also called perpendicular form. I have given graphs of four equations.
The equation of a line can be found if we know anyone of
The four equations in the graph are
y = 3x + 4, y = x + 7 , y = 2x− 5 and y = −x + 3.
Let A(x1, y1) and B(x2, y2) be two points. Then the four types of equations can be framed as follows:
The equation of a line can be found if we know anyone of
- two points
- one point and slope of the line
- slope of the line and y intercept
- angle made by the perpendicular and its length to the line from the origin
- two points form: (y−y1)/(x−x1) = (y2−y1)/(x2−x1)
- point-slope form: (y−y1) = m (x−x1)
- slope-intercept form: y = mx + c
- normal form: x cos α + y sin α = p
The four equations in the graph are
y = 3x + 4, y = x + 7 , y = 2x− 5 and y = −x + 3.
Let A(x1, y1) and B(x2, y2) be two points. Then the four types of equations can be framed as follows:
- As the slope of the line will be constant. So, if a variable point is (x,y) then
(y−y1)/(x−x1) = (y2−y1)/(x2−x1) - If the value of the slope is m then we can substitute,
(y2−y1)/(x2−x1) = m
and get the equation of the line as
(y−y1)/(x−x1) = m
(y−y1)= m (x−x1) - Expanding (y−y1)=m(x−x1)
y−y1= mx − mx1
y = mx−(mx1− y1)
As [−(mx1 − y1)] is a constant and can be substituted for c
y = mx + c
From the above equation we get y = c when x = 0. Hence, c is the y intercept. - Normal form is found by considering the angle which the perpendicular from the origin to the line makes with the x-axis and its length.
In the figure the angle is α and the length of the perpendicular is p. Equation of line in two point form is
(x1, y1) ≡ (a,0) and (x2, y2) ≡ (0,b)
Using the two point form
(y−y1)/(x−x1) = (y2−y1)/(x2−x1)
y/(x − a) = b/(− a)
−ay = bx − ab
bx + ay = ab
x/a + y/b = 1
a = p sec α
b = p cosec α
x/(p sec α) + y/(p cosec α) = 1
x cos α + y sin α = p
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