Write the equation x+5y-2= 0 in normal form. Then, find the length of the normal and the length and the angle makes with the positive x-axis.

Question

asked Aug 1, 2020 73.0k views

2 Answers

Hllau · Answer 1 · 2020-08-08T01:38:07+0000

Answer with explanation:

≡The given equation of line is :

x + 5 y -2=0------------(1)

⇒A line having equation, Ax +By +C=0, can be written in normal form as Follows:

$\rightarrow (Ax)/(√(A^2+B^2))+ (By)/(√(A^2+B^2))+ (C)/(√(A^2+B^2))=0$

⇒Length of Normal

=|(C)/(√(A^2+B^2))|

Let, A=Angle made by line with positive Direction of X axis.

$\sin A=(A)/(√(A^2+B^2))\\\\ \cos A=(B)/(√(A^2+B^2))$

⇒Line, 1 in normal form can be written as:

$x + 5 y=2\\\\\rightarrow(x)/(√(1^2+5^2))+(5 y)/(√(1^2+5^2))=(2)/(√(1^2+5^2))\\\\\rightarrow(x)/(√(26))+(5 y)/(√(26))=(2)/(√(26))$

⇒Length of Normal

$=(2)/(√(26))\\\\=0.40\text{approx}$

⇒Writing equation of line in slope intercept form

$5 y= -x +2\\\\y=(-x)/(5)+(2)/(5)$

⇒Comparing with general slope intercept form of line:

y = m x +c,

or, y = x tan A +c

$m=\tan A=(-1)/(5)$