¼x + y = 2 transformed to equation
Answers
Answer:
We know that if Ax + By = C where C > 0
then the equation of the line in normal form is given by
\frac{A}{\sqrt{A^2+B^2}} x+\frac{B}{\sqrt{A^2+B^2}}y=\frac{C}{\sqrt{A^2+B^2}}
A
2
+B
2
A
x+
A
2
+B
2
B
y=
A
2
+B
2
C
if \frac{A}{\sqrt{A^2+B^2}} = \cos\alpha
A
2
+B
2
A
=cosα
\frac{B}{\sqrt{A^2+B^2}} = \sin\alpha
A
2
+B
2
B
=sinα , and
\frac{C}{\sqrt{A^2+B^2}} = p
A
2
+B
2
C
=p
Then the equation becomes
\cos\alpha x+\sin\alpha x y=pcosαx+sinαxy=p
Given equation
x + y - 2 = 0
or, x + y = 2
Here, A = 1, B = 1 and C = 2
Dividing the equation by \sqrt{1^2+1^2}=\sqrt{2}
1
2
+1
2
=
2
\frac{1}{\sqrt{2} } x+\frac{1}{\sqrt{2} } y=\frac{2}{\sqrt{2} }
2
1
x+
2
1
y=
2
2
or, \cos(\frac{\pi}{4}) x+\sin(\frac{\pi}{4})y=\sqrt{2}cos(
4
π
)x+sin(
4
π
)y=
2
which is the required equation in normal form