Answer:
[tex]y=\frac{1}{7}x+\frac{\sqrt{5} }{7}\:or\: y=0.14x+0.32[/tex]
Step-by-step explanation:
Hi there,
1) According to Analytical Geometry, parallel lines have the same slope. In this case:
[tex]x-7y=\sqrt5 \Rightarrow-7y=\sqrt5-x \Rightarrow y=\frac{\sqrt{5}}{7}+\frac{x}{7}[/tex]
The slope is
[tex]m=\frac{1}{7}[/tex] or [tex]m=0.14x[/tex]
2) Since we have the slope [tex]m=\frac{1}{7}[/tex] and one x coordinate: -1.41 ([tex](-\sqrt2)[/tex] and the y-coordinate has not been given. Let's find precisely where this point is located at.
Plugging values.
[tex]y=\frac{\sqrt{5}}{7}+\frac{1}{7}(-\sqrt{2})\Rightarrow y=\frac{2.25}{7}+\frac{1}{7}*-1.41\:y=0.32-0.20\:y=0.12\\(-1.41,0.12)[/tex]
Finding 'b'
[tex]0.12=0.14(-1.41)+b\\0.12=-0.20+b\\b=0.32[/tex]
3) Then
[tex]y=\frac{1}{7}x+\frac{\sqrt{5} }{7}\:or\: y=0.14x+0.32[/tex]
This equation is parallel to
[tex]x-7y=\sqrt5 \Rightarrow y=0.14x-0.32[/tex]
And the point
[tex](-\sqrt2, 0.12)[/tex]
Graph:
Red line: Original equation given
Green one: parallel found.
