Answer:
The equation of line is [tex]y=\frac{1}{10}x+\frac{13}{2}[/tex]
Step-by-step explanation:
To find : What is the equation of a line perpendicular to the line [tex]y=-10x+1[/tex] passing through point (5,7) ?
Solution :
We know that,
When two lines are perpendicular then slope of one line is negative reciprocal of another line.
Let the required line [tex]y=mx+b[/tex]
A line perpendicular to the line [tex]y=-10x+1[/tex]
Slope of the line is [tex]m_1=-10[/tex]
Slope of the perpendicular line is [tex]m=-(\frac{1}{m_1}=\frac{1}{10}[/tex]
The passing points of the equation is (5,7)
Substitute in the formula,
[tex]7=\frac{1}{10}(5)+b[/tex]
[tex]7=\frac{1}{2}+b[/tex]
[tex]b=7-\frac{1}{2}[/tex]
[tex]b=\frac{13}{2}[/tex]
Now, substitute [tex]m=\frac{1}{10}[/tex] and [tex]b=\frac{13}{2}[/tex] in required line.
[tex]y=\frac{1}{10}x+\frac{13}{2}[/tex]
Therefore, The equation of line is [tex]y=\frac{1}{10}x+\frac{13}{2}[/tex]
